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Zigmanuir [339]
4 years ago
10

The area of a certain rectangle is 288 yd2. The perimeter is 68 yd. What are the dimensions of the rectangle?

Mathematics
1 answer:
arsen [322]4 years ago
8 0

Answer:

The rectangle is 16*18

Step-by-step explanation:

The area of a rectangle is length * width and the perimeter is 2 times length * width

A = l*w

P = 2(l+w)

Replacing A and P

288 = l*w

68 = 2(l+w) => 34 = l+w => 34 - w = l

replacing l in the area

288 = (34 - w) w

w^2 - 34w + 288 = 0

(w - 16)(w - 18)

w = 16

w = 18

replacing in 34 - w = l

you get that when w = 16, l = 18 and when w = 18, l = 16

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The expression -22x + 3y = 6 in the form ax + by + c = 0 is -2x + 3y - 6 = 0

<h3>Linear Equations</h3>

The given equation is:

-2x + 3y = 6

Comparing the equation above with ax + by + c = 0

Rewrite -2x + 3y = 6 in the form ax + by + c = 0

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a = -2, b = 3, c = -6

Therefore, the expression-22x + 3y = 6 in the form ax + by + c = 0 is -2x + 3y - 6 = 0

Learn more on linear equations here: brainly.com/question/14323743

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Step-by-step explanation:

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