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Mathematics, 04.05.2021 01:00 violetagamez2

If f(x)=ax-b and f^-1(x+1)=x/2+2 for each value of R then what must be the values of a anf b?​

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 \cos( \frac{4\pi}{5} ) + i \:  sin( \frac{4\pi}{5}) \\ \\ = e {}^{i \frac{4\pi}{5} } \\ \\ \ \\ so \\ \\ \sqrt[3]{8 \:  ( \cos( \frac{4\pi}{5} ) + isin \frac{4\pi}{5 } ) } \\ \\ \\ \\ 2 \:  \:  \sqrt[3]{e {}^{ \frac{i4\pi}{5} } } \\ \\ 2 \:  e {}^{ \frac{i4\pi}{15} }

Use de moivre’s theorem to compute the following: i attempted to do it but i got the wrong answer a
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square root of 36/6 = 2.449489743

square root of 16/6 = 1.632993162

square root of 6 = 2.449489743

the answer is square root of 16/6 because the square root of 36/6 is the same answer as square root of 6 = 2.449489743

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the function makes a strait line horizontally with no x intercept but a y intercept at 9.

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If f(x)=ax-b and f^-1(x+1)=x/2+2 for each value of R then what must be the values of a anf b?​...
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