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Quadratic Word Problems WS 2 Solutions #110 inches #2 #3 #4 #5 #6 15 cm by 7 cm a. 156.25 ftb. 6.25 secc. 0.55 & 5.70 sec 72 units 96 cm Height is 5.25.

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Presentation on theme: "Quadratic Word Problems WS 2 Solutions #110 inches #2 #3 #4 #5 #6 15 cm by 7 cm a. 156.25 ftb. 6.25 secc. 0.55 & 5.70 sec 72 units 96 cm Height is 5.25."— Presentation transcript:

1 Quadratic Word Problems WS 2 Solutions #110 inches #2 #3 #4 #5 #6 15 cm by 7 cm a. 156.25 ftb. 6.25 secc. 0.55 & 5.70 sec 72 units 96 cm Height is 5.25 m or 5 m and 25 cm Distance is 3.79 meters or 3 m and 79 cm

2 About 10 inches #1 2x + 15 2x + 20 0.8188 ft x 12 in/ft

3 #2 The dimensions of the rectangle are 15 cm by 7 cm Perimeter = 44 cm Area = 105 sq. cm Width = W Length = L 2L + 2W = 44 L + W = 22 L = 22 - W (L)(W) = 105 (22 - W)(W) = 105 22W – W 2 = 105 W 2 – 22W + 105 = 0 (W - 15)(W - 7) = 105

4 #3ah = -16t 2 + 100t First find Axis of Symmetry Plug it in to get the vertex Solving for h will give the maximum point (or highest point)

5 #3b Why does this work? h = -16t 2 + 100t Alternate solution 0 = -16t 2 + 100t 0 = -4t(4t – 25) t = 0 or 6.25 Two ways to solve …

6 #3ch = -16t 2 + 100t 50 = -16t 2 + 100t 16t 2 - 100t + 50 = 0 8t 2 - 50t + 25 = 0

7 #4 x x + 6 x x + 9

8 Volume = L x W x H Width = w Length = 3w 512 = (3w - 4)(w - 4)(2) w - 43w - 4 256 = (3w - 4)(w – 4) 256 = 3w 2 – 16w + 16 0 = 3w 2 – 16w - 240 0 = (w – 12)(3w + 20) Width = 12 cm Perimeter = 96 cm

9 3 meters Maximum height (at vertex) height when I hit the water? Distance from board when entering water? Mr. Lomas (at least the way I remember me) …. #6

10 Maximum height of 525 cm

11 #7 Equation: y = x 2 - 10x + 15 Direct: Up Width: Standard L of S: x = 5 Vertex: (5, -10) y-int: (0, 15) Roots:

12 #8 Equation: y = -2x 2 + 8 Direct: Down Width: Narrow L of S: x = 0 Vertex: (0, 8) y-int: (0, 8) Roots: 2 or -2

13 #9 Equation: y = (1/4)x 2 - 2x - 5 Direct: Up Width: Wide L of S: x = 4 Vertex: (4, -9) y-int: (0, -5) Roots: 10 or -2


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