Presentation on theme: "MTH-4111 Pretest B 1. Solve the following system of equations: 5 marks"— Presentation transcript:
1 MTH-4111 Pretest B1. Solve the following system of equations:5 marksThis is a disjoint system and therefore there is no solution (no points of intersection between the line and the parabola).
2 QUADRATIC EQUATION: y Zeros: (5,0), (25,0) Other point: (0,2) 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?QUADRATIC EQUATION:Zeros: (5,0), (25,0)Other point: (0,2)y = a(x – x1)(x – x2)y = a(x – 5)(x – 25)y = a(x2 – 30x + 125)2 = a(02 – 30(0) + 125)2 = 125axy30 m8 mLINEAR EQUATION:Zeros: (30,8), (40,0)5(y - 0) = -4(x – 40)5y = -4x + 160
3 The chain is attached to the boat 34 The chain is attached to the boat 34.44m to the right of the pier and 4.45m above the water surface.xy34.44 m4.45 m
4 h g D) D) A) B) C) 3. Two functions are described below. f(x) = mx + b where m < 0g(x) = ax2 + c where a > 0 and b = c.Which of the following graphs represents the function operation, f – g?f(x) = -1x + bg(x) = 1x2 + bf – g = (-x + b) – (1x2 + b)= -x + b - x2 – b= - x2 - x (A)A)A)B)C)D)hgFunctions g and h are represented graphically to the right.Identify the graph below that corresponds to g • h.g(x) = -1xh(x) = 1xg h = (-1x) (1x)= - x2 (D)D)D)A)B)C)
5 Point C is on the vertical line x = 3 so its x-value is 3. (3, ? ) 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?y = 2x + 3CDPoint 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) + 3y = = 9 C (3,9)B(3, -5)A(-5, -7)Perimeter of the parallelogram = 2(8.25) + 2(14)== 44.5 units5 marks
6 6. Prove that the following quadrilateral is a rhombus. C (5, -4)D (-1, -2)10 marks
7 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.Length of BD:C (6, -2)B (2, 5)D (-4, -1)A (-3, 3)Length of altitude from A:(-3,3) to side BD.Slope of diagonal BD:Equation of BD:Area of ΔABD
8 Equation of BD: Length of altitude from C:(6,-2) to side BD. Area of ΔCBDArea of Quadrilateral ABCD
9 The area of square ABCD is twice the area of square EFGH. 8. Complete the demonstration of the following proposition using geometric analysis.5 marksThe area of square ABCD is twice the area of square EFGH.yxA (0,0)B (a,0)C (a,a)D (0,a)EFGHCONCLUSION:HYPOTHESIS:STATEMENTSJUSTIFICATIONS1. The coordinates of E and F are:1. Midpoint Formula:2. Length of the sides of both squares:2. Distance Formula:
10 STATEMENTS JUSTIFICATIONS yxA (0,0)B (a,0)C (a,a)D (0,a)EFGH3. Area of both squares:3. Area of a Square Formula:4. Relation between Areas of both squares:
11 9. Complete the demonstration of following proposition: The following diagonals of the regular pentagon form an isosceles triangle with a vertex angle of 36º.5 marksAEDCB36ºHYPOTHESIS:CONCLUSION:STATEMENTSJUSTIFICATIONSCongruent sides of a regular pentagonCongruent sides of a regular pentagonCongruent angles of a regular pentagonTheorem 16 (SAS Congruency Theorem)Theorem 39aTheorem 45Congruent angles of a regular pentagonTheorem 7Theorem 5
12 ACDF and ABHG are similar trapezoids and the ratio of their areas is 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 marksEADCBFHG12 cm4 cm12 cm6.67 cm8 cm20 cm
13 Let x = base of a parallelogram Long base of trapezoid GHEF = 20 - x CBFHG12 cm4 cm12 cm6.67 cm8 cm20 - xxTrapezoid GHEF is equivalent to parallelogram BCDE.EADCBFHGThe base of parallelogram BCDE is 6.67 cm.6.67 cm
14 D E H F G I ΔFDE is equivalent to rhombus IFHG. ΔFDE ~ ΔFHG 38°ΔFDE is equivalent to rhombus IFHG.ΔFDE ~ ΔFHGFind the perimeter of ΔFDE.OTheorem 28Theorem 23Theorem 28Theorem 2110 marks
15 H F G D E F H F G I H F G D E F Theorem 50a 10 cm Congruent sides of a rhombus.HFGITheorem 50a8.121 cm8.121 cm8.121 cmHFGDEF8.121 cm11.48 cm8.121 cm11.48 cm
16 πr2h = πr2h π(3)2(10) = π(4)2h = (10)(10)h 90π = 16πh = 100h 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 marks10 cm6 cma) Volumeshort glass = Volumetall glassπr2h = πr2hπ(3)2(10) = π(4)2h90π = 16πhb) Volumerect prism = lwh= (10)(10)h= 100h8 cmThe volume in each glass is 90π. This makes a total of 180π poured from the rectangular prism.Volumerect prism = 100h180π = 100h= 100hh = cmShe must fill the glass to a height of cm.The juice was cm inside the container.
17 Lateral Areacylinder = 2πrh = 2π(7.5)(20) = 300π = 942.5 cm2 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.h2 = a2 + b2===h = cm (slant height)20 cm7.5 cm20 cm15 cm20 cm15 cmLateral Areacylinder = 2πrh= 2π(7.5)(20)= 300π= cm2Areacircle = πr2= π(7.5)2= 56.25π= cm2Lateral Areacone = πrs= π(7.5)(21.36)= 160.2π= cm2Closed end of hatTotal area of cylindrical hats open at one end= = cm2100 m2 = 100 (10000 cm2)= cm2# of cylinder-shaped hats = ÷= 893 hats# of cone-shaped hats = ÷ 503.3= 1986 hats
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