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3 years ago

the perimeter of a rectangle is 232 inches. the width is 36 inches. what is the length of the rectangle?

Mathematics

1 answer:

hram777 [196]3 years ago

7 0

Answer:

80 inches

Step-by-step explanation:

Perimeter= 2l + 2w

232= 2l + 2(36)

232= 2(l+36)

232/2 = (l+36)

116 = l+36

116-36=l

80=l

Hence, Length is 80 inches

Hope this helps!

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Step-by-step explanation:

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For the diagonal line of the first triangle, the coordinates are (0, 0) and (2, 1)

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For the diagonal line of the large triangle, the coordinates are (2, 1) and (6, 3)

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What are the domain and range of f(x) = |x – 3 | + 6? Domain: {x | x is all real numbers} Range: {y | y ≥ 6} Domain: {x | x ≥ 3}

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Find the area of the regular polygon

Y_Kistochka [10]

Answer:

A = 374.123 ft^2

Step-by-step explanation:

First, lets calculate the perimeter:

Perimeter (p) = side length (s) * number of sides (n)

$p = s * n$

$p = 12 * 6$

$p = 72$

Next, lets find the apothem, which is the shortest length from any side to the middle. It's like the radius in a circle, but more complicated.

Apothem (a) = side length (s) / ( 2 * tan(180/number of sides (n)) )

$a = \frac{s}{2 * tan (\frac{180}{n} )}$

$a = \frac{12}{2 * tan (\frac{180}{6} )}$

$a = \frac{12}{2 * \frac{\sqrt{3} }{3}}$

$a = \frac{12}{\frac{2\sqrt{3} }{3}}$

$a = \frac{12*3}{2\sqrt{3}}$

$a = \frac{6*3}{\sqrt{3}}$

$a = \frac{18}{\sqrt{3}}$

Now, finally, to find the area of a regular polygon, we use the following equation:

Area (A) = ( apothem (a) * perimeter (p) ) / 2

$A = \frac{a * p}{2}$

$A = \frac{\frac{18}{\sqrt{3} } * 72}{2}$

$A = \frac{18}{\sqrt{3}} * 36$

$A = \frac{640}{\sqrt{3}}$

Turning into a decimal:

$A = 374.123 ft ^2$

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