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Luda [366]
4 years ago
10

Solve for x. (e^x-e^-x)/((e^x+e^-x)=t

Mathematics
2 answers:
Sergio [31]4 years ago
8 0
\dfrac{e^x-e^{-x}}{e^x+e^{-x}}=t\\\\
\dfrac{e^{2x}-1}{e^{2x}+1}=t\\\\
e^{2x}-1=t(e^{2x}+1)\\\\
e^{2x}-1=e^{2x}t+t\\\\
e^{2x}-e^{2x}t=t+1\\\\
e^{2x}(1-t)=t+1\\\\
e^{2x}=\dfrac{t+1}{1-t}\\\\
e^{2x}=-\dfrac{t+1}{t-1}\\\\
2x=\ln\left(-\dfrac{t+1}{t-1}\right)\\\\
x=\dfrac{\ln\left(-\dfrac{t+1}{t-1}\right)}{2}\qquad t\in(-1,1)




sweet-ann [11.9K]4 years ago
4 0

Answer:

d on edge

Step-by-step explanation:

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<u>ANSWER </u>


x=\frac{1}{2} and y=\frac{1}{4}



<u>EXPLANATION</u>


Given;

4x-12y=1

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The augmented matrix of the two linear equation is given by;


\left[\begin{array}{ccc}4&-12|&-1\\6\:&\:\:\:4|&4\end{array}\right]


We now perform row operations;

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This gives us


\left[\begin{array}{ccc}1&-3|&\frac{-1}{4}\\6\:&\:\:\:4|&4\end{array}\right]



R_2-6R_1 \rightarrow R_2


\left[\begin{array}{ccc}1&-3|&\frac{-1}{4}\\0\:&\:\:\:22|&\frac{11}{2}\end{array}\right]


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\left[\begin{array}{ccc}1&-3|&\frac{-1}{4}\\0\:&\:\:\:1|&\frac{1}{4}\end{array}\right]


R_1+3R_2 \rightarrow R_1


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Hence x=\frac{1}{2} and y=\frac{1}{4}




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