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hodyreva [135]
3 years ago
6

Find the expression that represents the length of the hypotenuse of a right triangle whose legs measure m^2-n^2 and 2mn

Mathematics
1 answer:
My name is Ann [436]3 years ago
7 0
\bf c^2=a^2+b^2\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
------\\
a=m^2-n^2\\
b=2mn
\end{cases}\\\\\\ c^2=(m^2-n^2)^2+(\underline{2mn})^2
\\\\\\
c^2=m^4-2m^2n^2+n^4+\underline{2^2m^2n^2}
\\\\\\
c^2=m^4-2m^2n^2+n^4+{4m^2n^2}
\\\\\\
c^2=m^4+2m^2n^2+n^4\impliedby \textit{perfect square trinomial}
\\\\\\
c^2=(m^2+n^2)^2\implies c=\sqrt{(m^2+n^2)^2}\implies c=m^2+n^2
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Umnica [9.8K]

Answer:

x = -2 —> y = 4

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<h3>X= –2</h3>

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y = ( \frac{1}{2} ) ^{x}  \\ y = ( \frac{1}{2} ) ^{ 0}  =1

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