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How do I solve for a?? 114=(180)(a^-2)+34 it is suppose to be 3/2 any help on how to get that?
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\(\begin{array} \ 115 &= 180(a^{-2}) +34 \\ 115-34 &=180(a^{-2}) \\ 81 &= 180(a^{-2}) \\ 0.45 &=a^{-2} \\ \end{array}\)
might want to start with \(115=\frac{180}{a^2}+34\) but it makes no difference,
where did u get 115?
Yeah, right, excellent advice ;D
Oh, it's 114.. \(\begin{array} \ 80 \div 180&= (a^{-2}) \\ \frac{4}{9} &=1 \div a^2 \\ 1 \div \frac{4}{9} &= a^2\\ \sqrt{2.25} &= a \\ \frac{3}{2} &= a\end{array}\)
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thanks! i understand now. yall ROCK
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