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Quadratic Word Problems WS 2 Solutions. #1. 10 inches. 15 cm by 7 cm. #2. a. 156.25 ft b. 6.25 sec c. 0.55 & 5.70 sec. #3. #4. 72 units. #5. 96 cm. #6. Height is 5.25 m or 5 m and 25 cm Distance is 3.79 meters or 3 m and 79 cm. #1. 2x + 15. 2x + 20. 0.8188 ft x 12 in/ft.
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Quadratic Word Problems WS 2 Solutions #1 10 inches 15 cm by 7 cm #2 a. 156.25 ft b. 6.25 sec c. 0.55 & 5.70 sec #3 #4 72 units #5 96 cm #6 Height is 5.25 m or 5 m and 25 cm Distance is 3.79 meters or 3 m and 79 cm
#1 2x + 15 2x + 20 0.8188 ft x 12 in/ft About 10 inches
#2 Perimeter = 44 cm Area = 105 sq. cm Width = W 2L + 2W = 44 L + W = 22 Length = L L = 22 - W The dimensions of the rectangle are 15 cm by 7 cm (L)(W) = 105 (22 - W)(W) = 105 22W – W2 = 105 W2 – 22W + 105 = 0 (W - 15)(W - 7) = 105
#3a h = -16t2 + 100t First find Axis of Symmetry Plug it in to get the vertex Solving for h will give the maximum point (or highest point)
Two ways to solve … #3b h = -16t2 + 100t Why does this work? Alternate solution 0 = -16t2 + 100t 0 = -4t(4t – 25) t = 0 or 6.25
#3c h = -16t2 + 100t 50 = -16t2 + 100t 16t2 - 100t + 50 = 0 8t2 - 50t + 25 = 0
#4 x x x + 6 x + 9
Width = w w - 4 3w - 4 Length = 3w Volume = L x W x H 512 = (3w - 4)(w - 4)(2) 256 = (3w - 4)(w – 4) Perimeter = 96 cm 256 = 3w2 – 16w + 16 0 = 3w2 – 16w - 240 0 = (w – 12)(3w + 20) Width = 12 cm
#6 Mr. Lomas (at least the way I remember me) …. Maximum height (at vertex) height when I hit the water? 3 meters Distance from board when entering water?
#7 Equation: y = x2 - 10x + 15 Direct: Up Width: Standard L of S: x = 5 Vertex: (5, -10) y-int: (0, 15) Roots:
#8 Equation: y = -2x2 + 8 Direct: Down Width: Narrow L of S: x = 0 Vertex: (0, 8) y-int: (0, 8) Roots: 2 or -2
#9 Equation: y = (1/4)x2 - 2x - 5 Direct: Up Width: Wide L of S: x = 4 Vertex: (4, -9) y-int: (0, -5) Roots: 10 or -2