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Arlecino [84]
1 year ago
8

Find the length of the third side. If necessary, write in simplest radical form.DV895

Mathematics
1 answer:
arlik [135]1 year ago
8 0

In order to solve the missing side for a right triangle, we can use the Pythagorean theorem

a^2+b^2=c^2

then, we rewrite the expression for on of the sides different from the hypotenuse

\begin{gathered} a^2=c^2-b^2 \\ a=\sqrt[]{c^2-b^2} \end{gathered}

replace with the values

\begin{gathered} a=\sqrt[]{(\sqrt[]{89})^2-5^2} \\ a=\sqrt[]{89-25} \\ a=\sqrt[]{64} \\ a=8 \end{gathered}

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3 years ago
A sailboat’s sail has three sides that are all the same length. Each side measures x−10 units. The sail’s perimeter is 63 units.
Scilla [17]

Answer:

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Step-by-step explanation:

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How do you do this problem? I need to know how you found the answer.
Alexxandr [17]

to get the equation of a line, we simply need two points, say for the Red one ... notice in the graph the lines passes through (0,2) and (-1,6), so let's use those


\bf (\stackrel{x_1}{0}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{-1}~,~\stackrel{y_2}{6}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{6-2}{-1-0}\implies \cfrac{4}{-1}\implies -4 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-2=-4(x-0) \\\\\\ y-2=-4x\implies \blacktriangleright y=-4x+2 \blacktriangleleft


now, for the Blue one, say let's use hmmm it passes through (0,2) and (1.6)


\bf (\stackrel{x_1}{0}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{1}~,~\stackrel{y_2}{6}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{6-2}{1-0}\implies \cfrac{4}{1}\implies 4 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-2=4(x-0) \\\\\\ y-2=4x\implies \blacktriangleright y=4x+2 \blacktriangleleft

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bezimeni [28]

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Step-by-step explanation:

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