1.
If no changes are made, the school has a revenue of :
625*400$/student=250,000$
2.
Assume that the school decides to reduce n*20$.
This means that there will be an increase of 50n students.
Thus there are 625 + 50n students, each paying 400-20n dollars.
The revenue is:
(625 + 50n)*(400-20n)=12.5(50+n)*20(20-n)=250(n+50)(20-n)
3.
check the options that we have,
a fee of $380 means that n=1, thus
250(n+50)(20-n)=250(1+50)(20-1)=242,250 ($)
a fee of $320 means that n=4, thus
250(n+50)(20-n)=250(4+50)(20-4)=216,000 ($)
the other options cannot be considered since neither 400-275, nor 400-325 are multiples of 20.
Conclusion, neither of the possible choices should be applied, since they will reduce the revenue.
7x - 2y = -3
14x + y = 14
14x -14x + y = 14-14x
Y = 14 - 14x
7x -2y = -3
7x - 2(14 - 14x) = -3
7x -28 + 28x = -3
7x + 28x - 28 = -3
35x - 28 = -3
35x - 28 + 28 = -3 + 28
35x = 25
35x/35 = 25/35
X = 25/35 = 5/7.
Y = 14 - 14x
Y = 14 - 14(5/7)
Y = 14 - 10
Y = 4.
The solution P(5/7,4).
<h3>Answer:</h3><h3>The equation for the slope of a line, given two coordinate points, (x1,y1) , (x2,y2), is (y2-y1)/(x2-x1) </h3><h3> So we get slope = (1-(-5)) / (4-1) = 6/3 = 2</h3><h3>A is your correct answer.</h3>
Hope this helps! please consider marking my answer the brainliest :)
Formula for exponential growth is:
A = P(1+r)^t
Where A is the resulting amount, P is the initial value, r is the rate, and t is the time passed.
Filling this out for this situation gets you:
13400 = 10000(1+r)^6
Solving for r:
Divide 13400 by 10000
1.34 = (1+r)^6
Take the 6th root of both sides...
1 + r = 1.05
No more work is needed, since you aren’t solving for r.
Full equation:
V = 10000(1.05)^t
Answer C is correct
