Answer: 12 students
Step-by-step explanation:
Let X and Y stand for the number of students in each respective class.
We know:
X/Y = 2/5, and
Y = X+24
We want to find the number of students, x, that when transferred from Y to X, will make the classes equal in size. We can express this as:
(Y-x)/(X+x) = 1
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We can rearrange X/Y = 2/5 to:
X = 2Y/5
The use this value of X in the second equation:
Y = X+24
Y =2Y/5+24
5Y = 2Y + 120
3Y = 120
Y = 40
Since Y = X+24
40 = X + 24
X = 16
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Now we want x, the number of students transferring from Class Y to Class X, to be a value such that X = Y:
(Y-x)=(X+x)
(40-x)=(16+x)
24 = 2x
x = 12
12 students must transfer to the more difficult, very early morning, class.
Hi there!
![\large\boxed{(-\infty, \sqrt[3]{-4}) \text{ and } (0, \infty) }](https://tex.z-dn.net/?f=%5Clarge%5Cboxed%7B%28-%5Cinfty%2C%20%5Csqrt%5B3%5D%7B-4%7D%29%20%5Ctext%7B%20and%20%7D%20%280%2C%20%5Cinfty%29%20%7D)
We can find the values of x for which f(x) is decreasing by finding the derivative of f(x):

Taking the derivative gets:

Find the values for which f'(x) < 0 (less than 0, so f(x) decreasing):
0 = -8/x³ - 2
2 = -8/x³
2x³ = -8
x³ = -4
![x = \sqrt[3]{-4}](https://tex.z-dn.net/?f=x%20%3D%20%5Csqrt%5B3%5D%7B-4%7D)
Another critical point is also where the graph has an asymptote (undefined), so at x = 0.
Plug in points into the equation for f'(x) on both sides of each x value to find the intervals for which the graph is less than 0:
f'(1) = -8/1 - 2 = -10 < 0
f'(-1) = -8/(-1) - 2 = 6 > 0
f'(-2) = -8/-8 - 2 = -1 < 0
Thus, the values of x are:
![(-\infty, \sqrt[3]{-4}) \text{ and } (0, \infty)](https://tex.z-dn.net/?f=%28-%5Cinfty%2C%20%5Csqrt%5B3%5D%7B-4%7D%29%20%5Ctext%7B%20and%20%7D%20%280%2C%20%5Cinfty%29)
I think it is -5,-4,-1/4, 1.75, 6.2, 9, 18 1/2, 21
Answer:.......
Step-by-step explanation:
Answer:
yes
Step-by-step explanation: