Answer:
yes It is relatively easy to determine whether an equation is a function by solving for y. When you are given an equation and a specific value for x, there should only be one corresponding y-value for that x-value. For example, y = x + 1 is a function because y will always be one greater than x.
Step-by-step explanation:
Answer:
Step-by-step explanation:
1.5
Solve 180=10x+10 to find your answer (:
Answer:
Step-by-step explanation:
![volume = \frac{4}{3} \pi {r}^{3} \\ \\ 288\pi = \frac{4}{3} \pi {r}^{3} \\ \\ 3(288\pi) = 4\pi {r}^{3} \\ \\ \frac{3(288\pi)}{4\pi} = {r}^{3} \\ \\ 216 = {r}^{3} \\ \\ r = \sqrt[3]{216} \\ r = 6](https://tex.z-dn.net/?f=volume%20%3D%20%20%5Cfrac%7B4%7D%7B3%7D%20%5Cpi%20%7Br%7D%5E%7B3%7D%20%20%5C%5C%20%20%5C%5C%20288%5Cpi%20%3D%20%20%5Cfrac%7B4%7D%7B3%7D%20%5Cpi%20%7Br%7D%5E%7B3%7D%20%5C%5C%20%20%5C%5C%203%28288%5Cpi%29%20%3D%204%5Cpi%20%7Br%7D%5E%7B3%7D%20%5C%5C%20%20%5C%5C%20%20%5Cfrac%7B3%28288%5Cpi%29%7D%7B4%5Cpi%7D%20%20%3D%20%20%7Br%7D%5E%7B3%7D%20%20%5C%5C%20%20%5C%5C%20216%20%3D%20%20%7Br%7D%5E%7B3%7D%20%20%5C%5C%20%20%5C%5C%20r%20%3D%20%20%5Csqrt%5B3%5D%7B216%7D%20%20%5C%5C%20r%20%3D%206)
