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3 years ago

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A parking lot in the shape of a trapezoid has an area of 12,0521 square meters the length of the other base is 108.6 meters. Wha

t is the width of the parking lot

1 answer:

andreev551 [17]3 years ago

5 0

Please consider the complete question.

A parking lot in the shape of a trapezoid has an area of 12,052.1 square meters. The length of one base is 82.4 meters and the length of the base is 108.6 meters. What is the width of the parking lot

We will use area of trapezoid to formula to solve our given problem.

$\text{Area of trapezoid}=\frac{1}{2}(a+b)\times h$ , where

a and b represents parallel sides of trapezoid,

h = height of trapezoid.

Width of the trapezoid will be equal to height.

Upon substituting our given values in above formula, we will get:

$12,052.1=\frac{1}{2}(82.4+108.6)\times h$

$12,052.1\cdot 2=2\cdot \frac{1}{2}(82.4+108.6)\times h$

$24104.2=(191)\times h$

$\frac{24104.2}{191}=\frac{(191)\times h}{191}$

$126.2=h$

Therefore, the width of the parking lot is 126.2 meters.

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Step-by-step explanation:

We are given the function:

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Recall that by the definition of inverse functions:

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Let k⁻¹(-4) = <em>x, </em>where <em>x</em> is an unknown value. Then by definition, k(x) must equal -4.

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$\displaystyle k(x) = \frac{x+3}{x+9} = -4$

Solve for <em>x: </em>

$\displaystyle \begin{aligned} \frac{x+3}{x+9} &= -4 \\ \\ x+3 &= -4(x+9) \\ \\ x+3 &= -4x - 36 \\ \\ x &= -\frac{39}{5} \end{aligned}$

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And the variance is given by, $\sigma ^2=\frac{q}{p^2}$

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$\sigma ^2=\frac{q}{p^2}$

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= $\mu = \frac{1}{0.001}$

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