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3 years ago

The perimeter of a rectangular photograph is 28 inches. If the width of the photo is 5 inches, how tall is the photo?

Mathematics

2 answers:

svetoff [14.1K]3 years ago

8 0

Answer:

9 inches.

Step-by-step explanation:

Since the width of the rectangle is 5 inches, We have to times it by 2 which is 10.

Then deduct it from the perimeter of 28, which equals to 18. We have to divide it by 2 so we can find out how tall each side is. 18 divided by 2=9.

Hope that helps.

Advocard [28]3 years ago

6 0

Answer:

it is 9 inches tall

Step-by-step explanation:

p=2L + 2W

28=2L + 2 x 5

28=2L +10

28-10=2L

18=2L

L=9

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$\dfrac{5x^4-7x^3-12x^2+6x+21}{(x-3)(x^2-2)^2}=\dfrac{a_1}{x-3}+\dfrac{a_2x+a_3}{x^2-2}+\dfrac{a_4x+a_5}{(x^2-2)^2}$
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When $x=\sqrt2$ or $x=-\sqrt2$ , you're left with

$\begin{cases}17-8\sqrt2=(\sqrt2a_4+a_5)(\sqrt2-3)&\text{for }x=\sqrt2\\17+8\sqrt2=(-\sqrt2a_4+a_5)(-\sqrt2-3)\end{cases}\implies\begin{cases}-5+\sqrt2=\sqrt2a_4+a_5\\-5-\sqrt2=-\sqrt2a_4+a_5\end{cases}$

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$5x^4-7x^3-12x^2+6x+21=3x^4-11x^2-8x+27+(a_2x+a_3)(x-3)(x^2-2)$
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So the partial fraction decomposition is

$\dfrac3{x-3}+\dfrac{2x-1}{x^2-2}+\dfrac{x-5}{(x^2-2)^2}$

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