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1. Solve the following system of equations: 5 marks MTH-4111 Pretest B This is a disjoint system and therefore there is no solution (no points of intersection between the line and the parabola).

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2. A ship is moored to a pier by a chain. The chain hangs in the shape of a parabola part of which is submerged. The chain sinks below the water surface 5 meters from the pier and re-emerges 25 meters from the pier. The chain is attached to the top of the pier which is 2 meters above water level. The bow of the ship is an oblique line. The top of the bow is 30 meters from the pier and 8 meters above the surface of the water. The bow enters the water 40 meters from the pier. At what point does the chain attach to the bow of the ship? x y 30 m 8 m QUADRATIC EQUATION: Zeros: (5,0), (25,0) Other point: (0,2) y = a(x – x 1 )(x – x 2 ) y = a(x – 5)(x – 25) y = a(x 2 – 30x + 125) 2 = a(0 2 – 30(0) + 125) 2 = 125a LINEAR EQUATION: Zeros: (30,8), (40,0) 5(y - 0) = -4(x – 40) 5y = -4x + 160

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The chain is attached to the boat 34.44m to the right of the pier and 4.45m above the water surface. x y 34.44 m 4.45 m

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D) A) 3. Two functions are described below. f(x) = mx + b where m < 0 g(x) = ax 2 + c where a > 0 and b = c. Which of the following graphs represents the function operation, f – g? f(x) = -1x + b g(x) = 1x 2 + b f – g = (-x + b) – (1x 2 + b) = -x + b - x 2 – b = - x 2 - x (A) D) C) B)A) 4.Functions g and h are represented graphically to the right. Identify the graph below that corresponds to g h. h g g(x) = -1x h(x) = 1x g h = (-1x) (1x) = - x 2 (D) B)A) D) C)

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5.The equations for the diagonals of a parallelogram are indicated in the diagram that is provided. A (-5, -7) and B (3, -5) are two of the vertices of the parallelogram. What is the perimeter of the parallelogram? A(-5, -7) B(3, -5) 5 marks y = 2x + 3 C D Point C is on the vertical line x = 3 so its x- value is 3. (3, ? ) Vertex C is also on the diagonal y = 2x + 3. y = 2(3) + 3 y = 6 + 3 = 9C (3,9) Perimeter of the parallelogram = 2(8.25) + 2(14) = 16.5 + 28 = 44.5 units

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6.Prove that the following quadrilateral is a rhombus. B (3, 2) A (-3, 4) C (5, -4) D (-1, -2) 10 marks

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7.The vertices of an irregular quadrilateral are as follows: A (-3, 3), B (2, 5), C (6, -2) and D (-4, -1). Calculate its area. C (6, -2) B (2, 5) D (-4, -1) A (-3, 3) Slope of diagonal BD: Equation of BD: Length of BD: Length of altitude from A:(-3,3) to side BD. Area of ΔABD

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Length of altitude from C:(6,-2) to side BD. Area of ΔCBD Area of Quadrilateral ABCD Equation of BD: C (6, -2) B (2, 5) D (-4, -1) A (-3, 3)

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1. Midpoint Formula: 8.Complete the demonstration of the following proposition using geometric analysis. The area of square ABCD is twice the area of square EFGH. HYPOTHESIS: CONCLUSION : y x A (0,0)B (a,0) C (a,a)D (0,a) E F G H STATEMENTSJUSTIFICATIONS 5 marks 1. The coordinates of E and F are: 2. Length of the sides of both squares:2. Distance Formula:

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STATEMENTSJUSTIFICATIONS y x A (0,0)B (a,0) C (a,a)D (0,a) E F G H 3. Area of both squares:3. Area of a Square Formula: 4. Relation between Areas of both squares:

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9.Complete the demonstration of following proposition: The following diagonals of the regular pentagon form an isosceles triangle with a vertex angle of 36º. HYPOTHESIS:CONCLUSION: STATEMENTSJUSTIFICATIONS 5 marks A E D C B 36º Congruent sides of a regular pentagon Congruent sides of a regular pentagon Congruent angles of a regular pentagon Theorem 16 (SAS Congruency Theorem) Theorem 39a Theorem 7 Congruent angles of a regular pentagon Theorem 45 Theorem 5

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10.ACDF and ABHG are similar trapezoids and the ratio of their areas is. Trapezoid GHEF is equivalent to parallelogram BCDE. Given that the height of trapezoid ACDF is 12 cm and its large base is 20 cm, what is the length of the base of parallelogram BCDE? 10 marks E AD C BF HG 12 cm 4 cm 8 cm 6.67 cm 12 cm 20 cm

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E A D C B F HG 12 cm 4 cm 8 cm 6.67 cm x20 - x 12 cm Let x = base of a parallelogram Long base of trapezoid GHEF = 20 - x E A D C B F HG Trapezoid GHEF is equivalent to parallelogram BCDE. The base of parallelogram BCDE is 6.67 cm. 6.67 cm

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11.ΔFDE is equivalent to rhombus IFHG. ΔFDE ~ ΔFHG Find the perimeter of ΔFDE. 10 marks DE H F G I 38° O Theorem 28Theorem 23Theorem 21 Theorem 28

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DE F H F G 10 cm H F G I 8.121 cm Congruent sides of a rhombus. Theorem 50a DE F H F G 8.121 cm 11.48 cm

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12.A mother is sharing a some Tropicana orange juice between her 2 sons. However she only has 2 glasses which are both cylindrical yet have different shapes. She tries to assure her sons that she will share the juice equally between them. She fills one glass having a diameter of 6 cm to a height of 10 cm. The other glass has a diameter of 8 cm. a) How high must she fill it to keep her promise? b) The juice comes from a container that is a rectangular prism with a height of 19 cm whose base is a square with 10 cm sides. If after serving the juice, the container is empty, How high was the juice inside the container? 10 marks 8 cm 10 cm 6 cm a) Volume short glass = Volume tall glass πr 2 h = πr 2 h π(3) 2 (10) = π(4) 2 h 90π = 16πh She must fill the glass to a height of 5.625 cm. b) Volume rect prism = l w h = (10)(10)h = 100h The volume in each glass is 90π. This makes a total of 180π poured from the rectangular prism. Volume rect prism = 100h 180π = 100h 565.49 = 100h h = 5.655 cm The juice was 5.655 cm inside the container.

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h 2 = a 2 + b 2 = 7.5 2 + 20 2 = 56.25 + 400 = 456.25 h = 21.36 cm (slant height) 13. A company called Party Poppers plans to produce cardboard party hats for the New Years’ Eve celebrations. They have 2 different proposals. A cone-shaped hat with a diameter of 15 cm and a height of 20 cm and a cylindrical-shaped hat with the same diameter and height. If they have 100 square meters of material to work with, how many hats can they make for each proposal? HINT: Remember the hats will be open at one end. 20 cm 15 cm 7.5 cm 20 cm Lateral Area cone = πrs = π(7.5)(21.36) = 160.2π = 503.3 cm 2 100 m 2 = 100 (10000 cm 2 ) = 1 000 000 cm 2 # of cone-shaped hats = 1 000 000 ÷ 503.3 = 1986 hats Lateral Area cylinder = 2πrh = 2π(7.5)(20) = 300π = 942.5 cm 2 Area circle = πr 2 = π(7.5) 2 = 56.25π = 176.7 cm 2 20 cm 15 cm Total area of cylindrical hats open at one end = 942.5 + 176.7 = 1119.2 cm 2 # of cylinder-shaped hats = 1 000 000 ÷ 1119.2 = 893 hats Closed end of hat

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