Find the perimeter of the parallelogram with these vertices.

Mathematics

1 answer:

blondinia [14]2 years ago

4 0

Check the picture below.

so we can say that two sides are "4" each in length, since opposite sides are equal, let's find how long the slanted sides are.

$~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-4}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{5})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ d=\sqrt{[3 - (-4)]^2 + [5 - 2]^2}\implies d=\sqrt{(3+4)^2+3^2} \\\\\\ d=\sqrt{49+9}\implies d=\sqrt{58} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{\Large Perimeter}}{4~~ + ~~4~~ + ~~\sqrt{58}~~ + ~~\sqrt{58}\implies 8+2\sqrt{58}}$

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$6.\\\text{We know}\ 9=3^2.\ \text{Therefore}\\\\9^2=(3^2)^2\\\\\text{use}\ (a^n)^m=a^{nm}\\\\9^2=(3^2)^2=3^{2\cdot2}=3^4\\\\\boxed{3^4=9^2}\\=================$

$7.\\METHOD\ 1:\\\\\sqrt{7^4}=\sqrt{7^{2\cdot2}}\\\\\text{use}\ (a^n)^m=a^{nm}\\\\=\sqrt{(7^2)^2}\\\\\text{use}\ \sqrt{a^2}=a\ \text{for}\ a\geq0\\\\=\boxed{7^2}\\\\METHOD\ 2:\\\\\text{use}\ a^\frac{m}{n}=\sqrt[n]{a^m}\\\\\sqrt{7^4}=7^\frac{4}{2}=\boxed{7^2}\\=================$

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