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Arturiano [62]
3 years ago
6

The length of a rectangle is 55 inches more than the width. the perimeter is 4242 inches. find the length and the width of the r

ectangle
Mathematics
1 answer:
faust18 [17]3 years ago
6 0
L = W + 55
2L + 2W = 4242

L + W = 2121
(W + 55) + W = 2121
2W + 55 = 2121

W = 1033 inches           L = 1088 inches
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A similar problem is given at brainly.com/question/24855677

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Check the picture below.

so the rhombus has the diagonals of AC and BD, now keeping in mind that the diagonals bisect each, namely they cut each other in two equal halves, let's find the length of each.

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\bf ~~~~~~~~~~~~\textit{distance between 2 points}&#10;\\\\&#10;B(\stackrel{x_1}{-2}~,~\stackrel{y_1}{6})\qquad &#10;D(\stackrel{x_2}{4}~,~\stackrel{y_2}{0})\qquad \qquad BD=\sqrt{[4-(-2)]^2+[0-6]^2}&#10;\\\\\\&#10;BD=\sqrt{(4+2)^2+(-6)^2}\implies BD=\sqrt{6^2+6^2}&#10;\\\\\\&#10;BD=\sqrt{6^2(2)}\implies \boxed{BD=6\sqrt{2}}

that simply means that each triangle has a side that is half of 10√2 and another side that's half of 6√2.

namely, each triangle has a "base" of 3√2, and a "height" of 5√2, keeping in mind that all triangles are congruent, then their area is,

\bf \stackrel{\textit{area of the four congruent triangles}}{4\left[ \cfrac{1}{2}(3\sqrt{2})(5\sqrt{2}) \right]\implies 4\left[ \cfrac{1}{2}(15\cdot (\sqrt{2})^2) \right]}\implies 4\left[ \cfrac{1}{2}(15\cdot 2) \right]&#10;\\\\\\&#10;4[15]\implies 60

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3 years ago
Read 2 more answers
Simplify. x/4x+x^2
djyliett [7]

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if you're wondering about the restriction of x ≠ -4, is due to that would make the fraction with a denominator of 0 and thus undefined.

3 0
3 years ago
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