You would have to solve for x the plug in the answer for c into the next equation to find your answer
I believe the equation is
![4 \sqrt[4]{2x} + 6 \sqrt[4]{2x}](https://tex.z-dn.net/?f=4%20%5Csqrt%5B4%5D%7B2x%7D%20%2B%206%20%20%5Csqrt%5B4%5D%7B2x%7D%20)
In this case, you would simplify it by adding them together.
![4 \sqrt[4]{2x} + 6 \sqrt[4]{2x}](https://tex.z-dn.net/?f=4%20%5Csqrt%5B4%5D%7B2x%7D%20%2B%206%20%20%5Csqrt%5B4%5D%7B2x%7D%20)
=
![10 \sqrt[4]{2x}](https://tex.z-dn.net/?f=10%20%20%5Csqrt%5B4%5D%7B2x%7D%20)
And can even be changed to an exponential equation:
1.50s + 0.75f = 116.25
2f = s
1.50(2f) + 0.75f = 116.25
3f + 0.75f = 116.25
3.75f = 116.25
f = 116.25/3.75
f = 31...31 friday papers
2f = s
2(31) = s
62 = s <=== 62 sunday papers
First, note that

is always positive (except for x=0), so

must be always negative.
Thus, the only plausible graphs are 1 and 3 since they are below the x-axis.
Now,

and

are only defined for x≥0, because only for these x'es we can take the square root.
Note that the third graph has domain (-infinity, 0], so it is not the right one, while 1 is ok.
Answer: first graph