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Nana76 [90]
2 years ago
15

Find the length and perimeter of a rectangle if its width is 19 m and its area is 475 m?.

Mathematics
1 answer:
Allisa [31]2 years ago
4 0

The length of the rectangle is: 25 m

The perimeter of the rectangle is: 88 m

<h3>What is the Area and Perimeter of a Rectangle?</h3>

Area of a rectangle = (length)(width).

Perimeter of a rectangle = 2(length + width).

Give the following:

Width of rectangle = 19 m

Area of rectangle = 475 m²

Find the Length using the area formula:

475 = (length)(19)

475/19 = length

Length of the rectangle = 25 m

Find the perimeter of the rectangle

Perimeter of the rectangle = 2(25 + 19)

Perimeter of the rectangle = 2(44)

Perimeter of the rectangle = 88 m

Thus, the length of the rectangle is: 25 m

The perimeter of the rectangle is: 88 m

Learn more about the area and perimeter of rectangle on:

brainly.com/question/24571594

#SPJ1

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Step-by-step explanation:

1)  Rewrite {y}^{2}-2y+1y  in the form{a}^{2}-2ab+{b}^{2}, where a = y and b = 1.

\frac{y}{{y}^{2}-2(y)(1)+{1}^{2}}+\frac{6}{{y}^{2}+6y-7}

2)  Use Square of Difference: {(a-b)}^{2}={a}^{2}-2ab+{b}^{2}.

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1 - Ask: Which two numbers add up to 6 and multiply to -7?

-1 and 7

2 - Rewrite the expression using the above.

(y-1)(y-7)

Outcome/Result: \frac{y}{(y-1)^2} +\frac{6}{(y-1)(y+7)}

4) Rewrite the expression with a common denominator.

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5)  Expand.

\frac{{y}^{2}+7y+6y-6}{{(y-1)}^{2}(y+7)}

6) Collect like terms.

\frac{{y}^{2}+(7y+6y)-6}{{(y-1)}^{2}(y+7)}

7) Simplify  {y}^{2}+(7y+6y)-6y  to  {y}^{2}+13y-6y

\frac{{y}^{2}+13y-6}{{(y-1)}^{2}(y+7)}

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<h3>What is Month-end review process?</h3>

This is usually done by the accounting department of an organization and are involved in payrolling.

Outstanding transactions and accounts are also reviewed so as to ensure accuracy.

Read more about Month-end review process here brainly.com/question/24134045

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