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Anna007 [38]
3 years ago
10

backyard garden rectangular in shape. determine the best length and width(in feet). how many feet of fencing would be needed? us

e the formula P=2L+2W
Mathematics
1 answer:
olga2289 [7]3 years ago
8 0
L = building side
W = non-building side

P = 2W + L = 42 (note only one L because the other L is the building itself)

Solve for L:

L = 42 - 2W


Area = l*w
Area = (42-2W)W = 42W - 2W2

Let area be y, so y = -2W2 +42W

Note this is a parabola pointing down because the coefficient of the W2 is negative. That makes the vertex the maximum for which we are searching.

Vertex of this parabola is at W=-b/2a, if the quadratic is aW2 + bW + c = 0

a = -2
b = 42

W = -42/(2*-2) = -42/-4 = 10.5

W = 10.5
L = 42-2(10.5) = 42-21 = 21

Area = L*W = 21 * 10.5

A = 220.5 ft2
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Given

\frac{10q^5w^7}{2w^3}.\ \frac{4\left(q^6\right)^2}{w^{-5}}

Required

Determine the equivalent expression

\frac{10q^5w^7}{2w^3}.\ \frac{4\left(q^6\right)^2}{w^{-5}}

Simplify the first fraction

\frac{5q^5w^7}{w^3}.\ \frac{4\left(q^6\right)^2}{w^{-5}}

Apply law of indices on the first fraction;

5q^5w^{7-3}.\ \frac{4\left(q^6\right)^2}{w^{-5}}

5q^5w^4.\ \frac{4\left(q^6\right)^2}{w^{-5}}

\ \frac{5q^5w^4*4\left(q^6\right)^2}{w^{-5}}

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Hence:

\frac{10q^5w^7}{2w^3}.\ \frac{4\left(q^6\right)^2}{w^{-5}} =20q^{17}w^9

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