加拿大安大略省十年级数学试卷英文
国际学校数学十年级试卷
1. The prime factorisation form of 156
A) 2 ´ 3´ 13
B) 22 ´ 32 ´ 13
2. The logarithemic form of 8x = 2
Chapter Real numbers
2.
Sets
3.
Polynomials
Weightage % 48% 20% 32% 100%
Marks 12 5 8 25
Table : 3 Weightage to difficulty level
S. No.
Difficulty level
Weightage % Marks
A) logx 8 = 2
B) log 2 = x 8
p 3. The q form of the decimal 0.875
7
A) 23
7
B) 16
C) 22 ´ 3 ´ 13 C) log 2 = 8
x
3
C) 8
() D) 2 ´ 32 ´ 13
() D) log x = 8
2
7
D) 22
()
3
1.
Easy
24%
06
2.
Average
3.
Difficulty
52%
13
24%
06
100%
25
Table : 4 Weightage to type of questions
S. No. Type of Questions
Weightage % No. of Questions Marks
加拿大数学10年级练习第一部分几何
1. ABCD is an isosceles trapezoid, and ≅. Which statement is NOT true?•ΔABY≅ΔBAX•≅•≅•∠DXB≅∠DYB2. It appears from the name of the HL Theorem that you actually need to know only two parts of a triangle in orderto prove two triangles congruent. Is this the case?•Yes, you only need to know the hypotenuse and a leg of a triangle.•No, you actually need to know two sides and an angle, because the triangle must be a right triangle.•No, you actually need to know three sides of the triangle.•No, you actually need to know two angles and a side.3. Complete the proof.Given: ≅, ∠1 ≅∠2, and ≅.Prove: ΔCEF≅ΔAEF.∠BEC≅∠DEA by vertical angles. ΔBEC≅ΔDEA by (a)_____. Then by CPCTC, ≅. ≅by the Reflexive Property. So ΔCEF≅ΔAEF by (b)_____.• a. SAS; b. SAS• a. AAS; b. SSS• a. ASA; b. SSS• a. AAS; b. HL4. Complete the proof.Given: ≅, ∠1 ≅∠2Prove: ΔBEA≅ΔDEC≅and ∠1 ≅∠2, so ΔBCA≅ΔDAC by SAS. Then, since (a)_____, ≅, ≅. ∠BEA≅∠DEC by (b)_____, so ΔBEA≅ΔDEC by (c)_____.• a. CPCTC; b. vertical angles; c. SAS• a. CPCTC; b. vertical angles; c. AAS• a. CPCTC; b. vertical angles; c. SSS• a. SAS; b. vertical angles; c. SSS5. What additional information can be used to prove the triangles congruent by the HL Theorem?•m∠BCE = 90•AB > AC•≅•≅6. Suppose ΔCED≅ΔDBC. If m∠EDC = 63 and m∠DBC = 82, what is m∠DCE?•63•82•145•357. If ∠A≅∠D and ∠C≅∠F, which statement would NOT prove that ΔABC≅ΔDEF?••∠B≅∠E•≅•none of these8. Determine what information you would need to know in order to use the SSS Congruence Postulate to show that thetriangles are congruent.•∠BAD≅∠CDB•≅•∠ADB≅∠CBD•≅9. Suppose ΔBCA≅ΔECD. Which statement is NOT necessarily true?•≅•∠A≅∠D•≅•∠BCA≅∠DCE10. In which triangles could you efficiently prove Δ1 ≅Δ2 using the HL Theorem?•II only•III only•II and III•I only11. Complete the proof.Given: ≅, ∠1 ≅∠2, and ≅.Prove: ΔCEF≅ΔAEF.∠BEC≅∠DEA with vetical angles. ΔBEC≅ΔDEA by (a)_____. Then by (b)_____, ≅. ≅by the Reflexive Property. So ΔCEF≅ΔAEF by (c)_____.• a. AAS; b. CPCTC; c. SSS• a. SAS; b. CPCTC; c. SSS• a. AAS; b. CPCTC; c. SAS• a. SSS; b. CPCTC; c. ASA12. In the paper airplane, ABCD≅EFGH, m∠B = m∠BCD = 90, and m∠BAD = 140. Find m∠GHE.•130•90•40•14013. Find the value of x.•x = –2•x = 9•x = 21•none of these14. Explain how you can use SSS, SAS, ASA, or AAS with CPCTC to prove that ∠D≅∠B.•≅and ∠ACB≅∠ACD. By the Symmetric Property, ≅. By SAS, ΔABC≅ΔADC, so by CPCTC ∠D≅∠B.•≅and ∠ACB≅∠ACD. By the Reflexive Property, ≅. By ASA, ΔABC≅ΔADC, so by CPCTC ∠D≅∠B.•≅and ∠ACB≅∠ACD. By the Reflexive Property, ≅. By SAS, ΔABC≅ΔADC, so by CPCTC ∠D≅∠B.•≅and ∠ACB≅∠ACD. By the Reflexive Property, ≅. By SSS, ΔABC≅ΔADC, so by CPCTC ∠D≅∠B.15. Complete the proof.Given: bisects ∠URS and bisects ∠UTS.Prove: ΔURT≅ΔSRT.•Reflexive property•definition of angle bisector•HL Theorem•CPCTC16. Complete the proof.Given: ∠RSQ≅∠TSQ, ∠RQS≅∠TQS.Prove: ≅.∠RSQ≅∠TSQ is given, as is ∠RQS≅∠TQS. By the Reflexive Property, ≅.ΔSRQ≅ΔSTQ by (a)_____, so ≅by (b)_____.• a. ASA; b. CPCTC• a. HL; b. CPCTC• a. SSS; b. CPCTC• a. SAS; b. CPCTC17. Determine which triangles are congruent by AAS using the information in the diagram below.•ΔABF≅ΔEDF•ΔADC≅ΔEBC•ΔABE≅ΔEDA•ΔABE≅ΔCBE18. Complete the proof.Given: bisects ∠EBC and bisects ∠ECC.Prove:ΔEBD≅ΔCBD.•Same-Side Interior Angles Theorem•given•SSS postulate•Triangle Inequality Theorem19. ΔABD≅ΔCBD. Name the theorem or postulate that justifies the congruence.•SAS•AAS•ASA•none of these1. Determine whether each quadrilateral can be a parallelogram. If not, write impossible.a. Two adjacent angles are right angles, but the quadrilateral is not a rectangle.b. All of the angles are congruent.• a. impossible; b. parallelogram• a. parallelogram; b. parallelogram• a. parallelogram; b. impossible• a. impossible; b. impossible2. Which statement is true?•All rectangles are squares.•All quadrilaterals are squares.•All quadrilaterals are parallelograms.•All parallelograms are quadrilaterals.3. Which statement can be used to determine whether quadrilateral XYZW must be a parallelogram?•≅and ≅•≅and ≅•≅and ≅•XW = WZ and XY = YZ4. Choose the best name for the parallelogram and find the measures of the numbered angles.•Square; all numbered angles are equal to 45°.•Rhombus; all numbered angles are equal to 115°.•Rhombus; all numbered angles are equal to 25°.•Square; all numbered angles are equal to 50°.5. Given: quadrilateral ABCD with A(–2, 3), B(2, –4), C(9, 0), D(5, 7). Then ABCD is a rectangle because•the slopes of the sides in pairs are negative reciprocals.•the product of the slopes of the diagonals is –1.•the figure has four vertices.•opposite sides have the same slope.6. Given the parallelogram below, find coordinates for P, without using any new variables.•(a–c, b)•(a + c, b)•(c, b)•(c, a)7. Find the values of the variables for the rectangle. Then find the lengths of the sides.•x = 7, y = 5; side lengths: 70, 45•x = 5, y = 7; side lengths: 33, 94•x = 5, y = 7; side lengths: 50, 63•x = 7, y = 5; side lengths: 45, 458. Determine whether the quadrilateral is a parallelogram. Explain.≅and ≅•Yes; if two opposite sides are congruent, then the quadrilateral is a parallelogram.•No; if the diagonals of a quadrilateral bisect each other, this is not enough to prove that the quadrilateral is a parallelogram.•No; if two opposite sides are congruent, this is not enough to prove that the quadrilateral is a parallelogram.•Yes; if the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram.9. Can coordinate geometry be used to prove that opposite sides and in quadrilateral EFGH are congruent?•Yes; use the Distance Formula to show that the diagonals are congruent.•No; you can only show that EF is parallel to GH by using coordinate geometry.•Yes; use the Distance Formula between vertices E and F, and between vertices G and H.•No; you can only find the slopes of EF and GH by using coordinate geometry.10. A square WXYZ has the vertices W(b, b), X(b, –b), Y(–b, –b), and Z(–b, b). Which vertex is in QuadrantII?•W•Z•X•Y11. ∠J and ∠M are base angles of isosceles trapezoid JKLM. If m∠J = 21x + 4, m∠K = 12x– 8, and ∠M = 14x+ 10, find the value of x.•• 2•–•–912. Suppose you are using coordinate geometry to prove that quadrilateral WXYZ is a square. Explain why no twosides should be parallel to the y-axis.•The x-axis would intersect two sides of the square, so the coordinates of the corners would not be clear.•Sides which are parallel to the y-axis would have an undefined slope, so you cannot prove numerically that these sides are parallel.•Points on the sides which are parallel to the y-axis could have any y-value.•The sides of the square which are parallel to the y-axis could be easily confused with the y-axis.13. Find AM if PN = 8 and AO = 5.•8• 3•13• 514. Given square ABCD, where A = (0, a), B = (a, a), C = (0, 0), and D = (a, 0). To prove that the diagonal ADis times the length of side CD, first use _____ to find that = a and = a. Therefore, the ratio= , or .•the definition of isosceles triangle ACD•the definition of right angle C•the Distance Formula•the definition of the origin C = (0, 0)15. A farmer is building a fence for his yard. He is considering two designs, which are shown below. Explain whythe quadrilaterals formed by the horizontal rails and the slanting boards are parallelograms in both designs.•The horizontal rails are parallel to each other. A parallelogram has exactly one pair of parallel sides, so both quadrilaterals are parallelograms.•The horizontal rails are congruent to each other. The slanting boards are all congruent to each other. A parallelogram has two pairs of adjacent sides, but opposite sides are not congruent, so both quadrilaterals are parallelograms.•The horizontal rails are congruent to each other. The slanting boards are all congruent to each other. A parallelogram has four congruent sides, so both quadrilaterals are parallelograms.•The horizontal rails are parallel to each other. The identical slanting boards all slant at the same angle, so the sides are parallel. A parallelogram has both pairs of sides parallel, so both quadrilaterals are parallelograms.16. A rhombus is centered on the origin. One side of the rhombus goes through the points (a, 0) and (0, b). Whatare possible coordinates for one of the other sides?•(–a, 0), (a, 0)•(b, 0), (0, –a)•(0, –b), (0, b)•(–a, 0), (0, –b)17. If a quadrilateral is a parallelogram, then its opposite sides are _____.•perpendicular•adjacent•congruent•none of these18. Find the value of each variable in the parallelogram. m∠1 = 10x, m∠2 = x + y, and m∠3 = 18z.•x = 9, y = 81, z = 5•x = 18, y = 167, z = 5•x = 18, y = 162, z = 10•x = 9, y = 86, z = 019. Complete ≅ _____ for parallelogram EFGH. Then state a definition or theorem as the reason.•; because the angles of a parallelogram bisect each other•; because the diagonals of a parallelogram bisect each other•; because the diagonals of a parallelogram bisect each other•; because the angles of a parallelogram bisect each other20. Which of the following sets of points represents a line segment in Quadrant III and its reflection in the x-axis?(Use the positive numbers a, b, c, d for the coordinates of the endpoints).•(a, b), (c, d); reflection (–a, b), (–c, d)•(–a, –b), (–c, –d); reflection (–a, b), (–c, d)•(a, b), (c, d); reflection (a, –b), (c, –d)•(–a, –b), (–c, –d); reflection (a, –b), (c, –d)1. Solve for a and b.•a = , b =•a = , b =•a = , b =•a = , b =2. State whether ΔADB∼ΔCDB, and if so, identify the theorem that proves the triangles similar.•yes, SSS∼•yes, AA∼•yes, SAS∼•no3. ABCDE∼GHJDF. Complete the congruence and proportion statements.a.∠H≅b.=• a.B; b.AE• a.E; b.DC• a.E; b.AE• a.B; b.DC4. Write a similarity statement for the two triangles.•ΔVUT∼ΔWXY•ΔTVU∼ΔWXY•ΔTUV∼ΔWXY•ΔTUV∼ΔWYX5. Find the geometric mean of 48 and 3.•9•25.5•12•166. The extendable ramp shown below is used to move crates of fruit to loading docks of different heights. When thehorizontal distance AB is 4 feet, the height of the loading dock, BC, is 3 feet. What is the height of the loading dock, DE?•7 ft•9 ft•11 ft7. Find the geometric mean of 20 and 5.•10• 4•12.5•258. In movies and television, the ratio of the width of the screen to the height is called the aspect ratio. Televisionscreens usually have an aspect ratio of 4 : 3, while movie screens usually have an aspect ratio of 1.85 : 1. However, if a movie is made for television in "Letterbox" format, it retains the 1.85 : 1 aspect ratio and fills in the top and bottom parts of the screen with black bars. What would be the height of a movie in "Letterbox" format on a television screen that measures 25 inches along its diagonal? (Hint: First find the width and height of the television screen.)•13.51 in.•10.81 in.•15 in.•8.12 in.9. Use the diagram to determine the height of the tree.•264 ft•72 ft•60 ft•80 ft10. The two rectangles are similar.Which is a correct proportion between corresponding sides?•=•=•=11. Use the Side-Splitter Theorem to find x given that || .•18•12•24• 612. Find OM if bisects ∠NLM, LM = 14, NO = 3, and LN = 4. Round your answer to the nearest hundredth, ifnecessary.•12.27•18.67•0.86•10.513. There is a law that the ratio of the width to length for the American flag should be 10 : 19. Which dimensionsare NOT in the correct ratio?•20 by 38 in.•50 by 95 ft•20 by 44 ft•100 by 190 ft14. If one measurement of a golden rectangle is 6.8 inches, which could be the other measurement?•8.418 in.•11.002 in.• 1.618 in.• 5.182 in.15. If one measurement of a golden rectangle is 8.2 inches, which could be the other measurement?•9.818 in.• 6.582 in.• 1.618 in.• 5.068 in.16. Solve = .•20•19•15•2417. Find OM if bisects ∠NLM, LM =15, NO = 5, and LN = 11. Round your answer to the nearest hundredth, ifnecessary.•33• 6.82• 3.67•8.5918. The width of a golden rectangle is 3 m, which is shorter than the length. What is the length?• 1.85 m• 2.32 m• 3.64 m• 4.85 m19. Find and simplify the ratio of the length to the width of the rectangle.••••20. ΔBGH∼ΔSWQ. What are the pairs of corresponding sides?•BG and SQ, BH and SW, GH and WQ•BG and GB, SQ and QS, GH and HG•BG and SW, BH and SQ, GH and WQ•BG and WQ, BH and SW, GH and SQ1. A building near Atlanta, Georgia, is 181 feet tall. On a particular day at noon it casts a 204-foot shadow. Whatis the sun's angle of elevation at that time?•41.6°•27.5°•62.5°•48.4°2. In right triangle ΔABC, sin A = . What is cos A?••••none of these3. Find the ratio for cos x.••••14. Find the value of x to the nearest meter.•46 m•40 m•35 m•36 m5. Compare the quantity in Column A with the quantity in Column B. The diagram may not be drawn to scale.•The quantity in Column A is greater.•The quantity in Column B is greater.•The two quantities are equal.•The relationship cannot be determined on the basis of the information given.6. How many of these triples could be sides of a right triangle: (27, 36, 45), (12, 17, 20), (24, 32, 40), (14,48, 50)?• 4 triples• 3 triples• 2 triples• 1 triple7. Find the ratio for cos x.•••2•8. Find a third number of the Pythagorean triple that includes 72 and 75.•9•21•37•1049. Find the measure of the marked acute angle to the nearest degree.•62°•28°•61°•118°10. In ΔABC, ∠A is a right angle and m∠B = 60. If AB = 20 ft, find BC. If necessary, round your answer tothe nearest tenth.•10 ft.•40 ft•20 ft•ft11. Find the value of the variable to the nearest hundredth.• 5.28 cm•0.32 cm• 3.13 cm• 5.12 cm12. Find the length of the leg of the right triangle. Leave your answer in simplest radical form.•48••288•13. In ΔABC, ∠A is a right angle and m∠B = 45. If AB = 20 ft, find BC.•10 ft•20 ft•40 ft•20 ft14. Which direction bearing is shown?•19° north of east•19° north of west•19° south of east•19° south of west15. Leslie used the diagram to compute the distance from Ferris to Dunlap to Butte. How much shorter is the distancedirectly from Ferris to Butte than the distance Leslie found?•123 mi•87 mi•36 mi•84 mi16. Which vector has a direction of 31° east of north?••••17. Find the value of x to the nearest tenth.•14.4• 6.3•7.8• 3.118. Find the value of x.•3•6•12• 619. Find the value of x to the nearest tenth.•7.8•18.3•33.0•8.920. Find the value of x to the nearest integer when tan x = 1.483.•58•56•55•571. Which type of isometry is the equivalent of two reflections across two vertical lines?•translation•rotation•glide reflection•none of these2. A section of a tessellated plane is shown below. Which types of symmetry does the tessellated plane have?•glide-reflectional symmetry•translational, rotational, and glide-reflectional symmetry•rotational symmetry•translational and reflectional symmetry3. A blueprint for a house has a scale of 1 : 30. A wall in the blueprint is 7 in. What is the length of the actualwall?•210 ft•21 ft•17.5 ft•none of these4. What is the image of the point (4, –2) after a rotation 270° clockwise about the origin?•(–4, 2)•(4, 2)•(–2, 4)•(2, 4)5. Which graph shows a triangle and its reflection image in the x-axis?••••6. Write a rule to describe a reflection over the y-axis.•(x, y) → (–x, y)•(x, y) → (–x, –y)•(x, y) → (x, –y)•(x, y) → (y, x)•reflectional•rotational•rotational and reflectional•none of these8. Find the image of O(0, 0) after two reflections, first in y = 4, and then in x = –7.•(7, –4)•(–14, 8)•(8, –14)•(–4, 7)9. A section of a tessellated plane is shown below. Which types of symmetry does the tessellated plane have?•rotational symmetry•translational and rotational symmetry•reflectional symmetry•glide-reflectional symmetry•rotational•reflectional•rotational and reflectional•none of these11. Describe the translation 7 units to the left, 12 units up using a vector.•)–12, 7*•)12, –7*•)–7, 12*•)7, –12*12. If the figure has rotational symmetry, find the angle of rotation about the center that results in an imagethat matches the original figure.•120°•90°•72°•It has no rotational symmetry.13. Find the glide reflection image of the solid triangle for the translation )–6, –3* and reflection in y = –1.••••14. If a point P(1, –2) is reflected across the line x = 3, what are the coordinates of its reflection image?•(1, –4)•(1, 8)•(–7, –2)•(5, –2)15. Which translation from thin-lined figure to thick-lined figure is given by the vector )6, 6*?••••16. The dotted triangle is a dilation image of the solid triangle. What is the scale factor?•• 2• 3•17. Find the image of C under the translation described by each vector.a.)4, 5*b.)11, –8*• a.A; b.B• a.B; b.A• a.E; b.D• a.D; b.E18. What is the image of the point (–3, 4) after a rotation of 90° counterclockwise about the point (–3, 0)?•(–7, 0)•(1, 0)•(0, –3)•(–3, –4)19. Which letter has rotational symmetry?• E•X•J•T20. Use scalar multiplication to find the image of the quadrilateral for a dilation with center (0, 0) and scalefactor 2. Graph the quadrilateral and its image.••••。
加拿大数学10年级练习第三部分
加拿大数学10年级练习第三部分1.writey=(x?6)(x?4)(x+2)instandardform.y=x3+48y=x3？8x2+4x+48y=x3？2x2+4x？8y=x3？8x2？8x+242。
因子X4？8x2+15。
（x？3）2（x？5）2(x?3)(x+3)(x?5)(x+5)(x?(x?)（x+）2（x？)(x?)2)（十）+)3.solvex3+5x2？9x？45=0通过绘图。
5,3? 5.3,3? 3,5,3? 5,94.evaluate.60,480504845.找到x3+4x2+3x+12=0的方向。
?x=?4,±ix=?4,x=？4，±3ix=4，±i6.classify2x3?2xbydegreeandbynumberofterms.cubicmonomialquadraticbinomialcubicbinomialcubictrinomial7.使用合成除法进行（5x3？6？5x）÷（10+5x）除法。
x2？2x+3，r24x2？3x，r44x2？2x+3，r？36x2？3x，r？56eagraphingcalculatortofindapolynomialfunctiontomodelthedata.f(x)=4.68x3+8.87x2+4.25x+6.32f(x)=4.68x2+8.87x+6.32f（x）=？4.68x3？8.87x2？4.25倍？6.32f（x）=0.4468x3+8.87x2+4.25x+6.329。
第三因素？813（q？3）（q2+3q+9）（q+3）（q2+3q+9）？第三季度（第二季度+第三季度+9季度） (q3)(q2+3q+9)10.表中显示了自1987年以来杂交棉树木取代树木的数量。
找到一个Cubic函数来对数据进行建模，并使用它来估计1998年种植的棉花数量。
t（x）=0.3x3+0.3x+0.4；167.2Housandtreest（x）=0.3x3+0.3x2+0.1x+0.1；167.2Housandtreest（x）=0.1x3+0.3x2+0.3x+0.1；1728万树（x）=0.1x3+0.1x+0.4；1728万树11。
amc10英语真题及答案新课标
amc10英语真题及答案新课标AMC10是美国数学竞赛（American Mathematics Competitions）的10年级级别，面向10年级及以下的学生。
以下是一套模拟的AMC10英语真题及答案，请注意，这只是一个示例，并非真实的AMC10题目。
AMC10 英语真题及答案Part 1: Multiple Choice1. The word "innovate" is most closely related to which of the following?A. InnovatorB. InventionC. InventionistD. InnovatoryAnswer: D. Innovatory2. Which of the following is the correct spelling of the word meaning "to make something new"?A. InnovateB. InnovateC. InnovateD. InnovateAnswer: A. Innovate3. The phrase "a leap of faith" is used to describe:A. A large jumpB. A risky decisionC. A new religionD. A sudden increaseAnswer: B. A risky decision4. In the sentence "The company is looking to streamline its operations," the word "streamline" means:A. To make more expensiveB. To make more efficientC. To make more complicatedD. To make more visibleAnswer: B. To make more efficient5. The word "altruistic" is an antonym for:A. SelfishB. AltruismC. AltruisticallyD. AltruistAnswer: A. SelfishPart 2: Fill in the Blanks6. The scientist was awarded the Nobel Prize for his _______ contributions to the field of physics.Answer: innovative7. The _______ of the old building was a significantachievement for the preservation society.Answer: renovation8. The _______ of the new policy was met with mixed reactions from the public.Answer: implementation9. The _______ of the company's profits was due to a series of successful marketing campaigns.Answer: increase10. The _______ of the ancient ruins provided valuable insights into the history of the civilization.Answer: excavationPart 3: Reading ComprehensionRead the following passage and answer the questions that follow.Passage:In recent years, there has been a significant increase in the number of people who are interested in sustainable living. This trend has led to the development of various eco-friendly products and practices. One such practice is the use of solar panels to generate electricity. Solar panels are becoming more popular due to their ability to harness the power of the sun and convert it into usable energy.Questions:11. What is the main topic of the passage?Answer: Sustainable living and the use of solar panels.12. Why are solar panels becoming more popular?Answer: Because they can harness the power of the sun and convert it into usable energy.13. What is the trend mentioned in the passage?Answer: An increase in the number of people interested in sustainable living.Part 4: Vocabulary in Context14. The _______ of the old factory was a major concern for the environmentalists.Answer: pollution15. The _______ of the new technology was celebrated by the scientific community.Answer: advancement16. The _______ of the endangered species was a top priority for the conservation organization.Answer: preservation17. The _______ of the ancient artifact provided evidence ofa previously unknown civilization.Answer: discovery18. The _______ of the new policy was met with skepticism by some members of the community.Answer: enforcement请注意，AMC10是一个数学竞赛，通常不包含英语题目。
国际学校十年级数学期末测试卷(英语)
2. If cos q = 0.6691306, then q = a) 48° b) 42° c) 33.8° d) 22.9°
3. If DABC ~ DPQR, then a) AB = PQ b) A = P c) a = b d) c = r
Final Practice Exam
4. The cost of 4 L of milk is $7.35. The unit cost per litre is a) $3.68 b) $1.84 c) $2.45 d) $0.735
5. A 10% discount off an item priced at $30.00 would save you a) $3.00 b) $27.00 c) $10.00 d) $1.00
8. A rotation of a shape is a) a slide b) a translation c) a turn d) a mirror image
9. Which statement is true about angles with parallel lines with a transversal? a) All angles are 60° b) Interior alternate angles are complementary c) Corresponding angles are supplementary d) Exterior alternate angles are congruent
Final Practice Exam
3
4
Grade 10 Essential Mathematics
加拿大数学十年级(高一)省考模拟试题A
PART A: MULTIPLE-CHOICE QUESTIONS (calculator not permitted)
Value: 12 marks
Suggested Time: 30 minutes Allowable Time: 40 minutes
Page 1
Use the following graph to answer question 2.
y
10
5
–10
–5
–5
x
5
10
–10
2. Which of the following equations describes the linear relation graphed above?
Do not access your calculator until directed by the supervisor. At the end of the 40 minutes, the supervisor will give you permission to access your calculator.
where n is the number of students attending the prom. Calculate the cost of renting the hall if 70 students attend. A. $108 B. $500 C. $780 D. $970
6. Which of the following statements are true?
Contents: 31 pages 54 multiple-choice questions (maximum of 54 marks) 6 numeric-response questions (maximum of 6 marks)
加拿大安大略地区十学年数学试卷英文
GRADE 10 PRINCIPLES OF MATHEMATICS (ACADEMIC)MPM 2DTotal Marks:INSTRUCTIONS:1. Calculators may be used.2. Read all instructions carefully in order to maximize your mark.A/C[K] Part A – Multiple Choice 25 Marks (25 questions * 1 mark each)For each of the following questions in this section, circle the letterrepresenting the correct answer.1. A linear system of two equations that has one solution represents twolines that are:a) parallel b) coincident c) intersecting d) none of these2. The midpoint of RS is M(8, -1). If point S has coordinates (11, 4) what are the coordinates of point R ? a) (3, -6)b) (15, -6)c) (5, -6)d) (3, 9)3. The midpoint of the line segment with end points A(-8, 8) and B(6, 4) is: a) (0, 10)b) (1, 2)c) (7, 2)d) (-1, 6)4. The equation of a horizontal line passing through the point (4, 2) is: a) 2=xb) 4=yc) 2=yd) 4=x5. The equation of a line with a slope of 5=m and a y intercept of 8 is: a) 85+=x yb) 85+-=x yc) 85--=x yd) 58+=x y6. The slopes of 2 lines are -7 and 71. These lines are said to be:a) parallelb) perpendicularc) coincidentd) none of these7. The slope of a line segment passing through 2 points (10,- 4) and (-2, -16) is: a) 1b) 2c) -1d) -28. The length of a line segment with end points (-6, 7) and (-1, -5) is: a) 12b) 5c) 13d) 1699. The diameter of a circle whose equation is 28922=+y x is:a) 15b) 16c)17d) none of these10. T he equation of a circle with a centre of (0, 0) that also passes through the point (-8, -6) is: a) 1022=+y xb) 10022=+y xc) 1422=+y xd) 4822=-y x11. T he y-intercept of the line 01052=+-y x is: a) 2b) -2c) 10d) 512. T he slope of the line 0124=-+y x is: a) 2b) -2c) 1d) 013. I f (-3, y) is a solution to the equation 132=+y x , what is the value of y ? a) 3b) 6c) 5d) 814. T he product ()()z y x z y x 323243-- is equal to:a) 2612z xyb) 26412z y xc) 2612z xy -d) 00412z y x15. A simplified expression for ()()n m n m ----52 is: a) m 7b) n m 27+c) m 3-d) n m 27-16. A simplified expression for 242927abcbc a -- is:a) ac 3 b) abc 3 c) 23acd) 223c a17. T he slope of the line, which is perpendicula r to the line, 084=+-y x is: a) -4 b) 4 c) 1 d) -118. T he shortest distance from the point (2, -3) to the line 4-=x is: a) 5 b) 3 c) 2 d) 619. T he value of the polynomial 8542+-a a when 3-=a is: a) 59 b) 44 c) 13 d) 2920. W hich of the following is not a function : a)()()(){}7,6,5,4,3,2b) 22x y =c) 22y x =d) ()()(){}3,8,3,7,2,621. T he range of the relation whose equation is 52--=x y is: a) 5-≤y b) 5≤y c) 5-≥y d) 5≥y22. T he vertex of the parabola ()642--=x y is:a) ()6,4- b) ()6,4- c) ()4,6- d) ()4,6-23. T he equation of the axis of symmetry of the parabola ()5242+--=x y is:a) 5=x b) 5-=x c) 2=xd) 2-=x 24. A parabola with a vertex of ()3,2 and a stretch factor of 41- (relative to2x y =) would have an equation of: a) ()32412+--=x y b) ()32412++-=x y c)()23412-+-=x y d) ()23412++-=x y25 The parabola k x y +-=24 passes through the point ()3,2-. T he value of k is: a) -19b) 11c) 13d) 19A/CPart B – Short AnswersFor each of the questions in this section, write your answers in the spaces provided . Use the foolscap provided for any rough work. Show details of calculations wherever requested.1. In the accompanying diagram, state each of the following: (4 Marks) [K] a) domain: __________ (1 Mark)[K]b)range: __________ (1 Mark)[C]c)Is the relation a function? Justify your answer. (2 marks)[A] 2. The x-intercepts of the parabola 2892-y are: __________ and=x__________.(Show your work) (2 Marks)[A] 3. The roots of the quadratic equation 032=1710x are: __________ and+-x__________.(Show your work) (3 Marks)[A] 4. Write the equation of the parabola with a vertex of (4, 23) if it passesthrough the point (-1, -2): (Show your work) (3 Marks)____________________[T] 5. A line passes through 2 points (1, 4) and (2,-4). Calculate the slope of the line. Also show the equation of the line in the form 0ByAx. (Show+=+Cyour work) (4 Marks)____________________ ____________________Slope Equation[K] 6. The Tangent of 45 is: __________ (1 Mark)[A] 7. a) In the accompanying diagram, the two triangles are similar. What is thevalue of x?(Show your work) (2 Marks) Array=x__________[T] b) If the area of the smaller triangle is 8 cm 2, what is the area of the larger triangle?(Show your work) ( 2 Marks)Area = __________[K]8 Given that sin A = 21, find A ∠ (to the nearest degree) __________ (1Mark)[A] 9. In the accompanying right triangle , find the value of x to one decimal place.(Show your work) (2 Marks)=x ________[A] 10. U se the SINE LAW to find the value of side x to one decimal place.(Show your work) (2 Marks)3028︒x56︒42︒x30x = ________[A] 11. U se the COSINE LAW to find the value of side x to one decimal place.(Show your work) (2 Marks)x = ________[T] 12. F actor each of the following to the fullest extent possible: (4 Questions * 2 marks each)a) y x my mx 22--+________________________b) 31142--x x________________________c) 2416916y x -________________________56︒2030xd) 2225rs-________________________ r+9s30A/C Part C – Full Solutions RequiredFor each of the questions in this section, full solutions are required.Record your answers in the spaces provided. Use the foolscap providedfor any rough work.[A] 1. Solve the linear system using the elimination method. Remember tofind values for both x and y. (5 Marks)22+yx5=-yx=32-21[C] Explain what the solution above represents geometrically. How do youknow that the solution you arrived at is the correct answer? (2 Marks)[A]2. Expand and simplify the polynomial ()()()21432+-+-x x x . (4 Marks)[T]3. Find the equation of the line perpendicular to the line 088=-+y x and passing through the point (-4, 1). (4 Marks)[T] 4. From the window of one building, a man finds that the angle ofelevation to the top of a second building is 47︒ and the angle ofdepression to the bottom of the same building is 58︒. The buildings are 60 m apart. Find the height of the 2nd building to the nearest metre.A diagram is required. (6 Marks)[T] 5. ABC has vertices A(1, 7), B(-5, 3) and C(3, -1). Determine the equation for AE, the altitude from vertex A to the opposite side BC.(5 Marks)6. The hypotenuse of a right triangle is 26 cm. The sum of the other twosides is 34 cm. (9 Marks)[T] a) Find the length of the other two sides of the triangle. (3 Marks)[T] b) Find the measure of the other two angles. Round to the nearest degree. (3 Marks)[C] c) Describe a situation where you would be able to use knowledge ofthe Pythagorean theorem in a practical, real life situation. (3 Marks)[T] 7. A rectangular skating rink measures 20m by 20m. It has been decided to increase the area of the rink by a factor of 4. Determine how mucheach side should be extended. Assume that each side is extendedby the same amount. (6 Marks)[C]What is the significance of keeping the skating rink in the shape of a square? Justify your answer. (3 Marks)[A]8. a) Solve 35122+=d d using the quadratic formula. (2 Marks)[A] b) Solve 03122=-x by factoring. Check your solutions. (2 Marks)。
加拿大数学十年级(高一)省考模拟试题B汇编
You have Examination Booklet Form B. In the box above #1 on your Answer Sheet, fill in the bubble as follows.
Exam Booklet Form/ A B C D E F G H Cahier d’examen
A. decreased by 15 km h B. decreased by 5 km h C. increased by 15 km h D. increased by 11 km h
3. Solve the following system of equations:
4x + 2y = 8 3x + y = 1
13. Marbles are placed in a jar one at a time. Which graph below best represents the total mass of the jar and marbles as the marbles are added?
A.
B.
Total Mass
INSTRUCTIONS: No calculator may be used for this part of the examination. For each question, select the best answer and record your choice on the blue Answer Sheet provided.
Foundations of Mathematics and Pre-Calculus 10 – 2010–2011 Sample B
Page 5
PART B: MULTIPLE-CHOICE QUESTIONS (calculator permitted)
加拿大数学10年级练习第四部分几何
1. Find the ratio of the perimeter of the larger rectangle to the perimeter of the smaller rectangle.••••2. Find the area of the triangle with ∠A = 41, b = 5 ft, and c = 4 ft. Round your answer to two decimal places.• 6.56 ft2•10.00 ft2•7.55 ft2•13.12 ft23. Find the area of kite ABCD if BD = 48 cm, AB = 25 cm, and BC = 26. The kite is not drawn to scale.•289 cm2•70 cm2•816 cm2•408 cm24. The diameter of a basketball rim is 18 inches. A standard basketball has a circumference of 30 inches. Abouthow much room is there between the ball and the rim in a shot in which the ball goes in exactly in the center of the rim?•none of these5. Find the area.•718 square units•545 square units•534.5 square units•701 square units6. Name the major arc and find its measure.•m = 275•m = 170•m = 85•m = 2757. Find the area of the regular polygon. Round your answer to the nearest tenth.•110.2 in.28. Name the minor arc and find its measure.•m = 275•m = 85•m = 170•m = 2759. Find the area of ΔABC. The figure is not drawn to scale.•26.06 cm2•28.00 cm2•22.73 cm2•24.95 cm210. Find the circumference of the circle. Use as an approximation of π.•11 cm•5 cm•6 cm11. If a dart hits the target at random, what it the probability that it will land in the unshaded region?••••12. Find the area of the circle. Use π = 3.14 and round to the nearest hundredth.•91.56 m2•16.96 m2• 5.72 m2•22.89 m213. Find the area of a regular pentagon with side 6 cm.•45.2 cm2•123.9 cm214. Find the probability that an object falling randomly on the figure will land in the shaded area.•0.32•0.36•0.5•0.2615. Find the area.•102.9 yd2•205.8 yd2•35.35 yd2•105.5 yd216. Two concentric circles have radii of 14 cm and 24 cm. Find the probability to the nearest thousandth that apoint chosen at random from the circles is located outside the smaller circle and inside the larger one.•0.066•0.58317. A slide that is inches by inches is projected onto a screen that is 3 feet by 7 feet, filling the screen.What will be the ratio of the area of the slide to its image on the screen?• 1 : 112• 1 : 2304• 2 : 4205• 1 : 12,54418. Find the area of a regular octagon with perimeter 48 cm.•188.1 cm2•190.5 cm2•347.6 cm2•173.8 cm219. Dorothy ran 6 times around a circular track that has a diameter of 47 m. Approximately how far did she run?Use π = 3.14 and round your answer to the nearest meter.•885 m•1328 m•443 m•1734 m20. Find the area.•10.26 cm21. Find the volume of the cylinder in terms of π.•24π in.3•48π in.3•56π in.3•288π in.32. Find the volume of the cylinder in terms of π.•287π in.3•275π in.2•275π in.3•287π in.23. A sphere has a volume of 288π ft3. Find the surface area of the sphere.•864π ft2•48π ft2•144π ft2•96π ft24. The volumes of two similar solids are 2197 m3 and 64 m3. The surface area of the larger one is 845 m2. What is•64 m2•320 m2•80 m2•none of these5. Use a net to find the surface area of the prism.•114 m2•240 m2•290 m2•145 m26. Find the surface area of a sphere that has a diameter of 4 cm.•64π cm2•π cm3•4π cm2•16π cm27. Find the lateral area and the surface area of the cone. Use 3.14 for πand round the answer to the nearest hundredth.The diagram is not to scale.•lateral area: 733.33 ft2; surface area: 690.80 ft2•lateral area: 690.80 ft2; surface area: 1004.80 ft2•none of these8. If the ratio of the radii of two spheres is 7 : 2, what is the ratio of the surface areas of the two spheres?•7 : 2•7r2π : 2r2π•49 : 4•343 : 89. Use a net to find the surface area of the prism.•465 m2•720 m2•918 m2•930 m210. Find the height of the cylinder to the nearest tenth of an inch.•94.6 in.•96.6 in.• 4.3 in.• 4.1 in.11. Use formulas to find the lateral area and the surface area of the prism. Show your answer to the nearest hundredth.•63.00 m2; 567.00 m2•36.00 m2; 1134.00 m2•479.22 m2; 533.22 m2•542.22 m2; 596.22 m212. Find the volume of the prism.•942 m3•38 m3•945 m3•1890 m313. The volumes of two similar solids are 1331 m3 and 343 m3. The surface area of the larger one is 484 m2. Whatis the surface area of the smaller one?•343 m2•1372 m2•196 m2•none of these14. Find the surface area of the sphere.•648π m2•72π m2•324π m2•1296π m215. Find the volume of the prism.•40.5 m3•162 m3•9 m3•81 m316. Find the surface area of the solid. Round to the nearest square foot.•36 ft2•68 ft217. Use formulas to find the surface area of the prism. Show your answer to the nearest hundredth.•75.42 cm2•170.16 cm2•86.94 cm2•69.52 cm218. Which figure is a net for a cube?••••19. Cylinder A has radius 1 and height 4 and cylinder B has radius 2 and height 4. Find the ratio of the volumesof the two cylinders.• 1 : 4• 1 : 120. Neil had a job helping a jeweler. He had the assignment of counting the faces, vertices, and edges on the emeralds.On the first emerald, Neil counted 9 faces and 16 edges. He quickly realized he didn't have to count the vertices.How many vertices were there?•10 vertices•7 vertices•8 vertices•9 vertices1. Find the value of x if AB = 20, BC = 12, and CD = 13. (not drawn to scale)•18.8•16.5•13.4•14.92. Find the measure of each variable if m∠A = 22 and m = 97. (not drawn to scale)•53; 210•53; 105•75; 210•75; 1053. In the plane of lines X and Y, what is the locus of points equidistant from lines X and Y?•line A•line D•line B•line C4. A small messenger company can only deliver within a certain distance from the company. On the graph below, thecircular region represents that part of the city where the company delivers, and the center of the circle represents the location of the company. Which equation represents the boundary for the region where the company delivers?•(x + 3)2 + (y– 1)2 = 49•(x + 3)2 + (y– 3)2 = 98•(x + 1)2 + (y– 3)2 = 98•(x + 3)2 + (y– 3)2 = 495. A low-watt radio station can be heard only within a certain distance from the station. On the graph below, thecircular region represents that part of the city where the station can be heard, and the center of the circle represents the location of the station. Which equation represents the boundary for the region where the station can be heard?•(x– 4)2 + (y– 5)2 = 50•(x + 5)2 + (y– 5)2 = 59•(x + 5)2 + (y + 4)2 = 25•(x + 5)2 + (y– 5)2 = 256. Find the center and radius of (x– 8)2 + (y + 7)2 = 64.•(–8, 7); 8•(–7, 8); 64•(8, –7); 8•(–7, –8); 87. is tangent to circle O at B. How close to the circle is point A? (The diagram is not to scale.)• 3• 4.5• 6•7.58. Compare the quantity in Column A with the quantity in Column B.•The quantity in Column A is greater.•The quantity in Column B is greater.•The two quantities are equal.•The relationship cannot be determined from the information given.9. Solve for x.•22•710. Find the value of x.•14.6•8.1•9.4•13.411. Write the standard equation for the circle with center (14, –48) that passes through (0, 0).•(x– 14)2 + (y + 48)2 = 2500•(x + 14)2 + (y– 48)2 = 2500•(x + 14)2 + (y– 48)2 = 50•(x– 14)2 + (y + 48)2 = 5012. Find the measure of ∠BAC.•30°•150°13. Find the value of x.•79•39•99•15914. In space, which description fits the locus of points 3 cm from ?•an open cylinder of diameter 6 cm•an open cylinder of radius 3 cm and two hemispheres of diameter 6 cm each•an open cylinder of radius 3 cm and height 6 cm•an open cylinder of diameter 6 cm and two spheres of radius 3 cm each15. and are tangent to circle O and bisects ∠BPA. If m∠AOC= 68°, how much greater is m∠BCO thanm∠OAD? (The diagram is not to scale.)•22°•112°16. Write the standard equation for the circle with center (6, –8) that passes through (0, 0).•(x + 6)2 + (y– 8)2 = 0•(x– 6)2 + (y + 8)2 = 0•(x + 6)2 + (y– 8)2 = 100•(x– 6)2 + (y + 8)2 = 10017. Find the center and radius of (x + 3)2 + (y + 8)2 = 169.•(3, 8); 13•(–8, 3); 13•(–3, –8); 13•(–8, –3); 16918. , , and are all tangent to circle O. If JA= 8, AL= 13, and CK= 11, what is the perimeter of ΔJKL?(The diagram is not to scale.)•32•64•45•5319. If m = 38, what is m∠YAC?•109°•52°20. Write the standard equation for the circle with center (–16, 30) that passes through (0, 0).•(x + 16)2 + (y– 30)2 = 34•(x + 16)2 + (y– 30)2 = 1156•(x– 16)2 + (y + 30)2 = 1156•(x– 16)2 + (y + 30)2 = 34。
加拿大数学十年级(高一)省考模拟试题A
PART A: MULTIPLE-CHOICE QUESTIONS (calculator not permitted)
Value: 12 marks
Suggested Time: 30 minutes Allowable Time: 40 minutes
INSTRUCTIONS: No calculator may be used for this part of the examination. For each question, select the best answer and record your choice on the blue Answer Sheet provided.
This is the end of Part A (calculator not permitted).
If there is some time left, you have two options: i) Make sure you have answered all the questions. You will not be able to go back to this section at the end of 40 minutes. ii) You may proceed to the rest of the examination without the use of a calculator; there are many questions that do not require a calculator. Make sure you flag any questions you skip to remember to go back to them later.
加拿大数学10年级练习第四部分几何电子教案
加拿大数学10年级练习第四部分几何1. Find the ratio of the perimeter of the larger rectangle to the perimeter of the smaller rectangle.••••2. Find the area of the triangle with ∠A = 41, b = 5 ft, and c = 4 ft. Round your answer to two decimalplaces.• 6.56 ft2•10.00 ft2•7.55 ft2•13.12 ft23. Find the area of kite ABCD if BD = 48 cm, AB = 25 cm, and BC = 26. The kite is not drawn to scale.•289 cm2•70 cm2•816 cm2•408 cm24. The diameter of a basketball rim is 18 inches. A standard basketball has a circumference of 30 inches.About how much room is there between the ball and the rim in a shot in which the ball goes in exactly in the center of the rim?• 4.2 in.• 4.78 in.•none of these5. Find the area.•718 square units•545 square units•534.5 square units•701 square units6. Name the major arc and find its measure.•m = 275•m = 170•m = 85•m = 2757. Find the area of the regular polygon. Round your answer to the nearest tenth.•40.0 in.2•67.6 in.2•110.2 in.28. Name the minor arc and find its measure.•m = 275•m = 85•m = 170•m = 2759. Find the area of ΔABC. The figure is not drawn to scale.•26.06 cm2•28.00 cm2•22.73 cm2•24.95 cm210. Find the circumference of the circle. Use as an approximation of π.•11 cm•5 cm•6 cm11. If a dart hits the target at random, what it the probability that it will land in the unshaded region?••••12. Find the area of the circle. Use π = 3.14 and round to the nearest hundredth.•91.56 m2•16.96 m2• 5.72 m2•22.89 m213. Find the area of a regular pentagon with side 6 cm.•76.6 cm2•45.2 cm2•123.9 cm214. Find the probability that an object falling randomly on the figure will land in the shaded area.•0.32•0.36•0.5•0.2615. Find the area.•102.9 yd2•205.8 yd2•35.35 yd2•105.5 yd216. Two concentric circles have radii of 14 cm and 24 cm. Find the probability to the nearest thousandth thata point chosen at random from the circles is located outside the smaller circle and inside the larger one.•0.066•0.583•0.01717. A slide that is inches by inches is projected onto a screen that is 3 feet by 7 feet, filling thescreen. What will be the ratio of the area of the slide to its image on the screen?• 1 : 112• 1 : 2304• 2 : 4205• 1 : 12,54418. Find the area of a regular octagon with perimeter 48 cm.•188.1 cm2•190.5 cm2•347.6 cm2•173.8 cm219. Dorothy ran 6 times around a circular track that has a diameter of 47 m. Approximately how far did she run?Use π = 3.14 and round your answer to the nearest meter.•885 m•1328 m•443 m•1734 m20. Find the area.•10.26 cm2•61.56 cm21. Find the volume of the cylinder in terms of π.•24π in.3•48π in.3•56π in.3•288π in.32. Find the volume of the cylinder in terms of π.•287π in.3•275π in.2•275π in.3•287π in.23. A sphere has a volume of 288π ft3. Find the surface area of the sphere.•864π ft2•48π ft2•144π ft2•96π ft24. The volumes of two similar solids are 2197 m3 and 64 m3. The surface area of the larger one is 845 m2. Whatis the surface area of the smaller one?•64 m2•320 m2•80 m2•none of these5. Use a net to find the surface area of the prism.•114 m2•240 m2•290 m2•145 m26. Find the surface area of a sphere that has a diameter of 4 cm.•64π cm2•π cm3•4π cm2•16π cm27. Find the lateral area and the surface area of the cone. Use 3.14 for π and round the answer to the nearesthundredth. The diagram is not to scale.•lateral area: 733.33 ft2; surface area: 690.80 ft2•lateral area: 6908.00 ft2; surface area: 1004.80 ft2•lateral area: 690.80 ft2; surface area: 1004.80 ft2•none of these8. If the ratio of the radii of two spheres is 7 : 2, what is the ratio of the surface areas of the twospheres?•7 : 2•7r2π : 2r2π•49 : 4•343 : 89. Use a net to find the surface area of the prism.•465 m2•720 m2•918 m2•930 m210. Find the height of the cylinder to the nearest tenth of an inch.•94.6 in.•96.6 in.• 4.3 in.• 4.1 in.11. Use formulas to find the lateral area and the surface area of the prism. Show your answer to the nearesthundredth.•63.00 m2; 567.00 m2•36.00 m2; 1134.00 m2•479.22 m2; 533.22 m2•542.22 m2; 596.22 m212. Find the volume of the prism.•942 m3•38 m3•945 m3•1890 m313. The volumes of two similar solids are 1331 m3 and 343 m3. The surface area of the larger one is 484 m2.What is the surface area of the smaller one?•343 m2•1372 m2•196 m2•none of these14. Find the surface area of the sphere.•648π m2•72π m2•324π m2•1296π m215. Find the volume of the prism.•40.5 m3•162 m3•9 m3•81 m316. Find the surface area of the solid. Round to the nearest square foot.•36 ft2•32 ft217. Use formulas to find the surface area of the prism. Show your answer to the nearest hundredth.•75.42 cm2•170.16 cm2•86.94 cm2•69.52 cm218. Which figure is a net for a cube?••••19. Cylinder A has radius 1 and height 4 and cylinder B has radius 2 and height 4. Find the ratio of thevolumes of the two cylinders.• 1 : 4• 5 : 620. Neil had a job helping a jeweler. He had the assignment of counting the faces, vertices, and edges on theemeralds. On the first emerald, Neil counted 9 faces and 16 edges. He quickly realized he didn't have to count the vertices. How many vertices were there?•10 vertices•7 vertices•8 vertices•9 vertices1. Find the value of x if AB = 20, BC = 12, and CD = 13. (not drawn to scale)•18.8•16.5•13.4•14.92. Find the measure of each variable if m∠A = 22 and m = 97. (not drawn to scale)•53; 210•53; 105•75; 210•75; 1053. In the plane of lines X and Y, what is the locus of points equidistant from lines X and Y?•line A•line D•line B•line C4. A small messenger company can only deliver within a certain distance from the company. On the graph below,the circular region represents that part of the city where the company delivers, and the center of the circle represents the location of the company. Which equation represents the boundary for the region where the company delivers?•(x + 3)2 + (y– 1)2 = 49•(x + 3)2 + (y– 3)2 = 98•(x + 1)2 + (y– 3)2 = 98•(x + 3)2 + (y– 3)2 = 495. A low-watt radio station can be heard only within a certain distance from the station. On the graph below,the circular region represents that part of the city where the station can be heard, and the center of the circle represents the location of the station. Which equation represents the boundary for the region where the station can be heard?•(x– 4)2 + (y– 5)2 = 50•(x + 5)2 + (y– 5)2 = 59•(x + 5)2 + (y + 4)2 = 25•(x + 5)2 + (y– 5)2 = 256. Find the center and radius of (x– 8)2 + (y + 7)2 = 64.•(–8, 7); 8•(–7, 8); 64•(8, –7); 8•(–7, –8); 87. is tangent to circle O at B. How close to the circle is point A? (The diagram is not to scale.)• 3• 4.5• 6•7.58. Compare the quantity in Column A with the quantity in Column B.•The quantity in Column A is greater.•The quantity in Column B is greater.•The two quantities are equal.•The relationship cannot be determined from the information given.9. Solve for x.•22•710. Find the value of x.•14.6•8.1•9.4•13.411. Write the standard equation for the circle with center (14, –48) that passes through (0, 0).•(x– 14)2 + (y + 48)2 = 2500•(x + 14)2 + (y– 48)2 = 2500•(x + 14)2 + (y– 48)2 = 50•(x– 14)2 + (y + 48)2 = 5012. Find the measure of ∠BAC.•30°•150°13. Find the value of x.•79•39•99•15914. In space, which description fits the locus of points 3 cm from ?•an open cylinder of diameter 6 cm•an open cylinder of radius 3 cm and two hemispheres of diameter 6 cm each•an open cylinder of radius 3 cm and height 6 cm•an open cylinder of diameter 6 cm and two spheres of radius 3 cm each15. and are tangent to circle O and bisects ∠BPA. If m∠AOC= 68°, how much greater is m∠BCOthan m∠OAD? (The diagram is not to scale.)•22°•112°16. Write the standard equation for the circle with center (6, –8) that passes through (0, 0).•(x + 6)2 + (y– 8)2 = 0•(x– 6)2 + (y + 8)2 = 0•(x + 6)2 + (y– 8)2 = 100•(x– 6)2 + (y + 8)2 = 10017. Find the center and radius of (x + 3)2 + (y + 8)2 = 169.•(3, 8); 13•(–8, 3); 13•(–3, –8); 13•(–8, –3); 16918. , , and are all tangent to circle O. If JA = 8, AL = 13, and CK = 11, what is the perimeter ofΔJKL? (The diagram is not to scale.)•32•64•45•5319. If m = 38, what is m∠YAC?•142°•71°•109°•52°20. Write the standard equation for the circle with center (–16, 30) that passes through (0, 0).•(x + 16)2 + (y– 30)2 = 34•(x + 16)2 + (y– 30)2 = 1156•(x– 16)2 + (y + 30)2 = 1156•(x– 16)2 + (y + 30)2 = 34。
2010EuclidSolution
1.(a)Solution1Since3x=27,then3x+2=3x32=27·9=243.Solution2Since3x=27and27=33,then x=3.Therefore,3x+2=35=243.(b)Since2531359x=2731459,then x=27314592531359=2231=12.(c)The lines y=x+2and y=−12x+2both pass through the point B on the y-axis.Since the y-intercept of the line y=x+2is2,then B has coordinates(0,2). Next,wefind the x-intercepts of each of the two lines by setting y=0.If y=x+2and y=0,then x+2=0or x=−2,so A has coordinates(−2,0).If y=−12x+2and y=0,then0=−12x+2or12x=2,and so x=4.Thus,C has coordinates(4,0).Since BO and AC are perpendicular,then we can treat AC as the base of ABC and BO as its height.Note that BO=2and AC=4−(−2)=6.Therefore,the area of ABC is12×AC×BO=12×6×2=6.2.(a)Let r,g and b be the masses of the red,green and blue packages,respectively.We are told that r+g+b=60,r+g=25,and g+b=50.Subtracting the second equation from thefirst,we obtain b=60−25=35.Substituting into the third equation,we obtain g=50−b=50−35=15.Therefore,the mass of the green package is15kg.(b)Suppose that a palindrome p is the sum of the three consecutive integers a−1,a,a+1.In this case,p=(a−1)+a+(a+1)=3a,so p is a multiple of3.The largest palindromes less than200are191,181,171.Note that191and181are not divisible by3,but171is divisible by3.One way to check these without using a calculator is to use the test for divisibility by3:A positive integer is divisible by3if and only if the sum of its digits is divisibleby3.Therefore,191and181cannot be the sum of three consecutive integers.The integer171can be written as56+57+58,so171is the largest palindrome less than 200that is the sum of three consecutive integers.(c)Solution1Since(x+1)(x−1)=8,then x2−1=8or x2=9.Thus,(x2+x)(x2−x)=x(x+1)x(x−1)=x2(x+1)(x−1)=9(8)=72.Solution2Since(x+1)(x−1)=8,then x2−1=8or x2=9,so x=±3.If x=3,then(x2+x)(x2−x)=(32+3)(32−3)=(9+3)(9−3)=12(6)=72.If x=−3,then(x2+x)(x2−x)=((−3)2+(−3))((−3)2−(−3))=(9−3)(9+3)=72.In either case,(x2+x)(x2−x)=72.3.(a)Solution1Bea spends60minutesflying from H to F,30minutes at F,45minutesflying from F to G,60minutes at G,and thenflies from G to H.Thus,her total time is 60+30+45+60=195minutes plus the length of time that it takes her to fly from G to H .Since Bea flies at a constant speed,then the ratio of the two distances equals the ratio of the corresponding times.Therefore,HF GF =60minutes 45minutes =43.Since F GH is right-angled at F ,then F GH must be similar to a 3-4-5triangle,and so HG GF =53.In particular,this means that the ratio of the times flying H to G and F to G is also 53.Thus,it takes her 5×45=75minutes to fly from G to H .In conclusion,Bea is away from her hive for 195+75=270minutes.Solution 2Bea spends 60minutes flying from H to F ,30minutes at F ,45minutes flying from F to G ,60minutes at G ,and then flies from G to H .Thus,her total time is 60+30+45+60=195minutes plus the length of time that it takes her to fly from G to H .Since Bea flies at a constant speed,then the ratio of the two distances equals the ratio of the corresponding times.Therefore,we can use the Pythagorean Theorem on the times to obtainTime G to H = (Time H to F )2+(Time F to G )2=√602+452=√5625=75minsince the time is positive.In conclusion,Bea is away from her hive for 195+75=270minutes.(b)Solution 1Since ∠OP B =90◦,then OP and P B are perpendicular,so the product of their slopes is −1.The slope of OP is 4−0p −0=4p and the slope of P B is 4−0p −10=4p −10.Therefore,we need 4p ·4p −10=−116=−p (p −10)p 2−10p +16=0(p −2)(p −8)=0and so p =2or p =8.Since each these steps is reversible,then OP B is right-angled precisely when p =2and p =8.Solution 2Since OP B is right-angled at P ,then OP 2+P B 2=OB 2by the Pythagorean Theorem.Note that OB =10since O has coordinates (0,0)and B has coordinates (10,0).Also,OP 2=(p −0)2+(4−0)2=p 2+16and P B 2=(10−p )2+(4−0)2=p 2−20p +116.Therefore,(p 2+16)+(p 2−20p +116)=1022p 2−20p +32=0p 2−10p +16=0and so(p−2)(p−8)=0,or p=2or p=8.Since each these steps is reversible,then OP B is right-angled precisely when p=2and p=8.4.(a)Suppose that Thurka bought x goats and y helicopters.Then19x+17y=201.Since x and y are non-negative integers,then19x≤201so x≤10.If x=10,then17y=201−19x=11,which does not have an integer solution because11 is not divisible by17.If x=9,then17y=201−19x=30,which does not have an integer solution.If x=8,then17y=201−19x=49,which does not have an integer solution.If x=7,then17y=201−19x=68,so y=4.Therefore,19(7)+17(4)=201,and so Thurka buys7goats and4helicopters.(We can check that x=0,1,2,3,4,5,6do not give values of y that work.)(b)Solution1Manipulating algebraically,(x+8)4=(2x+16)2(x+8)4−22(x+8)2=0(x+8)2((x+8)2−22)=0(x+8)2((x+8)+2)((x+8)−2)=0(x+8)2(x+10)(x+6)=0Therefore,x=−8or x=−10or x=−6.Solution2Manipulating algebraically,(x+8)4=(2x+16)2(x+8)4−22(x+8)2=0(x+8)2((x+8)2−22)=0(x+8)2(x2+16x+64−4)=0(x+8)2(x2+16x+60)=0(x+8)2(x+10)(x+6)=0Therefore,x=−8or x=−10or x=−6.Solution3Since(x+8)4=(2x+16)2,then(x+8)2=2x+16or(x+8)2=−(2x+16).From thefirst equation,x2+16x+64=2x+16or x2+14x+48=0or(x+6)(x+8)=0.From the second equation,x2+16x+64=−2x−16or x2+18x+80=0or (x+10)(x+8)=0.Therefore,x=−8or x=−10or x=−6.5.(a)Solution1We use the fact that g(x)=g(f(f−1(x))).Since f(x)=2x+1,then to determine f−1(x)we solve x=2y+1for y to get2y=x−1or y=12(x−1).Thus,f−1(x)=12(x−1).Since g(f(x))=4x2+1,theng(x)=g(f(f−1(x)))=g(f(12(x−1)))=4(12(x−1))2+1=4·14(x−1)2+1=(x−1)2+1=x2−2x+2Solution2We use the expressions for f(x)and g(f(x))to construct g(x).Since f(x)is linear and g(f(x))is quadratic,then it is likely that g(x)is also quadratic.Since f(x)=2x+1,then(f(x))2=4x2+4x+1.Since g(f(x))has no term involving x,then we subtract2f(x)(to remove the4x term) to get(f(x))2−2f(x)=(4x2+4x+1)−2(2x+1)=4x2−1 To get g(f(x))from this,we add2to get4x2+1.Therefore,g(f(x))=(f(x))2−2f(x)+2,and so an expression for g(x)is x2−2x+2.Solution3We use the expressions for f(x)and g(f(x))to construct g(x).Since f(x)is linear and g(f(x))is quadratic,then it is likely that g(x)is also quadratic.Suppose that g(x)=ax2+bx+c for some real numbers a,b,c.Theng(f(x))=g(2x+1)=a(2x+1)2+b(2x+1)+c=a(4x2+4x+1)+b(2x+1)+c=4ax2+(4a+2b)x+(a+b+c)Since we are told that g(f(x))=4x2+1,then we can compare coefficients to deduce that 4a=4and4a+2b=0and a+b+c=1.From thefirst equation,a=1.From the second equation,b=−2a=−2.From the third equation,c=1−a−b=2.Therefore,an expression for g(x)is x2−2x+2.(b)Solution1Since the sum of thefirst two terms is40and the sum of thefirst three terms is76,then the third term is76−40=36.Since the sum of thefirst three terms is76and the sum of thefirst four terms is130,then the fourth term is130−76=54.Since the third term is36and the fourth term is54,then the common ratio in the geo-metric sequence is5436=32.Therefore,thefifth term is54·3=81and the sixth term is81·3=243.Also,the second term is 36÷32=36·23=24and the first term is 24÷32=24·23=16.Thus,the first six terms of the sequence are 16,24,36,54,81,2432.Since the first term equals 24and the common ratio is 32,then the n th term in the sequence is 24 32 n −1=3n −12n −5.When n ≥6,this is a fraction whose numerator is odd and whose denominator is even,and so,when n ≥6,the n th term is not an integer.(An odd integer is never divisible by an even integer.)Therefore,there will be 5integers in the sequence.Solution 2Suppose that a is the first term and r is the common ratio between consecutive terms (so that ar is the second term,ar 2is the third term,and so on).From the given information,a +ar =40and a +ar +ar 2=76and a +ar +ar 2+ar 3=130.Subtracting the first equation from the second,we obtain ar 2=36.Subtracting the second equation from the third,we obtain ar 3=54.Since ar 3=54and ar 2=36,then r =ar 3ar 2=5436=32.Since ar 2=36and r =32,then a (32)2=36or 94a =36or a =49·36=16.Since a =16and r =3,then the first six terms of the sequence are 16,24,36,54,81,243.Since the first term equals 24and the common ratio is 32,then the n th term in the sequence is 24 32 n −1=3n −12n −5.When n ≥6,this is a fraction whose numerator is odd and whose denominator is even,and so,when n ≥6,the n th term is not an integer.(An odd integer is never divisible by an even integer.)Therefore,there will be 5integers in the sequence.6.(a)In a 30◦-60◦-90◦triangle,the ratio of the side opposite the 90◦to the side opposite the 60◦angle is 2:√3.Note that each of ABC , ACD , ADE , AEF , AF G ,and AGH is a 30◦-60◦-90◦triangle.Therefore,AH AG =AG AF =AF AE =AE AD =AD AC =AC AB =2√3.Thus,AH =2√3AG = 2√3 2AF = 2√3 3AE = 2√3 4AD = 2√3 5AC = 2√3 6AB .(In other words,to get from AB =1to the length of AH ,we multiply by the “scalingfactor”2√3six times.)Therefore,AH = 2√36=6427.(b)Solution 1Since AF D is right-angled at F ,then by the Pythagorean Theorem,AD =√AF 2+F D 2=√42+22=√20=2√5since AD >0.Let ∠F AD =β.Since ABCD is a rectangle,then ∠BAF =90◦−β.Since AF D is right-angled at F ,then ∠ADF =90◦−β.Since ABCD is a rectangle,then ∠BDC =90◦−(90◦−β)=β.BCTherefore, BF A, AF D,and DF E are all similar as each is right-angled and has either an angle ofβor an angle of90◦−β(and hence both of these angles). Therefore,ABAF=DADFand so AB=4(2√5)2=4√5.Also,F EF D=F DF Aand so F E=2(2)4=1.Since ABCD is a rectangle,then BC=AD=2√5,and DC=AB=4√5.Finally,the area of quadrilateral BCEF equals the area of DCB minus the area DF E. Thus,the required area is1(DC)(CB)−1(DF)(F E)=1(4√5)(2√5)−1(2)(1)=20−1=19Solution2Since AF D is right-angled at F,then by the Pythagorean Theorem,AD=√AF2+F D2=√42+22=√20=2√5since AD>0.Let∠F AD=β.Since ABCD is a rectangle,then∠BAF=90◦−β.Since BAF is right-angled at F, then∠ABF=β.Since AF D is right-angled at F,then∠ADF=90◦−β.Since ABCD is a rectangle,then∠BDC=90◦−(90◦−β)=β.BCLooking at AF D,we see that sinβ=F DAD=22√5=1√5,cosβ=AFAD=42√5=2√5, and tanβ=F DAF=24=12.Since AF=4and∠ABF=β,then AB=AFsinβ=41√5=4√5.Since F D=2and∠F DE=β,then F E=F D tanβ=2·12=1.Since ABCD is a rectangle,then BC=AD=2√5,and DC=AB=4√5.Finally,the area of quadrilateral EF BC equals the area of DCB minus the area DF E. Thus,the required area is12(DC)(CB)−12(DF)(F E)=12(4√5)(2√5)−12(2)(1)=20−1=197.(a)Using the facts that 9=32and 27=33,and the laws for manipulating exponents,wehave3x −1932x 2=273x −1(32)32x 2=333x −133x 2=333x −1+3x 2=33When two powers of 3are equal,their exponents must be equal sox −1+3x 2=3x 3−x 2+3=3x 2(multiplying by x 2)x 3−4x 2+3=0Since x =1satisfies the equation,then x −1is a factor of the left ing long division or synthetic division,we can factor this out to get (x −1)(x 2−3x −3)=0.Using the quadratic formula,the quadratic equation x 2−3x −3=0has rootsx =3± (−3)2−4(1)(−3)2=3±√212Therefore,the solutions to the original equation are x =1and x =3±√212.(b)To determine the points of intersection,we equate y values of the two curves and obtainlog 10(x 4)=(log 10x )3.Since log 10(a b )=b log 10a ,the equation becomes 4log 10x =(log 10x )3.We set u =log 10x and so the equation becomes 4u =u 3,or u 3−4u =0.We can factor the left side as u 3−4u =u (u 2−4)=u (u +2)(u −2).Therefore,u (u +2)(u −2)=0,and so u =0or u =−2or u =2.Therefore,log 10x =0or log 10x =−2or log 10x =2.Therefore,x =1or x =1100or x =100.Finally,we must calculate the y -coordinates of the points of intersection.Since one of theoriginal curves is y =(log 10x )3,we can calculate the corresponding values of y by using the fact that y =u 3.The corresponding values of y are y =03=0and y =(−2)3=−8and y =23=8.Therefore,the points of intersection are (1,0),(1,−8)and (100,8).8.(a)If Oi-Lam tosses 3heads,then George has no coins to toss,so cannot toss exactly 1head.If Oi-Lam tosses 2,1or 0heads,then George has at least one coin to toss,so can toss exactly 1head.Therefore,the following possibilities exist:∗Oi-Lam tosses 2heads out of 3coins and George tosses 1head out of 1coin∗Oi-Lam tosses 1head out of 3coins and George tosses 1head out of 2coins∗Oi-Lam tosses 0heads out of 3coins and George tosses 1head out of 3coinsWe calculate the various probabilities.If 3coins are tossed,there are 8equally likely possibilities:HHH,HHT,HTH,THH,TTH,THT,HTT,TTT.Each of these possibilities has probability 1 3=1.Therefore,∗the probability of tossing 0heads out of 3coins is 18∗the probability of tossing 1head out of 3coins is 38∗the probability of tossing 2heads out of 3coins is 38∗the probability of tossing 3heads out of 3coins is 18If 2coins are tossed,there are 4equally likely possibilities:HH,HT,TH,TT.Each of these possibilities has probability 12 2=14.Therefore,the probability of tossing 1head out of 2coins is 2=1.If 1coin is tossed,the probability of tossing 1head is 12.To summarize,the possibilities are∗Oi-Lam tosses 2heads out of 3coins (with probability 38)and George tosses 1head out of 1coin (with probability 12)∗Oi-Lam tosses 1head out of 3coins (with probability 38)and George tosses 1head out of 2coins (with probability 12)∗Oi-Lam tosses 0heads out of 3coins (with probability 18)and George tosses 1head out of 3coins (with probability 38)Therefore,the overall probability is 38·12+38·12+18·38=2764.(b)Suppose ∠P AR =x ◦and ∠QDR =y ◦.Since P R and P A are radii of the larger circle,then P AR is isosceles.Thus,∠P RA =∠P AR =x ◦.Since QD and QR are radii of the smaller circle,then QRD is isosceles.Thus,∠QRD =∠QDR =y ◦.In ARD ,the sum of the angles is 180◦,so x ◦+(x ◦+40◦+y ◦)+y ◦=180◦or 2x +2y =140or x +y =70.Therefore,∠CP D =x ◦+40◦+y ◦=(x +y +40)◦=110◦.9.(a)(i)Solution 1LS =cot θ−cot 2θ=cos θsin θ−cos 2θsin 2θ=sin 2θcos θ−cos 2θsin θsin θsin 2θ=sin(2θ−θ)sin θsin 2θ=sin θsin θsin 2θ=1sin 2θ=RSas required.Solution2LS=cotθ−cot2θ=cosθsinθ−cos2θsin2θ=cosθsinθ−cos2θ2sinθcosθ=2cos2θ−cos2θ2sinθcosθ=2cos2θ−(2cos2θ−1)sin2θ=1 sin2θ=RS as required.(ii)We use(i)to note that1sin8◦=cot4◦−cot8◦and1sin16◦=cot8◦−cot16◦and soon.Thus,S=1sin8◦+1sin16◦+1sin32◦+···+1sin4096◦+1sin8192◦=(cot4◦−cot8◦)+(cot8◦−cot16◦)+(cot16◦−cot32◦)+···+(cot2048◦−cot4096◦)+(cot4096◦−cot8192◦)=cot4◦−cot8192◦since the sum“telescopes”.Since the cotangent function has a period of180◦,and8100◦is a multiple of180◦, then cot8192◦=cot92◦.Therefore,S=cot4◦−cot92◦=cos4◦sin4◦−cos92◦sin92◦=cos4◦sin4◦−−sin2◦cos2◦=cos4◦2sin2◦cos2◦+sin2◦cos2◦=cos4◦+2sin22◦2sin2◦cos2◦=(1−2sin22◦)+2sin22◦sin4◦=1 sin4◦Therefore,α=4◦.(b)Solution 1We use the notation A =∠BAC ,B =∠ABC and C =∠ACB .We need to show that A <12(B +C ).Since the sum of the angles in ABC is 180◦,then B +C =180◦−A ,and so this inequality is equivalent to A <12(180◦−A )which is equivalent to 32A <90◦or A <60◦.So we need to show that A <60◦.We know that a <12(b +c ).Thus,2a <b +c and so 4a 2<b 2+c 2+2bc because all quantities are positive.Using the cosine law in ABC ,we obtain a 2=b 2+c 2−2bc cos A .Therefore,4a 2<b 2+c 2+2bc 4(b 2+c 2−2bc cos A )<b 2+c 2+2bc 4b 2+4c 2−8bc cos A<b 2+c 2+2bc 4b 2+4c 2−8bc cos A<b 2+c 2+2bc +3(b −c )2(since (b −c )2≥0)4b 2+4c 2−8bc cos A<b 2+c 2+2bc +3b 2−6bc +3c 24b 2+4c 2−8bc cos A<4b 2+4c 2−4bc −8bc cos A<−4bc cos A >12(since 8bc >0)Since 2a <b +c ,then a cannot be the longest side of ABC (that is,we cannot have a ≥b and a ≥c ),so A must be an acute angle.Therefore,cos A >12implies A <60◦,as required.Solution 2We use the notation A =∠BAC ,B =∠ABC and C =∠ACB .We need to show that A <12(B +C ).Since the sum of the angles in ABC is 180◦,then B +C =180◦−A ,and so this inequality is equivalent to A <12(180◦−A )which is equivalent to 32A <90◦or A <60◦.So we need to show that A <60◦.We know that a <12(b +c )which implies 2a <b +c .Using the sine law in ABC ,we obtain a sin A =b sin B =c sin C ,which gives b =a sin B sin A and c =a sin C sin A .Therefore,we obtain equivalent inequalities2a <b +c2a <a sin B sin A +a sin C sin A2a sin A <a sin B +a sin C(since sin A >0for 0◦<A <180◦)2sin A <sin B +sin C since a >0.Next,we use the trigonometric formula sin B +sin C =2sin B +C 2 cos B −C 2 .Since cos θ≤1for any θ,then sin B +sin C ≤2sin B +C 2 ·1=2sin B +C 2.Therefore,2sin A<sin B+sin C≤2sinB+C22sin A<2sinB+C22sin A<2sin180◦−A24sin(12A)cos(12A)<2sin(90◦−12A)2sin(12A)cos(12A)<cos(12A)Since0◦<A<180◦,then cos(1A)>0,so sin(1A)<1.Since2a<b+c,then a cannot be the longest side of ABC,so A must be an acute angle.Therefore,1A<30◦or A<60◦,as required.10.Denote the side lengths of a triangle by a,b and c,with0<a≤b≤c.In order for these lengths to form a triangle,we need c<a+b and b<a+c and a<b+c.Since0<a≤b≤c,then b<a+c and a<b+c follow automatically,so only c<a+b ever needs to be checked.Instead of directly considering triangles and sets of triangle,we can consider triples(a,b,c)and sets of triples(a,b,c)with the appropriate conditions.For each positive integer k≥3,we use the notation S k to denote the set of triples of positive integers(a,b,c)with0<a≤b≤c and c<a+b and a+b+c=k.In this case,c<a+b and a+b+c=k,so c+c<a+b+c=k,so2c<k or c<12k.Also,if0<a≤b≤c and a+b+c=k,then k=a+b+c≤c+c+c,so3c≥k or c≥13k.(a)Consider T(10),which is the number of elements in S10.We want tofind all possible triples(a,b,c)of integers with0<a≤b≤c and c<a+b and a+b+c=10.We need c<102=5and c≥103.Thus,c=4.Therefore,we need0<a≤b≤4and a+b=6.There are two possibilities:(a,b,c)=(2,4,4)or(a,b,c)=(3,3,4).Therefore,T(10)=2.Consider T(11).We want tofind all possible triples(a,b,c)of integers with0<a≤b≤c and c<a+b and a+b+c=11.We need c<112and c≥113.Thus,c=4or c=5.If c=4,we need0<a≤b≤4and a+b=7.There is only one possibility:(a,b,c)=(3,4,4).If c=5,we need0<a≤b≤5and a+b=6.There are three possibilities:(a,b,c)=(1,5,5)or(a,b,c)=(2,4,5)or(a,b,c)=(3,3,5). Therefore,T(11)=4.Consider T(12).We want tofind all possible triples(a,b,c)of integers with0<a≤b≤c and c<a+b and a+b+c=12.We need c<122and c≥123.Thus,c=4or c=5.If c=4,we need0<a≤b≤4and a+b=8. There is only one possibility:(a,b,c)=(4,4,4).If c=5,we need0<a≤b≤5and a+b=7.There are two possibilities:(a,b,c)=(2,5,5)or(a,b,c)=(3,4,5).Therefore,T(12)=3.(b)We show that T(2m)=T(2m−3)by creating a one-to-one correspondence between thetriples in S2m and the triples S2m−3.Note that S2m is the set of triples(a,b,c)of positive integers with0<a≤b≤c,with c<a+b,and with a+b+c=2m.Also,S2m−3is the set of triples(A,B,C)of positive integers with0<A≤B≤C,with C<A+B,and with A+B+C=2m−3.Consider a triple(a,b,c)in S2m and a corresponding triple(a−1,b−1,c−1).We show that(a−1,b−1,c−1)is in S2m−3:∗Since(a,b,c)is in S2m,then c<12(2m)=m.This means that b≤c≤m−1,soa=2m−b−c≥2.Therefore,a−1,b−1and c−1are positive integers since a,b and c are positive integers with2≤a≤b≤c.∗Since2≤a≤b≤c,then1≤a−1≤b−1≤c−1,so0<a−1≤b−1≤c−1.∗Since a+b+c=2m,then c=2m−(a+b)so a+b and c have the same parity.Since c<a+b,then c≤a+b−2.(In other words,it cannot be the case that c=a+b−1.)Therefore,c−1≤(a−1)+(b−1)−1;that is,c−1<(a−1)+(b−1).∗Since a+b+c=2m,then(a−1)+(b−1)+(c−1)=2m−3.Therefore,(a−1,b−1,c−1)is in S2m−3,since it satisfies all of the conditions of S2m−3.Note as well that two different triples in S2m correspond to two different triples in S2m−3.Thus,every triple in S2m corresponds to a different triple in S2m−3.Thus,T(2m)≤T(2m−3).Consider a triple(A,B,C)in S2m−3and a corresponding triple(A+1,B+1,C+1).We show that(A+1,B+1,C+1)is in S2m:∗Since(A,B,C)is in S2m−3,then A,B and C are positive integers,so A+1,B+1 and C+1are positive integers.∗Since0<A≤B≤C,then1<A+1≤B+1≤C+1,so0<A+1≤B+1≤C+1.∗Since C<A+B,then C+1<(A+1)+(B+1)−1so C+1<(A+1)+(B+1).∗Since A+B+C=2m−3,then(A+1)+(B+1)+(C+1)=2m.Therefore,(A+1,B+1,C+1)is in S2m.Note again that two different triples in S2m−3correspond to two different triples in S2m.Thus,every triple in S2m−3corresponds to a different triple in S2m.Therefore,T(2m−3)≤T(2m).Since T(2m)≤T(2m−3)and T(2m−3)≤T(2m),then T(2m)=T(2m−3).(c)We will use two important facts:(F1)T(2m)=T(2m−3)for every positive integer m≥3,and(F2)T(k)≤T(k+2)for every positive integer k≥3We proved(F1)in(b).Next,we prove(F2):Consider a triple(a,b,c)in S k and a corresponding triple(a,b+1,c+1).Weshow that the triple(a,b+1,c+1)is in S k+2:∗Since a,b and c are positive integers,then a,b+1and c+1are positiveintegers.∗Since0<a≤b≤c,then0<a≤b+1≤c+1.∗Since c<a+b,then c+1<a+(b+1).∗Since a+b+c=k,then a+(b+1)+(c+1)=k+2.Therefore,(a,b+1,c+1)is in S k+2.Note that,using this correspondence, different triples in S k correspond different triples in S k+2.Thus,every triple in S k corresponds to a different triple in S k+2.This proves that T(k)≤T(k+2). Suppose that n=N is the smallest positive integer for which T(n)>2010.Then N must be odd:If N was even,then by(F1),T(N−3)=T(N)>2010and so n=N−3would be an integer smaller than N with T(n)>2010.This contradicts the fact that n=N is the smallest such integer.Therefore,we want tofind the smallest odd positive integer N for which T(N)>2010. Next,we note that if we canfind an odd positive integer n such that T(n)>2010≥T(n−2),then we will have found the desired value of n:This is because n and n−2are both odd,and by property(F2),any smaller odd positive integer k will give T(k)≤T(n−2)≤2010and any larger odd positive integer m will give T(m)≥T(n)>2010.We show that N=309is the desired value of N by showing that T(309)>2010and T(307)≤2010.Calculation of T(309)We know that3093≤c<3092,so103≤c≤154.For each admissible value of c,we need to count the number of pairs of positive integers (a,b)with a≤b≤c and a+b=309−c.For example,if c=154,then we need a≤b≤154and a+b=155.This gives pairs(1,154),(2,153),...,(76,79),(77,78),of which there are77.Also,if c=153,then we need a≤b≤153and a+b=156.This gives pairs(3,153),...,(77,79),(78,78),of which there are76.In general,if c is even,then the minimum possible value of a occurs when b is as large as possible–that is,when b=c,so a≥309−2c.Also,the largest possible value of a occurs when a and b are as close to equal as possible. Since c is even,then309−c is odd,so a and b cannot be equal,but they can differ by1.In this case,a=154−12c and b=155−12c.Therefore,if c is even,there are(154−1c)−(309−2c)+1=3c−154possible pairs(a,b)and so32c−154possible triples.In general,if c is odd,then the minimum possible value of a occurs when b is as large as possible–that is,when b=c,so a≥309−2c.Also,the largest possible value of a occurs when a and b are as close to equal as possible.Since c is odd,then309−c is even,so a and b can be equal.In this case,a=12(309−c).Therefore,if c is odd,there are1(309−c)−(309−2c)+1=3c−307possible pairs(a,b)and so32c−3072possible triples.The possible even values of c are104,106,...,152,154(there are26such values)and the possible odd values of c are103,105,...,151,153(there are26such values).Therefore,T(309)= 32(104)−154+32(106)−154+···+32(154)−154+ 32(103)−3072+32(105)−3072+···+32(153)−3072=32(104+106+···+154)−26·154+32(103+105+···+153)−26·3072=32(103+104+105+106+···+153+154)−26·154−26·3072=32·12(103+154)(52)−26·154−26·3072=32(26)(257)−26·154−26·3072=2028Therefore,T(309)>2010,as required.Calculation of T(307)We know that307≤c<307,so103≤c≤153.For each admissible value of c,we need to count the number of pairs of positive integers (a,b)with a≤b≤c and a+b=307−c.This can be done in a similar way to the calculation of T(309)above.If n is even,there are32c−153possible triples.If n is odd,there are32c−3052possible triples.The possible even values of c are104,106,...,150,152(there are25such values)and the possible odd values of c are103,105,...,151,153(there are26such values).Therefore,T(307)= 32(104)−153+32(106)−153+···+32(152)−153+ 32(103)−3052+32(105)−3052+···+32(153)−3052=32(104+106+···+152)−25·153+32(103+105+···+153)−26·3052=32(103+104+105+106+···+152+153)−25·153−26·3052=3·1(103+153)(51)−25·153−26·305=32(51)(128)−25·153−26·3052=2002Therefore,T(307)<2010,as required.Therefore,the smallest positive integer n such that T(n)>2010is n=309.As afinal note,we discuss briefly how one could guess that the answer was near N=309.Consider the values of T(n)for small odd positive integers n.In(a),by considering the possible values of c from smallest(roughly13n)to largest(roughly12n),we saw that T(11)=1+3=4.If we continue to calculate T(n)for a few more small odd values of n we will see that:T(13)=2+3=5T(15)=1+2+4=7T(17)=1+3+4=8T(19)=2+3+5=10T(21)=1+2+4+5=12T(23)=1+3+4+6=14The pattern that seems to emerge is that for n odd,T(n)is roughly equal to thesum of the integers from1to14n,with one out of every three integers removed.Thus,T(n)is roughly equal to23of the sum of the integers from1to14n.Therefore,T(n)≈23·12(14n)(14n+1)≈23·12(14n)2≈148n2.It makes sense to look for an odd positive integer n with T(n)≈2010.Thus,we are looking for a value of n that roughly satisfies148n2≈2010orn2≈96480or n≈310.Since n is odd,then it makes sense to consider n=309,as in the solution above.。
加拿大数学10年级练习第二部分
1. Solve – + = 4. Check your solution.•5, –2• 5•6, –1• 62. Solve = . Check your solution.•–•8• 4•no solution3. Divide ÷ .••••4. Solve –= 1. Check your solution.•0•–7•–7, 0•6, –75. The graph of the function y = is on the left. The graph on the right is the same graph translated right threeunits and up three units. What is the equation of the graph on the right?•y = + 3•y = – 3•y = + 3•y = – 36. Divide (11x2 + 2x3 + 14 + 17x) ÷ (2x + 5).•x2 + 3x + 2 +•x2 + 3x + 1 +•x2– 3x– 1 +•x2– 3x– 2 +7. If no digit appears more than once, how many 2-digit numbers can be formed from the digits 2, 3, 5, 6?• 6• 4•12•658. Divide ÷ (x– 2).•••9. Divide (–3x3 + 4x–5) ÷ (x + 3).•–3x2 + 9x– 23 +•–3x2 + 13x + 39 –•–3x2 + 9x + 31 –•–3x2 + 13x– 44 +10. Divide ÷ .••••11. Multiply • .••••12. Simplify .•••13. Divide (6x2– 5x+ 2) ÷ (3x + 2).•2x + 3•2x– 3 –•2x + 3 –•2x– 3 +14. Simplify .•x2– 9••–x– 5•x + 515. Jamestown Builders has a development of new homes. There are six different floor plans, five exterior colors,and an option of either a two-car or a three-car garage. How many choices are there for one home?•30•90•60•5216. Divide ÷ (x + 6).••••17. Simplify .•••18. The rate of discount R, can be determined using the formula R = 1 –, where P is the regular price ofan item and D is the discount, or amount saved off the regular price. Find the rate of discount on a sweater witha regular price of $28.99 that is on sale for $10.99.•37.9%•$10.99•$18•62.1%19. Fifteen students volunteer for a committee. How many different eight-person committees can be chosen?•6435 committees•40,320 committees•5040 committees•259,459,200 committees20. Divide (2x2 + 4x3– 9 – 10x) ÷ (2x + 3).•2x2– 2x– 1 –•2x2 + 2x + 2 –•2x2 + 2x + 1 –•2x2– 2x– 2 –1. Identify the graph of 2x– 5y = –10.••••2. Write an equation for the translation of y = |x| left 0.3 units.•y = |x| – 0.3•y = |x| + 0.3•y = |x– 0.3|•y = |x + 0.3|3. Which is the graph of y = |x + 4| – 2?••••4. The rate of change is constant in the graph. Find the rate of change. Explain what the rate of change means forthe situation.•–200, value drops $200 every year•–, value drops $200 every 3 years•–20, value drops $20 every year•–60, value drops $60 every year5. Identify the graph of y = –|x– 5| by translating y = –|x|.••••6. Identify the graph of –10x + 5y = –50.••••7. A line passes through (–7, 1) and (4, 4). Identify the point-slope and slope-intercept form of an equation ofthe line.•point-slope: y– 1 = (x + 7); slope-intercept: y = x +•point-slope: y + 7 = (x– 1); slope-intercept: y = x–•point-slope: y– 1 = (x + 7); slope-intercept: y = x +•point-slope: y + 7 = (x– 1); slope-intercept: y = x–8. As water is poured into a tank, the volume was measured every minute. It produced the graph below. What was thevolume at three minutes?• 3.0• 6.0• 4.0•7.99. The table shows the time spent researching the stock market each week and the average weekly percent gain foran investor over one year. Use a trend line to predict the average weekly percent gain from researching the stock market for 20 hours per week.•about 10%•about 20%•about 0.5%•about 8%10. Write an equation for the translation of y = |x| down 0.75 units.•y = |x+ 0.75|•y = |x| + 0.75•y = |x| – 0.75•y = |x– 0.75|11. Identify the graph of y = |x| + 3.5 by translating y = |x|.••••12. Which is the graph of y = |x + 5| + 3?••••13. Graph the data for the times Shauna spent learning music for the violin and her scores at the annual solocompetitions. Find an equation of the trend line of the data.•y– 31 = 2.5(x– 10)•x– 31 = 2.5(y– 10)•y– 10 = 2.5(x– 21)•y + 31 = 2.5(x + 10)14. Find the slope of the line.• 6•–6••–15. Find the slope of the line.•–1• 5•–5•undefined16. Write the slope-intercept form of the equation for the line.•y = –x + 7•y = x– 7•y = –x– 7•y = x + 717. Find the slope of a line perpendicular to the graph of 5x + y = 7.•–5••–• 518. Describe the relationship between the average June temperatures and the latitude positions of cities on thescatter plot below.•The temperatures are about the same at all latitudes.•The lower the latitude, the lower the temperature.•The higher the latitude, the lower the temperature.•The higher the latitude, the higher the temperature.19. Find the slope of the line.•–•–3•0•undefined20. The rate of change is constant in the graph. Find the rate of change. Explain what the rate of change meansfor the situation.•–20, value drops $20 every year•–60, value drops $60 every year•–, value drops $200 every 3 years•–200, value drops $200 every year1. You decide to market your own custom computer software. You must invest $4305 on computer hardware, and spend$2.75 to buy and package each disk. If each program sells for $13.00, how many copies must you sell to break even?•421 copies•274 copies•273 copies•420 copies2. Solve by elimination.3x– 4y = 9–3x + 2y = 9•(–27, –9)•(–9, –9)•(–3, –6)•(3, 9)3. You decide to market your own custom computer software. You must invest $3255 on computer hardware, and spend$2.90 to buy and package each disk. If each program sells for $13.75, how many copies must you sell to break even?•195 copies•301 copies•196 copies•300 copies4. What is the graph of y > –6x + 3?••••5. Which ordered pair is a solution of the system?7x + 2y = –39–8x + 3y = –3•(–3, –9)•(–7, –11)•(–4, –7)•(–1, –10)6. Solve by elimination.4x–y = 43x + y = 10•(0, –4)•(12, 4)•(2, 4)•(4, 12)7. The length of a rectangle is 3 centimeters more than 3 times the width. If the perimeter of the rectangle is46 centimeters, what are the dimensions of the rectangle?•length = 5 cm; width = 18 cm•length = 13 cm; width = 5 cm•length = 13 cm; width = 8 cm•length = 18 cm; width = 5 cm8. Which point is a solution of y≥ 4x– 5?•(1, 1)•(2, 1)•(3, 0)•(3, 4)9. Which system has infinitely many solutions?•x + y = 3x + y = 4•2x + y = 54x + 2y = 10•2x + y = 74x– 2y = 14•3x–y = 2x– 3y = –210. Solve the system using substitution.f(x) = x– 4f(x) = x– 6•(3, –3)•(3, –4)•(6, –2)•(4, 6)11. A motorboat can go 16 miles downstream on a river in 20 minutes. It takes 30 minutes for this boat to go backupstream the same 16 miles. Find the speed of the boat in still water.•40 mi/h•32 mi/h•8 mi/h•48 mi/h12. Which system has no solution?•4x– 2y = 1y = 2x– 7•y = 2x + 2x– 2y = 1•y = –x + 1x–y = 1•3x–y = 3y = –3x + 313. A motorboat can go 8 miles downstream on a river in 20 minutes. It takes 30 minutes for this boat to go backupstream the same 8 miles. Find the speed of the current.•20 mi/h•16 mi/h•24 mi/h• 4 mi/h14. Find the solution of the system.3y = –x + 2y = –x + 9•(3, 6)•(20, –4)•(10, –1)•(–1, 8)15. Mrs. Huang operates a soybean farm outside Grinnell, Iowa. To keep her operating costs down, she buys many productsin bulk and transfers them to smaller containers for use on the farm. Often the bulk products are not the correct concentration and need to be custom mixed before Mrs. Huang can use them. One day she wants to apply herbicide toa large field. A solution of 67% herbicide is to be mixed with a solution of 46% herbicide to form 42 liters ofa 55% solution. How much of the 67% solution must she use?•34 L•23 L•18 L•35 L16. What is the graph of y≤ 2x– 2?••••17. Write x– 2y < –2 in slope-intercept form. Then graph the inequality.•y > x + 1•y < x + 1•y < –x + 1•y > –x + 118. Which graphing calculator screen shows the solution of the system?y = –x + 3y = 6x– 5••••19. Identify the graph of y≤ 3x– 3.••••20. Mike and Kim invest $10,000 in equipment to print yearbooks for schools. Each yearbook costs $5.75 to printand sells for $25. How many yearbooks must they sell before their business breaks even?•500 yearbooks•520 yearbooks•400 yearbooks•1,740 yearbooks1. This graph represents a translation of the graph of y = |x|. What is the equation of this graph?•y = |x− 2|•y = |x| − 2•y= |x + 2|•y = |x| + 22. Find an equation for the linear model of the situation below and use it to make a prediction. A train is travelingnorth at a constant rate. At 3:00 P.M. it is 55 miles north of a city. At 4:15 P.M. it is 80 miles north of the city.If d represents the distance in miles, and t represents the time in hours, how many miles north of the city will the train be at 5:45 P.M.?•d = 64t + 55; d = 231 miles•d = 80t + 55; d = 255 miles•d = 64t + 55; d = 96 miles•d = 100t + 55; d = 285 miles3. What are the domain and range of this function?•Domain: x > 0; Range: y > 0•Domain: x≥ 0; Range: y≥ 0•Domain: all real numbers; Range: all real numbers•Domain: positive integers; Range: positive integers4. What is the graph of y > |x− 3|?••••5. An electronics store makes a profit of $55 for every standard CD player sold and $77 for every portable CD playersold. The manager's goal is to make at least $385 a day on sales of both standard and portable CD players. If s represents the number of standard CD players sold, and p represents the number of portable CD players sold, which inequality and graph represent the manager's daily goal?•55s + 77p≤ 385•77s + 55p≥ 385•55s + 77p≥ 385•77s + 55p≤ 3856. What is an equation of the inequality?•y≤ |x− 1| − 2•y≥ |x + 1| − 2•y≥ |x− 1| − 27. What is the graph of y≥ |x + 1| − 1?••••8. What is the graph of y = −|3x|?•••9. What is the graph of y = |−x− 4|?•••10. What is the equation of y = |x| translated 1 unit up and 6 units to the left?•y = |x− 6| + 1•y = |x− 1| + 6•y = |x + 1| + 6•y = |x + 6| + 111. What is the graph of y = |x| − |2x|?••••not here12. The scatter plot below shows the weights in ounces of several kittens at various ages. What is the best equationof the trend line through these points?•y = 10x•y = 4x•y = x + 10•y = 4x + 1013. What is the standard form of the equation of the line whose slope is −2 and which contains the point (3, −5)?•2x + y = 1•2x + y = −8•2x + y = 11•2x + y = −714. Find f(−5) if f(x) = −8x− 1.•39•−41•−39•1215. The points in the table have been graphed in the scatter plot below. What is the best equation for the trendline, and the value of y when x is 16?•y = 1.25x + 1; y = 21•y = −1.25x + 1; y = 19•y = 2x + 1; y = 33•y = 0.8x + 1; y = 11.816. A linear model for the situation below passes through the origin.A manufacturer can produce 4200 gears in 8 hours. How many gears can be made in 1 hour?•525•262.5•1050•60017. What is an inequality for this graph?•y− 3x > −5•y− 3x≥ −5•y− 3x≤ −5•y− 3x < −518. Which set of points represents the graph of the function below?(−3, 2), (−1, −1), (2, 2), (4, 0)••••19. What is the graph of the line whose slope is and whose y-intercept is 4?••••20. The graph below models the burning of a candle. At time t = 0, the candle was 8 inches tall. If the candle burnsat a rate of 2 inches per hour, what equation models the height h of the candle after burning for t hours?•h = 2t + 8•h = 8t + 2•h = 8t− 2•h = −2t + 81. Which system of equations is the result of using the elimination method to solve the system ,and what is the solution?•; (1, 12)•; (11, 13)•; (2, )•; (13, 4)2. Which point lies in the plane represented by the equation 2x + 7y + 3z = 25?•(−6, −4, 5)•(3, −3, 3)•(−3, 3, −2)•(6, 1, −3)3. The equation P= 5x+ 3y represents the objective function for this linear programming model. What is the vertexof the feasible region shown below that results in the maximum value of P?•(12, 14)•(12, 2)•(14, 4)•(8, 12)4. Use elimination to find the solution of the system .•(−1, , 3)•(−1, 3, 2)•(1, 3, 0)•(−1, 2, 4)5. Find the values of x and y that maximize the objective function for this graph.Maximum for P = 3x + 2y•(0, 9)•(0, 0)•(5, 0)•(1, 8)6. Which system is inconsistent?••••7. A point (x, y, z) can be used to represent the location of a geographical point in space, where x representsthe latitude in degrees, y represents the longitude in degrees, and z represents the elevation in feet above or below sea level.Jerimoth Hill is the highest point of elevation in Rhode Island. It is 812 feet above sea level and lies at about 42° north latitude and 72° west longitude. What ordered triple represents the location of Jerimoth Hill?•(42°, −72°, −812)•(42°, −72°, 812)•(−72°, 42°, 812)•(−72°, 42°, −812)8. What is the graph of −3x + 5y +10z = 15?••••9. What steps would you use to plot (−3, −4, 0) in a three-dimensional coordinate system?•From the origin, move back 3 units and down 4 units.•From the origin, move back 3 units and left 4 units.•From the origin, move down 3 units and left 4 units.•From the origin, move back 3 units and up 4 units.10. Identify the graph of the system and its solution.•(3, 0)•(−3, 0)•infinite number of solutions•no solution11. What is the solution of this system?•(, )•(−2, −3)•(−14, −23)•(−3, −2)12. A small fish market sells only tuna and salmon. A tuna costs the fish market $0.75 per pound to buy and $2.53per pound to clean and package. A salmon costs the fish market $3.00 per pound to buy and $2.75 per pound to clean and package. The fish market makes $2.50 per pound profit for each tuna it sells and $2.80 per pound profit for each salmon it sells. The fish market owner can spend only $159.00 per day to buy fish and $197.34 per day to clean and package the fish.The graph below represents the feasible region for this linear programming model. What are the coordinates of the vertices of the feasible region, and what are the vales of t and s that maximize the objective function?; P = 2.50t + 2.80s•(0, 0), (0, 53), (71.76, 0), (46, 28); t = 46 and s = 28.•(0, 0), (53, 0), (0, 71.76), (28, 46); t = 28 and s = 46.•(0, 0), (0, 53), (71.76, 0), (28, 46); t = 28 and s = 46.•(0, 0), (0, 53), (71.76, 0), (46, 28); t = 0 and s = 53.13. Identify the graph of the system and its solution.•(−5, −4)•(, −)•(, −)•(−1, 8)14. Identify the graph that shows the solution to the system of inequalities.••••15. What is the maximum for the objective function for this graph?Maximum for P = 5x + 2y•90•80•250•56016. How could you solve this system using substitution?•Substitute for r in the equation t = 2r + 3.•Substitute for t in the equation t = 2r + 3.•Substitute 2r + 3 for t in the equation 5r− 4t = 6.•Substitute 2r + 3 for r in the equation 5r− 4t = 6.17. How could you solve this system by substitution?•Substitute 6 for c in the equation 3c + 5d = 2.•Substitute 6 for d into the equation 3c + 5d = 2.•Substitute for d into the equation d = 6.•Substitute 3(3c + 5d) for 6 in the equation d = 6.18. Identify the graph that shows the solution to the system of inequalities.••••19. Identify the graph that shows the solution to the system of inequalities.••••20. Locate the point (4, −3, 3) in a three-dimensional coordinate system.••••none of these。
加拿大数学10年级练习第四部分几何
1. Find the ratio of the perimeter of the larger rectangle to the perimeter of the smaller rectangle.∙∙∙∙2. Find the area of the triangle with ∠A = 41, b = 5 ft, and c = 4 ft. Round your answer to two decimal places. ∙ 6.56 ft2∙10.00 ft2∙7.55 ft2∙13.12 ft23. Find the area of kite ABCD if BD = 48 cm, AB = 25 cm, and BC = 26. The kite is not drawn to scale.∙289 cm2∙70 cm2∙816 cm2∙408 cm24. The diameter of a basketball rim is 18 inches. A standard basketball has a circumference of 30 inches. Abouthow much room is there between the ball and the rim in a shot in which the ball goes in exactly in the center of the rim?∙none of these5. Find the area.∙718 square units∙545 square units∙534.5 square units∙701 square units6. Name the major arc and find its measure.∙m = 275∙m = 170∙m = 85∙m = 2757. Find the area of the regular polygon. Round your answer to the nearest tenth.∙110.2 in.28. Name the minor arc and find its measure.∙m = 275∙m = 85∙m = 170∙m = 2759. Find the area of ΔABC. The figure is not drawn to scale.∙26.06 cm2∙28.00 cm2∙22.73 cm2∙24.95 cm210. Find the circumference of the circle. Use as an approximation of π.∙11 cm∙5 cm∙6 cm11. If a dart hits the target at random, what it the probability that it will land in the unshaded region?∙∙∙∙12. Find the area of the circle. Use π = 3.14 and round to the nearest hundredth.∙91.56 m2∙16.96 m2∙ 5.72 m2∙22.89 m213. Find the area of a regular pentagon with side 6 cm.∙45.2 cm2∙123.9 cm214. Find the probability that an object falling randomly on the figure will land in the shaded area.∙0.32∙0.36∙0.5∙0.2615. Find the area.∙102.9 yd2∙205.8 yd2∙35.35 yd2∙105.5 yd216. Two concentric circles have radii of 14 cm and 24 cm. Find the probability to the nearest thousandth that apoint chosen at random from the circles is located outside the smaller circle and inside the larger one.∙0.066∙0.58317. A slide that is inches by inches is projected onto a screen that is 3 feet by 7 feet, filling the screen.What will be the ratio of the area of the slide to its image on the screen?∙ 1 : 112∙ 1 : 2304∙ 2 : 4205∙ 1 : 12,54418. Find the area of a regular octagon with perimeter 48 cm.∙188.1 cm2∙190.5 cm2∙347.6 cm2∙173.8 cm219. Dorothy ran 6 times around a circular track that has a diameter of 47 m. Approximately how far did she run?Use π = 3.14 and round your answer to the nearest meter.∙885 m∙1328 m∙443 m∙1734 m20. Find the area.∙10.26 cm21. Find the volume of the cylinder in terms of π.∙24π in.3∙48π in.3∙56π in.3∙288π in.32. Find the volume of the cylinder in terms of π.∙287π in.3∙275π in.2∙275π in.3∙287π in.23. A sphere has a volume of 288π ft3. Find the surface area of the sphere.∙864π ft2∙48π ft2∙144π ft2∙96π ft24. The volumes of two similar solids are 2197 m3 and 64 m3. The surface area of the larger one is 845 m2. What is∙64 m2∙320 m2∙80 m2∙none of these5. Use a net to find the surface area of the prism.∙114 m2∙240 m2∙290 m2∙145 m26. Find the surface area of a sphere that has a diameter of 4 cm.∙64π cm2∙π cm3∙4π cm2∙16π cm27. Find the lateral area and the surface area of the cone. Use 3.14 for πand round the answer to the nearest hundredth.The diagram is not to scale.∙lateral area: 733.33 ft2; surface area: 690.80 ft2∙lateral area: 690.80 ft2; surface area: 1004.80 ft2∙none of these8. If the ratio of the radii of two spheres is 7 : 2, what is the ratio of the surface areas of the two spheres? ∙7 : 2∙7r2π : 2r2π∙49 : 4∙343 : 89. Use a net to find the surface area of the prism.∙465 m2∙720 m2∙918 m2∙930 m210. Find the height of the cylinder to the nearest tenth of an inch.∙94.6 in.∙96.6 in.∙ 4.3 in.∙ 4.1 in.11. Use formulas to find the lateral area and the surface area of the prism. Show your answer to the nearest hundredth.∙63.00 m2; 567.00 m2∙36.00 m2; 1134.00 m2∙479.22 m2; 533.22 m2∙542.22 m2; 596.22 m212. Find the volume of the prism.∙942 m3∙38 m3∙945 m3∙1890 m313. The volumes of two similar solids are 1331 m3 and 343 m3. The surface area of the larger one is 484 m2. Whatis the surface area of the smaller one?∙343 m2∙1372 m2∙196 m2∙none of these14. Find the surface area of the sphere.∙648π m2∙72π m2∙324π m2∙1296π m215. Find the volume of the prism.∙40.5 m3∙162 m3∙9 m3∙81 m316. Find the surface area of the solid. Round to the nearest square foot.∙36 ft2∙68 ft217. Use formulas to find the surface area of the prism. Show your answer to the nearest hundredth.∙75.42 cm2∙170.16 cm2∙86.94 cm2∙69.52 cm218. Which figure is a net for a cube?∙∙∙∙19. Cylinder A has radius 1 and height 4 and cylinder B has radius 2 and height 4. Find the ratio of the volumesof the two cylinders.∙ 1 : 4∙ 1 : 120. Neil had a job helping a jeweler. He had the assignment of counting the faces, vertices, and edges on the emeralds.On the first emerald, Neil counted 9 faces and 16 edges. He quickly realized he didn't have to count the vertices.How many vertices were there?∙10 vertices∙7 vertices∙8 vertices∙9 vertices1. Find the value of x if AB = 20, BC = 12, and CD = 13. (not drawn to scale)∙18.8∙16.5∙13.4∙14.92. Find the measure of each variable if m∠A = 22 and m = 97. (not drawn to scale)∙53; 210∙53; 105∙75; 210∙75; 1053. In the plane of lines X and Y, what is the locus of points equidistant from lines X and Y?∙line A∙line D∙line B∙line C4. A small messenger company can only deliver within a certain distance from the company. On the graph below, thecircular region represents that part of the city where the company delivers, and the center of the circle represents the location of the company. Which equation represents the boundary for the region where the company delivers?∙(x + 3)2 + (y– 1)2 = 49∙(x + 3)2 + (y– 3)2 = 98∙(x + 1)2 + (y– 3)2 = 98∙(x + 3)2 + (y– 3)2 = 495. A low-watt radio station can be heard only within a certain distance from the station. On the graph below, thecircular region represents that part of the city where the station can be heard, and the center of the circle represents the location of the station. Which equation represents the boundary for the region where the station can be heard?∙(x– 4)2 + (y– 5)2 = 50∙(x + 5)2 + (y– 5)2 = 59∙(x + 5)2 + (y + 4)2 = 25∙(x + 5)2 + (y– 5)2 = 256. Find the center and radius of (x– 8)2 + (y + 7)2 = 64.∙(–8, 7); 8∙(–7, 8); 64∙(8, –7); 8∙(–7, –8); 87. is tangent to circle O at B. How close to the circle is point A? (The diagram is not to scale.)∙ 3∙ 4.5∙ 6∙7.58. Compare the quantity in Column A with the quantity in Column B.∙The quantity in Column A is greater.∙The quantity in Column B is greater.∙The two quantities are equal.∙The relationship cannot be determined from the information given.9. Solve for x.∙22∙710. Find the value of x.∙14.6∙8.1∙9.4∙13.411. Write the standard equation for the circle with center (14, –48) that passes through (0, 0). ∙(x– 14)2 + (y + 48)2 = 2500∙(x + 14)2 + (y– 48)2 = 2500∙(x + 14)2 + (y– 48)2 = 50∙(x– 14)2 + (y + 48)2 = 5012. Find the measure of ∠BAC.∙30°∙150°13. Find the value of x.∙79∙39∙99∙15914. In space, which description fits the locus of points 3 cm from ?∙an open cylinder of diameter 6 cm∙an open cylinder of radius 3 cm and two hemispheres of diameter 6 cm each∙an open cylinder of radius 3 cm and height 6 cm∙an open cylinder of diameter 6 cm and two spheres of radius 3 cm each15. and are tangent to circle O and bisects ∠BPA. If m∠AOC= 68°, how much greater is m∠BCOthan m∠OAD? (The diagram is not to scale.)∙22°∙112°16. Write the standard equation for the circle with center (6, –8) that passes through (0, 0).∙(x + 6)2 + (y– 8)2 = 0∙(x– 6)2 + (y + 8)2 = 0∙(x + 6)2 + (y– 8)2 = 100∙(x– 6)2 + (y + 8)2 = 10017. Find the center and radius of (x + 3)2 + (y + 8)2 = 169.∙(3, 8); 13∙(–8, 3); 13∙(–3, –8); 13∙(–8, –3); 16918. , , and are all tangent to circle O. If JA= 8, AL= 13, and CK= 11, what is the perimeter of ΔJKL?(The diagram is not to scale.)∙32∙64∙45∙5319. If m = 38, what is m∠YAC?∙109°∙52°20. Write the standard equation for the circle with center (–16, 30) that passes through (0, 0). ∙(x + 16)2 + (y– 30)2 = 34∙(x + 16)2 + (y– 30)2 = 1156∙(x– 16)2 + (y + 30)2 = 1156∙(x– 16)2 + (y + 30)2 = 34。
加拿大数学10年级练习第一部分几何
1. ABCD is an isosceles trapezoid, and ≅. Which statement is NOT true?•ΔABY≅ΔBAX•≅•≅•∠DXB≅∠DYB2. It appears from the name of the HL Theorem that you actually need to know only two parts of a triangle in orderto prove two triangles congruent. Is this the case?•Yes, you only need to know the hypotenuse and a leg of a triangle.•No, you actually need to know two sides and an angle, because the triangle must be a right triangle.•No, you actually need to know three sides of the triangle.•No, you actually need to know two angles and a side.3. Complete the proof.Given: ≅, ∠1 ≅∠2, and ≅.Prove: ΔCEF≅ΔAEF.∠BEC≅∠DEA by vertical angles. ΔBEC≅ΔDEA by (a)_____. Then by CPCTC, ≅. ≅by the Reflexive Property. So ΔCEF≅ΔAEF by (b)_____.• a. SAS; b. SAS• a. AAS; b. SSS• a. ASA; b. SSS• a. AAS; b. HL4. Complete the proof.Given: ≅, ∠1 ≅∠2Prove: ΔBEA≅ΔDEC≅and ∠1 ≅∠2, so ΔBCA≅ΔDAC by SAS. Then, since (a)_____, ≅, ≅. ∠BEA≅∠DEC by (b)_____, so ΔBEA≅ΔDEC by (c)_____.• a. CPCTC; b. vertical angles; c. SAS• a. CPCTC; b. vertical angles; c. AAS• a. CPCTC; b. vertical angles; c. SSS• a. SAS; b. vertical angles; c. SSS5. What additional information can be used to prove the triangles congruent by the HL Theorem?•m∠BCE = 90•AB > AC•≅•≅6. Suppose ΔCED≅ΔDBC. If m∠EDC = 63 and m∠DBC = 82, what is m∠DCE?•63•82•145•357. If ∠A≅∠D and ∠C≅∠F, which statement would NOT prove that ΔABC≅ΔDEF?••∠B≅∠E•≅•none of these8. Determine what information you would need to know in order to use the SSS Congruence Postulate to show that thetriangles are congruent.•∠BAD≅∠CDB•≅•∠ADB≅∠CBD•≅9. Suppose ΔBCA≅ΔECD. Which statement is NOT necessarily true?•≅•∠A≅∠D•≅•∠BCA≅∠DCE10. In which triangles could you efficiently prove Δ1 ≅Δ2 using the HL Theorem?•II only•III only•II and III•I only11. Complete the proof.Given: ≅, ∠1 ≅∠2, and ≅.Prove: ΔCEF≅ΔAEF.∠BEC≅∠DEA with vetical angles. ΔBEC≅ΔDEA by (a)_____. Then by (b)_____, ≅. ≅by the Reflexive Property. So ΔCEF≅ΔAEF by (c)_____.• a. AAS; b. CPCTC; c. SSS• a. SAS; b. CPCTC; c. SSS• a. AAS; b. CPCTC; c. SAS• a. SSS; b. CPCTC; c. ASA12. In the paper airplane, ABCD≅EFGH, m∠B = m∠BCD = 90, and m∠BAD = 140. Find m∠GHE.•130•90•40•14013. Find the value of x.•x = –2•x = 9•x = 21•none of these14. Explain how you can use SSS, SAS, ASA, or AAS with CPCTC to prove that ∠D≅∠B.•≅and ∠ACB≅∠ACD. By the Symmetric Property, ≅. By SAS, ΔABC≅ΔADC, so by CPCTC ∠D≅∠B.•≅and ∠ACB≅∠ACD. By the Reflexive Property, ≅. By ASA, ΔABC≅ΔADC, so by CPCTC ∠D≅∠B.•≅and ∠ACB≅∠ACD. By the Reflexive Property, ≅. By SAS, ΔABC≅ΔADC, so by CPCTC ∠D≅∠B.•≅and ∠ACB≅∠ACD. By the Reflexive Property, ≅. By SSS, ΔABC≅ΔADC, so by CPCTC ∠D≅∠B.15. Complete the proof.Given: bisects ∠URS and bisects ∠UTS.Prove: ΔURT≅ΔSRT.•Reflexive property•definition of angle bisector•HL Theorem•CPCTC16. Complete the proof.Given: ∠RSQ≅∠TSQ, ∠RQS≅∠TQS.Prove: ≅.∠RSQ≅∠TSQ is given, as is ∠RQS≅∠TQS. By the Reflexive Property, ≅.ΔSRQ≅ΔSTQ by (a)_____, so ≅by (b)_____.• a. ASA; b. CPCTC• a. HL; b. CPCTC• a. SSS; b. CPCTC• a. SAS; b. CPCTC17. Determine which triangles are congruent by AAS using the information in the diagram below.•ΔABF≅ΔEDF•ΔADC≅ΔEBC•ΔABE≅ΔEDA•ΔABE≅ΔCBE18. Complete the proof.Given: bisects ∠EBC and bisects ∠ECC.Prove:ΔEBD≅ΔCBD.•Same-Side Interior Angles Theorem•given•SSS postulate•Triangle Inequality Theorem19. ΔABD≅ΔCBD. Name the theorem or postulate that justifies the congruence.•SAS•AAS•ASA•none of these1. Determine whether each quadrilateral can be a parallelogram. If not, write impossible.a. Two adjacent angles are right angles, but the quadrilateral is not a rectangle.b. All of the angles are congruent.• a. impossible; b. parallelogram• a. parallelogram; b. parallelogram• a. parallelogram; b. impossible• a. impossible; b. impossible2. Which statement is true?•All rectangles are squares.•All quadrilaterals are squares.•All quadrilaterals are parallelograms.•All parallelograms are quadrilaterals.3. Which statement can be used to determine whether quadrilateral XYZW must be a parallelogram?•≅and ≅•≅and ≅•≅and ≅•XW = WZ and XY = YZ4. Choose the best name for the parallelogram and find the measures of the numbered angles.•Square; all numbered angles are equal to 45°.•Rhombus; all numbered angles are equal to 115°.•Rhombus; all numbered angles are equal to 25°.•Square; all numbered angles are equal to 50°.5. Given: quadrilateral ABCD with A(–2, 3), B(2, –4), C(9, 0), D(5, 7). Then ABCD is a rectangle because•the slopes of the sides in pairs are negative reciprocals.•the product of the slopes of the diagonals is –1.•the figure has four vertices.•opposite sides have the same slope.6. Given the parallelogram below, find coordinates for P, without using any new variables.•(a–c, b)•(a + c, b)•(c, b)•(c, a)7. Find the values of the variables for the rectangle. Then find the lengths of the sides.•x = 7, y = 5; side lengths: 70, 45•x = 5, y = 7; side lengths: 33, 94•x = 5, y = 7; side lengths: 50, 63•x = 7, y = 5; side lengths: 45, 458. Determine whether the quadrilateral is a parallelogram. Explain.≅and ≅•Yes; if two opposite sides are congruent, then the quadrilateral is a parallelogram.•No; if the diagonals of a quadrilateral bisect each other, this is not enough to prove that the quadrilateral is a parallelogram.•No; if two opposite sides are congruent, this is not enough to prove that the quadrilateral is a parallelogram.•Yes; if the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram.9. Can coordinate geometry be used to prove that opposite sides and in quadrilateral EFGH are congruent?•Yes; use the Distance Formula to show that the diagonals are congruent.•No; you can only show that EF is parallel to GH by using coordinate geometry.•Yes; use the Distance Formula between vertices E and F, and between vertices G and H.•No; you can only find the slopes of EF and GH by using coordinate geometry.10. A square WXYZ has the vertices W(b, b), X(b, –b), Y(–b, –b), and Z(–b, b). Which vertex is in QuadrantII?•W•Z•X•Y11. ∠J and ∠M are base angles of isosceles trapezoid JKLM. If m∠J = 21x + 4, m∠K = 12x– 8, and ∠M = 14x+ 10, find the value of x.•• 2•–•–912. Suppose you are using coordinate geometry to prove that quadrilateral WXYZ is a square. Explain why no twosides should be parallel to the y-axis.•The x-axis would intersect two sides of the square, so the coordinates of the corners would not be clear.•Sides which are parallel to the y-axis would have an undefined slope, so you cannot prove numerically that these sides are parallel.•Points on the sides which are parallel to the y-axis could have any y-value.•The sides of the square which are parallel to the y-axis could be easily confused with the y-axis.13. Find AM if PN = 8 and AO = 5.•8• 3•13• 514. Given square ABCD, where A = (0, a), B = (a, a), C = (0, 0), and D = (a, 0). To prove that the diagonal ADis times the length of side CD, first use _____ to find that = a and = a. Therefore, the ratio= , or .•the definition of isosceles triangle ACD•the definition of right angle C•the Distance Formula•the definition of the origin C = (0, 0)15. A farmer is building a fence for his yard. He is considering two designs, which are shown below. Explain whythe quadrilaterals formed by the horizontal rails and the slanting boards are parallelograms in both designs.•The horizontal rails are parallel to each other. A parallelogram has exactly one pair of parallel sides, so both quadrilaterals are parallelograms.•The horizontal rails are congruent to each other. The slanting boards are all congruent to each other. A parallelogram has two pairs of adjacent sides, but opposite sides are not congruent, so both quadrilaterals are parallelograms.•The horizontal rails are congruent to each other. The slanting boards are all congruent to each other. A parallelogram has four congruent sides, so both quadrilaterals are parallelograms.•The horizontal rails are parallel to each other. The identical slanting boards all slant at the same angle, so the sides are parallel. A parallelogram has both pairs of sides parallel, so both quadrilaterals are parallelograms.16. A rhombus is centered on the origin. One side of the rhombus goes through the points (a, 0) and (0, b). Whatare possible coordinates for one of the other sides?•(–a, 0), (a, 0)•(b, 0), (0, –a)•(0, –b), (0, b)•(–a, 0), (0, –b)17. If a quadrilateral is a parallelogram, then its opposite sides are _____.•perpendicular•adjacent•congruent•none of these18. Find the value of each variable in the parallelogram. m∠1 = 10x, m∠2 = x + y, and m∠3 = 18z.•x = 9, y = 81, z = 5•x = 18, y = 167, z = 5•x = 18, y = 162, z = 10•x = 9, y = 86, z = 019. Complete ≅ _____ for parallelogram EFGH. Then state a definition or theorem as the reason.•; because the angles of a parallelogram bisect each other•; because the diagonals of a parallelogram bisect each other•; because the diagonals of a parallelogram bisect each other•; because the angles of a parallelogram bisect each other20. Which of the following sets of points represents a line segment in Quadrant III and its reflection in the x-axis?(Use the positive numbers a, b, c, d for the coordinates of the endpoints).•(a, b), (c, d); reflection (–a, b), (–c, d)•(–a, –b), (–c, –d); reflection (–a, b), (–c, d)•(a, b), (c, d); reflection (a, –b), (c, –d)•(–a, –b), (–c, –d); reflection (a, –b), (c, –d)1. Solve for a and b.•a = , b =•a = , b =•a = , b =•a = , b =2. State whether ΔADB∼ΔCDB, and if so, identify the theorem that proves the triangles similar.•yes, SSS∼•yes, AA∼•yes, SAS∼•no3. ABCDE∼GHJDF. Complete the congruence and proportion statements.a.∠H≅b.=• a.B; b.AE• a.E; b.DC• a.E; b.AE• a.B; b.DC4. Write a similarity statement for the two triangles.•ΔVUT∼ΔWXY•ΔTVU∼ΔWXY•ΔTUV∼ΔWXY•ΔTUV∼ΔWYX5. Find the geometric mean of 48 and 3.•9•25.5•12•166. The extendable ramp shown below is used to move crates of fruit to loading docks of different heights. When thehorizontal distance AB is 4 feet, the height of the loading dock, BC, is 3 feet. What is the height of the loading dock, DE?•7 ft•9 ft•11 ft7. Find the geometric mean of 20 and 5.•10• 4•12.5•258. In movies and television, the ratio of the width of the screen to the height is called the aspect ratio. Televisionscreens usually have an aspect ratio of 4 : 3, while movie screens usually have an aspect ratio of 1.85 : 1. However, if a movie is made for television in "Letterbox" format, it retains the 1.85 : 1 aspect ratio and fills in the top and bottom parts of the screen with black bars. What would be the height of a movie in "Letterbox" format on a television screen that measures 25 inches along its diagonal? (Hint: First find the width and height of the television screen.)•13.51 in.•10.81 in.•15 in.•8.12 in.9. Use the diagram to determine the height of the tree.•264 ft•72 ft•60 ft•80 ft10. The two rectangles are similar.Which is a correct proportion between corresponding sides?•=•=•=11. Use the Side-Splitter Theorem to find x given that || .•18•12•24• 612. Find OM if bisects ∠NLM, LM = 14, NO = 3, and LN = 4. Round your answer to the nearest hundredth, ifnecessary.•12.27•18.67•0.86•10.513. There is a law that the ratio of the width to length for the American flag should be 10 : 19. Which dimensionsare NOT in the correct ratio?•20 by 38 in.•50 by 95 ft•20 by 44 ft•100 by 190 ft14. If one measurement of a golden rectangle is 6.8 inches, which could be the other measurement?•8.418 in.•11.002 in.• 1.618 in.• 5.182 in.15. If one measurement of a golden rectangle is 8.2 inches, which could be the other measurement?•9.818 in.• 6.582 in.• 1.618 in.• 5.068 in.16. Solve = .•20•19•15•2417. Find OM if bisects ∠NLM, LM =15, NO = 5, and LN = 11. Round your answer to the nearest hundredth, ifnecessary.•33• 6.82• 3.67•8.5918. The width of a golden rectangle is 3 m, which is shorter than the length. What is the length?• 1.85 m• 2.32 m• 3.64 m• 4.85 m19. Find and simplify the ratio of the length to the width of the rectangle.••••20. ΔBGH∼ΔSWQ. What are the pairs of corresponding sides?•BG and SQ, BH and SW, GH and WQ•BG and GB, SQ and QS, GH and HG•BG and SW, BH and SQ, GH and WQ•BG and WQ, BH and SW, GH and SQ1. A building near Atlanta, Georgia, is 181 feet tall. On a particular day at noon it casts a 204-foot shadow. Whatis the sun's angle of elevation at that time?•41.6°•27.5°•62.5°•48.4°2. In right triangle ΔABC, sin A = . What is cos A?••••none of these3. Find the ratio for cos x.••••14. Find the value of x to the nearest meter.•46 m•40 m•35 m•36 m5. Compare the quantity in Column A with the quantity in Column B. The diagram may not be drawn to scale.•The quantity in Column A is greater.•The quantity in Column B is greater.•The two quantities are equal.•The relationship cannot be determined on the basis of the information given.6. How many of these triples could be sides of a right triangle: (27, 36, 45), (12, 17, 20), (24, 32, 40), (14,48, 50)?• 4 triples• 3 triples• 2 triples• 1 triple7. Find the ratio for cos x.•••2•8. Find a third number of the Pythagorean triple that includes 72 and 75.•9•21•37•1049. Find the measure of the marked acute angle to the nearest degree.•62°•28°•61°•118°10. In ΔABC, ∠A is a right angle and m∠B = 60. If AB = 20 ft, find BC. If necessary, round your answer tothe nearest tenth.•10 ft.•40 ft•20 ft•ft11. Find the value of the variable to the nearest hundredth.• 5.28 cm•0.32 cm• 3.13 cm• 5.12 cm12. Find the length of the leg of the right triangle. Leave your answer in simplest radical form.•48••288•13. In ΔABC, ∠A is a right angle and m∠B = 45. If AB = 20 ft, find BC.•10 ft•20 ft•40 ft•20 ft14. Which direction bearing is shown?•19° north of east•19° north of west•19° south of east•19° south of west15. Leslie used the diagram to compute the distance from Ferris to Dunlap to Butte. How much shorter is the distancedirectly from Ferris to Butte than the distance Leslie found?•123 mi•87 mi•36 mi•84 mi16. Which vector has a direction of 31° east of north?••••17. Find the value of x to the nearest tenth.•14.4• 6.3•7.8• 3.118. Find the value of x.•3•6•12• 619. Find the value of x to the nearest tenth.•7.8•18.3•33.0•8.920. Find the value of x to the nearest integer when tan x = 1.483.•58•56•55•571. Which type of isometry is the equivalent of two reflections across two vertical lines?•translation•rotation•glide reflection•none of these2. A section of a tessellated plane is shown below. Which types of symmetry does the tessellated plane have?•glide-reflectional symmetry•translational, rotational, and glide-reflectional symmetry•rotational symmetry•translational and reflectional symmetry3. A blueprint for a house has a scale of 1 : 30. A wall in the blueprint is 7 in. What is the length of the actualwall?•210 ft•21 ft•17.5 ft•none of these4. What is the image of the point (4, –2) after a rotation 270° clockwise about the origin?•(–4, 2)•(4, 2)•(–2, 4)•(2, 4)5. Which graph shows a triangle and its reflection image in the x-axis?••••6. Write a rule to describe a reflection over the y-axis.•(x, y) → (–x, y)•(x, y) → (–x, –y)•(x, y) → (x, –y)•(x, y) → (y, x)•reflectional•rotational•rotational and reflectional•none of these8. Find the image of O(0, 0) after two reflections, first in y = 4, and then in x = –7.•(7, –4)•(–14, 8)•(8, –14)•(–4, 7)9. A section of a tessellated plane is shown below. Which types of symmetry does the tessellated plane have?•rotational symmetry•translational and rotational symmetry•reflectional symmetry•glide-reflectional symmetry•rotational•reflectional•rotational and reflectional•none of these11. Describe the translation 7 units to the left, 12 units up using a vector.•)–12, 7*•)12, –7*•)–7, 12*•)7, –12*12. If the figure has rotational symmetry, find the angle of rotation about the center that results in an imagethat matches the original figure.•120°•90°•72°•It has no rotational symmetry.13. Find the glide reflection image of the solid triangle for the translation )–6, –3* and reflection in y = –1.••••14. If a point P(1, –2) is reflected across the line x = 3, what are the coordinates of its reflection image?•(1, –4)•(1, 8)•(–7, –2)•(5, –2)15. Which translation from thin-lined figure to thick-lined figure is given by the vector )6, 6*?••••16. The dotted triangle is a dilation image of the solid triangle. What is the scale factor?•• 2• 3•17. Find the image of C under the translation described by each vector.a.)4, 5*b.)11, –8*• a.A; b.B• a.B; b.A• a.E; b.D• a.D; b.E18. What is the image of the point (–3, 4) after a rotation of 90° counterclockwise about the point (–3, 0)?•(–7, 0)•(1, 0)•(0, –3)•(–3, –4)19. Which letter has rotational symmetry?• E•X•J•T20. Use scalar multiplication to find the image of the quadrilateral for a dilation with center (0, 0) and scalefactor 2. Graph the quadrilateral and its image.••••。
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GRADE 10 PRINCIPLES OF MATHEMATICS (ACADEMIC)MPM 2DTotal Marks:INSTRUCTIONS:1. Calculators may be used.2. Read all instructions carefully in order to maximize your mark.A/C [K] Part A – Multiple Choice 25 Marks (25 questions * 1 mark each)For each of the following questions in this section, circle the letter representing the correct answer.1. A linear system of two equations that has one solution represents two lines that are:a) parallel b) coincident c) intersecting d) none of these2. The midpoint of RS is M(8, -1). If point S has coordinates (11, 4) what are the coordinates of point Ra) (3, -6) b) (15, -6) c) (5, -6) d) (3, 9)3. The midpoint of the line segment with end points A(-8, 8) and B(6, 4) is:a) (0, 10) b) (1, 2) c) (7, 2) d) (-1, 6)4. The equation of a horizontal line passing through the point (4, 2) is:a) 2=x b) 4=y c) 2=y d) 4=x5. The equation of a line with a slope of 5=m and a y intercept of 8 is:a) 85+=x y b) 85+-=x y c) 85--=x y d) 58+=x y6. The slopes of 2 lines are -7 and 71. These lines are said to be:a) parallel b) perpendicular c) coincident d) none of these7.The slope of a line segment passing through 2 points (10,- 4) and (-2, -16) is:a) 1 b) 2 c) -1 d) -28. The length of a line segment with end points (-6, 7) and (-1, -5) is:a) 12 b) 5 c) 13 d) 1699. The diameter of a circle whose equation is 28922=+y x is:a) 15 b) 16 c)17 d) none of these10. The equation of a circle with a centre of (0, 0) that also passes through the point (-8, -6) is:a) 1022=+y x b) 10022=+y x c) 1422=+y x d) 4822=-y x11. The y-intercept of the line 01052=+-y x is:a) 2 b) -2 c) 10 d) 512. The slope of the line 0124=-+y x is:a) 2 b) -2 c) 1 d) 013. If (-3, y) is a solution to the equation 132=+y x , what is the value of ya) 3 b) 6 c) 5 d) 814. The product ()()z y x z y x 323243-- is equal to:a) 2612z xy b) 26412z y x c) 2612z xy - d) 00412z y x15. A simplified expression for ()()n m n m ----52 is:a) m 7 b) n m 27+ c) m 3- d) n m 27-16. A simplified expression for 242927abc bc a -- is:a) ac 3 b) abc 3 c) 23ac d) 223c a17. The slope of the line, which is perpendicula r to the line, 084=+-y x is:a) -4 b) 4 c) 1 d) -118. The shortest distance from the point (2, -3) to the line 4-=x is:a) 5 b) 3 c) 2 d) 619. The value of the polynomial 8542+-a a when 3-=a is:a) 59 b) 44 c) 13 d) 2920. Which of the following is not a function :a) ()()(){}7,6,5,4,3,2 b) 22x y = c) 22y x = d) ()()(){}3,8,3,7,2,621. The range of the relation whose equation is 52--=x y is:a) 5-≤y b) 5≤y c) 5-≥y d) 5≥y22. The vertex of the parabola ()642--=x y is:a) ()6,4- b) ()6,4- c) ()4,6- d) ()4,6-23. The equation of the axis of symmetry of the parabola ()5242+--=x y is:a) 5=x b) 5-=x c) 2=x d) 2-=x24. A parabola with a vertex of ()3,2 and a stretch factor of 41-(relative to 2x y =) would have an equation of:a) ()32412+--=x y b) ()32412++-=x y c)()23412-+-=x y d) ()23412++-=x y25 The parabola k x y +-=24 passes through the point ()3,2-. T he value ofk is:a) -19 b) 11 c) 13 d) 19A/CPart B – Short AnswersFor each of the questions in this section, write your answers in the spaces provided . Use the foolscap provided for any rough work. Show details of calculations wherever requested.1. In the accompanying diagram, state each of the following: (4 Marks)[K]a) domain: __________ (1 Mark) [K]b) range: __________ (1 Mark)[C]c) Is the relation a functionJustify your answer. (2 marks)[A]2. The x-intercepts of the parabola 2892-=x y are: __________ and __________.(Show your work) (2 Marks)[A] 3. The roots of the quadratic equation 0+x are: __________ and-x101732=__________.(Show your work) (3 Marks)[A] 4. Write the equation of the parabola with a vertex of (4, 23) if itpasses through the point (-1, -2): (Show your work) (3 Marks)____________________[T] 5. A line passes through 2 points (1, 4) and (2,-4). Calculate the slope of the line. Also show the equation of the line in the form 0Ax.By+=+C (Show your work) (4 Marks)____________________ ____________________Slope Equation45 is: __________ (1 Mark)[K] 6. The Tangent of ο[A] 7. a) In the accompanying diagram, the two triangles are similar. What isthe value of x(Show your work) (2 Marks)=x __________[T] b) If the area of the smaller triangle is 8 cm 2, what is the area ofthe larger triangle(Show your work) ( 2 Marks)Area = __________[K]8 Given that sin A = 21, find A ∠ (to the nearest degree) __________(1 Mark)[A] 9. In the accompanying right triangle , find the value of x to one decimal place.(Show your work) (2 Marks)=x ________[A] 10. Use the SINE LAW to find the value of side x to one decimal place. (Show your work) (2 Marks)x = ________3028x5642x30[A] 11. Use the COSINE LAW to find the value of side x to one decimal place.(Show your work) (2 Marks)x = ________[T] 12. Factor each of the following to the fullest extent possible: (4 Questions * 2 marks each) a) y x my mx 22--+________________________b) 31142--x x________________________c) 2416916y x -________________________d) 2225309s rs r +-________________________A/CPart C – Full Solutions RequiredFor each of the questions in this section, full solutions are required. Record your answers in the spaces provided. Use the foolscap provided for any rough work.[A] 1. Solve the linear system using the elimination method . Remember to find values for both x and y. (5 Marks)562030 x22x+y5=x=-y3212-[C] Explain what the solution above represents geometrically. How do youknow that the solution you arrived at is the correct answer (2 Marks) [A] 2. Expand and simplify the polynomial ()()()21+-xxx. (4 Marks)-2+43[T] 3. Find the equation of the line perpendicular to the line 0-x+y8=8 and passing through the point (-4, 1). (4 Marks)[T] 4. From the window of one building, a man finds that the angle of elevation to the top of a second building is 47 and the angle ofdepression to the bottom of the same building is 58. The buildingsare 60 m apart. Find the height of the 2nd building to the nearestmetre. A diagram is required. (6 Marks)[T] 5. ABC has vertices A(1, 7), B(-5, 3) and C(3, -1). Determine the equation for AE, the altitude from vertex A to the opposite side BC.(5 Marks)6. The hypotenuse of a right triangle is 26 cm. The sum of the othertwo sides is 34 cm. (9 Marks)[T] a) Find the length of the other two sides of the triangle. (3 Marks)[T] b) Find the measure of the other two angles. Round to the nearest degree. (3 Marks)[C] c) Describe a situation where you would be able to use knowledge ofthe Pythagorean theorem in a practical, real life situation. (3Marks)[T] 7. A rectangular skating rink measures 20m by 20m. It has been decided to increase the area of the rink by a factor of 4. Determine howmuch each side should be extended. Assume that each side isextended by the same amount. (6 Marks)[C] What is the significance of keeping the skating rink in the shape of a square Justify your answer. (3 Marks)[A] 8. a) Solve 35122+=d d using the quadratic formula. (2 Marks)[A] b) Solve 03122=-x by factoring. Check your solutions. (2 Marks)。