Answer: 4.5 hours
<u>Step-by-step explanation:</u>


Answer:
D. (-∞,4]
Step-by-step explanation:
The range is the y values
The lowest y values is negative infinity
The highest y values is 4
( - inf, 4]
We use the parentheses since we cannot get to negative infinity, the bracket since we reach 4
Answer:

Step-by-step explanation:
We can use formula (a-b)² = a² -2ab + b².
In our example a = 3c^4 and b = 5c^6
![(3c^{4} - 5c^{6})^{2} = [3c^{4} ]^{2} - 2*3c^{4} *5c^{6} + [5c^{6}]^{2}=\\=9c^{8} -30c^{10} + 25c^{12}](https://tex.z-dn.net/?f=%283c%5E%7B4%7D%20-%205c%5E%7B6%7D%29%5E%7B2%7D%20%3D%20%5B3c%5E%7B4%7D%20%5D%5E%7B2%7D%20-%202%2A3c%5E%7B4%7D%20%2A5c%5E%7B6%7D%20%2B%20%5B5c%5E%7B6%7D%5D%5E%7B2%7D%3D%5C%5C%3D9c%5E%7B8%7D%20-30c%5E%7B10%7D%20%2B%2025c%5E%7B12%7D)
I'm taking the liberty of editing your function <span>v = e5xey: It should be
</span>
<span>v = e^5x^ey, with " ^ " indicating exponentiation.
</span>
Did you mean e^(5x) or (e^5)x? I'll assume it's e^(5x).
The partial of v = e^(5x)e^y with respect to x is e^(5x)(5)*e^y, or 25x*e^y.
The partial of v = e^(5x)e^y with respect to y is e^(5x)e^y.
The answer to this question is 1:5