Answer: 31,365
Step-by-step explanation:
Given : The number of horror films = 15
The number of comedies = 18
Then , the number of different combinations of 4 movies can he rent if he wants at least two comedies is given by :-

Hence, the number of different combinations of 4 movies can he rent if he wants at least two comedies = 31,365
Plug three in wherever there is an x
y=.25(3)
y=.75
Y=2x-1
y=-3x+14
First you would substitue one of the y's for the other so
2x-1=-3x+14
no you would solve for x...
add one on both sides
2x=-3x+15
now add 3x to both sides which becomes
5x=15
divide 5 on both sides which gives you x=3
now to solve for y plug into any one of the equations(doesn't matter which one) from before 3 for x and solve
y=2(3)-1
y=6-1
y=5
And that is you answer y=5
Answer:
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The 1st choice is the correct one.
Isabel is correct because the y-intercept of Line B is (0, 9) and the value of y when x = 0 in Parabola A is 9.