You are graphing rectangle ABCD in the coordinate plane. The following are the three vertices of the rectangle:

Mathematics

1 answer:

professor190 [17]3 years ago

4 0

Has the same x-value of A
And the same y-value of B
So you’re coordinate is (2,6)

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A 20-ft by the 40-ft rectangular swimming pool is surrounded by a walkway of uniform width. If the total area of the walkway i

Jlenok [28]

<h3>Answer: 2 feet</h3>

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Explanation:

x = width of the walkway in feet

This is some positive real number.

The dimension of 20 feet bumps up to 20+2x when adding on x from both directions. Similarly, the 40 ft dimension becomes 40+2x

Refer to the diagram below.

The 20 ft by 40 ft pool is surrounded by a larger rectangle that is 20+2x ft by 40+2x ft

The pool itself is 20*40 = 800 sq ft. Add on the walkway area to get 800+256 = 1056 sq ft.

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Area = length*width

1056 = (20+2x)*(40+2x)

1056 = 20*40 + 20*2x + 2x*40 + 2x*2x ... FOIL rule

1056 = 800 + 40x + 80x + 4x^2

0 = 4x^2 + 40x + 80x + 800 - 1056

0 = 4x^2 + 120x - 256

4x^2 + 120x - 256 = 0

4(x^2 + 30x - 64) = 0

x^2 + 30x - 64 = 0

Let's use the quadratic formula to finish solving for x.

Plug in a = 1, b = 30, c = -64

$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\x = \frac{-30\pm\sqrt{(30)^2-4(1)(-64)}}{2(1)}\\\\x = \frac{-30\pm\sqrt{1156}}{2}\\\\x = \frac{-30\pm34}{2}\\\\x = \frac{-30+34}{2} \ \text{ or } \ x = \frac{-30-34}{2}\\\\x = \frac{4}{2} \ \text{ or } \ x = \frac{-64}{2}\\\\x = 2 \ \text{ or } \ x = -32\\\\$