using the law of cosines:
a^2 = b^2 + c^2 - 2*b*c*cos(A)
a = 90, b = 55, c = 50
90^2 = 55^2 + 50^2 - 2*55*50*cos(A)
8100 = 3025 + 2500 - 5500 * cos(A)
5500 * cos(A) = 3025 + 2500 - 8100
5500 * cos(A) = -2575
cos(A) = -103/220
A = arccos(-103/220)
A = 117.9 degrees
So if the driveway is 45 feet long and he clears 6 feet per hour, we can divide 45 by 6 to find out how many hours it will take for him to clear the driveway and that is essential
45 / 6 = 7.5
45 feet / 6 feet per hour = 7.5 hours to clear the driveway
So now we know that the graph will end when he clears the driveway
If we know he will finish after 7.5 hours then find the graph that ends at the 7.5 hour mark, and that is the last one
The area of each rect. would be xy, so the total area of the square would be 2xy, by definition of area of a square
1. First calculate the positive root of the quadratic using your preferred formula. I got a solution 
. That gives you the distance the ball traveled from d=0 (small hill). So the answer is A.
2. The domain is [0,2.121...]. The range is from 0 to the max height. The d for which the ball is the highest is given by the formula x = -b / 2a = -3.3/(-3.2) = 1.031... so the max is 1.902... So the range is [0,1.902...] So the answer is C.
The coefficent is 5, since it’s placed before the variable.
