Answer:
it would be just a little less to fit in
Step-by-step explanation:
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.
Answer:
$109 and 35%
Step-by-step explanation:
49.05 is the sale price and the discount is 55%. What’s the original price?
price/ 1- discount
49.05/ 1 - 0.55
49.05/ 0.45
109
for the other part
to find the discount subtract final price from original price
94 - 61.10 = 32.9
Which would be 35% in percent form.