<span>Let y= original side
So
A = LW
60 = (y+15)(y-13)
(y+15)(y-13) = 60
y^2 + 2y - 195 = 60
y^2 + 2y- 195 - 60 = 0
y = -17 or y= 15
Toss out the negative solution to get y= 15 as the only solution.
So the square has a side length of 15 meters.
The area of this square is 15^2 = 15*15 = 225 square meters.</span>
Step-by-step explanation:
Y-0 = 1/4(x-8)
4Y= x-8
4Y-x+8=0
Answer:
Step-by-step explanation:
From the given information:
r = 10 cos( θ)
r = 5
We are to find the the area of the region that lies inside the first curve and outside the second curve.
The first thing we need to do is to determine the intersection of the points in these two curves.
To do that :
let equate the two parameters together
So;
10 cos( θ) = 5
cos( θ) = 

Now, the area of the region that lies inside the first curve and outside the second curve can be determined by finding the integral . i.e









The diagrammatic expression showing the area of the region that lies inside the first curve and outside the second curve can be seen in the attached file below.
The answer is C. They need 10 more baskets to win the game.