Answer:
If there are N students, the total cost will be:
C(N) = $500 + $5*N.
And we want the amount that each student pays to be equal or less than $15.
then:
C(N)/N ≤ $15.
Then the inequality that represents this situation is:
($500 + $5*N)/N ≤ $15.
Now, let's solve this:
($500 + $5*N) ≤ $15*N
$500 ≤ $15*N - $5*N
$500 ≤ $10*N
$500/$10 ≤ N
50 ≤ N
So the minimum number of students needed is 50.
What in the world.. weirdest thing that i saw today
![\bf f(x)=x+3x^{\frac{2}{3}}\implies \cfrac{dy}{dx}=1+3\left(\frac{2}{3}x^{-\frac{1}{3}} \right)\implies \cfrac{dy}{dx}=1+\cfrac{2}{\sqrt[3]{x}} \\\\\\ \cfrac{dy}{dx}=\cfrac{\sqrt[3]{x}+2}{\sqrt[3]{x}}\implies 0=\cfrac{\sqrt[3]{x}+2}{\sqrt[3]{x}}\implies 0=\sqrt[3]{x}+2\implies -2=\sqrt[3]{x} \\\\\\ (-2)^3=x\implies \boxed{-8=x}\\\\ -------------------------------\\\\ 0=\sqrt[3]{x}\implies \boxed{0=x}](https://tex.z-dn.net/?f=%5Cbf%20f%28x%29%3Dx%2B3x%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%5Cimplies%20%5Ccfrac%7Bdy%7D%7Bdx%7D%3D1%2B3%5Cleft%28%5Cfrac%7B2%7D%7B3%7Dx%5E%7B-%5Cfrac%7B1%7D%7B3%7D%7D%20%20%5Cright%29%5Cimplies%20%5Ccfrac%7Bdy%7D%7Bdx%7D%3D1%2B%5Ccfrac%7B2%7D%7B%5Csqrt%5B3%5D%7Bx%7D%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7Bdy%7D%7Bdx%7D%3D%5Ccfrac%7B%5Csqrt%5B3%5D%7Bx%7D%2B2%7D%7B%5Csqrt%5B3%5D%7Bx%7D%7D%5Cimplies%200%3D%5Ccfrac%7B%5Csqrt%5B3%5D%7Bx%7D%2B2%7D%7B%5Csqrt%5B3%5D%7Bx%7D%7D%5Cimplies%200%3D%5Csqrt%5B3%5D%7Bx%7D%2B2%5Cimplies%20-2%3D%5Csqrt%5B3%5D%7Bx%7D%0A%5C%5C%5C%5C%5C%5C%0A%28-2%29%5E3%3Dx%5Cimplies%20%5Cboxed%7B-8%3Dx%7D%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0A0%3D%5Csqrt%5B3%5D%7Bx%7D%5Cimplies%20%5Cboxed%7B0%3Dx%7D)
now, f(0) = 0, and f(-8) is an imaginary value or no real value.
now, f(-10) will also give us an imaginary value
and f(1) = 4
so, doing a first-derivative test on 0, is imaginary to the left and positive on the right, and before and after 1, is positive as well, so f(x) is going up on those intervals.
however, f(0) is 0 and f(1) is higher up, so the absolute maximum will have to be f(1), and we can use f(0) as a minimum, and since it's the only one, the absolute minimum.
the other two, the endpoint of -10 and the critical point of -8, do not yield any values for f(x).
Answer:
The second one.
Step-by-step explanation: