The equation $y=2x+c$ is a linear equation. By definition, the independent term on this equation (that is, the number that is not being multiplied by $x$ ) is the y-intercept, which is a fancy way of saying "the point where the line crosses the y-axis".

By looking at the equation, we know that our y-intercept is c. By looking at the graph, we can see that the y-intercept is -3. Therefore, $c=-3$ and we get the complete version of our linear equation:

$y=2x-3$

Now, looking at the graph we can see that the point $(k,10)$ lies on the line of the equation, which means that the point is a solution to our equation. All we have to do is replace $x$ and $y$ by the values of the given point (which are $k$ and $10$ , respectively), and then solve for $k$ :

$10=2k-3\\10+3=2k\\13=2k\\2k=13\\k=\frac{13}{2}$

And we are done!

4 0

3 years ago

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

Sergio [31]

$\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)



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8 0

4 years ago

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Bananas are curved because they grow towards the sun ☀️

VLD [36.1K]

Hope you have a good day

5 0

3 years ago

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