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____ [38]
3 years ago
8

If u cut out an rectangular piece of wrapping paper that was 2 times as long and 3 times as wide as the box that she was wrappin

g. The box was 5 inches long and 4 inches wide what is the perimeter of the wrapping paper that she cut?
Mathematics
1 answer:
andrezito [222]3 years ago
5 0
The anwser is 44 in because 5×2=10 for the length and 4×3=12 for the width and since its a square the equation looks like this 12+12+10+10 and thats how you get your anwser p.s. I hope this makes sense XD
hope this helped and dont forget to hit that heart :)
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\bf \textit{distance between 2 points}\\ \quad \\&#10;\begin{array}{lllll}&#10;&x_1&y_1&x_2&y_2\\&#10;%  (a,b)&#10;&({{ -1}}\quad ,&{{ 2}})\quad &#10;%  (c,d)&#10;&({{ 2}}\quad ,&{{ 4}})\\&#10;&({{ 2}}\quad ,&{{ 4}})\quad &#10;%  (c,d)&#10;&({{ 3}}\quad ,&{{ -2}})\\&#10;&({{ 3}}\quad ,&{{ -2}})\quad &#10;%  (c,d)&#10;&({{ -2}}\quad ,&{{ -3}})\\&#10;&({{ -2}}\quad ,&{{ -3}})\quad &#10;%  (c,d)&#10;&({{ -1}}\quad ,&{{ 2}})&#10;\end{array}\qquad &#10;%  distance value&#10;d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}

\bf -------------------------------\\\\&#10;d=\sqrt{[2-(-1)]^2+(4-2)^2}\implies d=\sqrt{(2+1)^2+(2)^2}&#10;\\\\\\&#10;d=\sqrt{3^2+2^2}\implies \boxed{d=\sqrt{13}}\\\\&#10;-------------------------------\\\\&#10;d=\sqrt{(3-2)^2+(-2-4)^2}\implies d=\sqrt{1^2+(-6)^2}\implies \boxed{d=\sqrt{37}}\\\\&#10;-------------------------------\\\\&#10;d=\sqrt{(-2-3)^2+[-3-(-2)]^2}\implies d=\sqrt{(-5)^2+(-3+2)^2}&#10;\\\\\\&#10;d=\sqrt{(-5)^2+(-1)^2}\implies \boxed{d=\sqrt{26}}

\\\\&#10;-------------------------------\\\\&#10;d=\sqrt{[-1-(-2)]^2+[2-(-3)]^2}\implies d=\sqrt{(-1+2)^2+(2+3)^2}&#10;\\\\\\&#10;d=\sqrt{(1)^2+(5)^2}\implies \boxed{d=\sqrt{26}}

so, those are their lengths, sum them all up, that's the polygon's perimeter.

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