This question s incomplete, the complete question is;
The Watson family and the Thompson family each used their sprinklers last summer. The Watson family's sprinkler was used for 15 hours. The Thompson family's sprinkler was used for 30 hours.
There was a combined total output of 1050 of water. What was the water output rate for each sprinkler if the sum of the two rates was 55L per hour
Answer:
The Watson family sprinkler is 40 L/hr while Thompson family sprinkler is 15 L/hr
Step-by-step explanation:
Given the data in the question;
let water p rate for Watson family and the Thompson family sprinklers be represented by x and y respectively
so
x + y = 55 ----------------equ1
x = 55 - y ------------------qu2
also
15x + 30y = 1050
x + 2y = 70 --------------equ3
input equ2 into equ3
(55 - y) + 2y = 70
- y + 2y = 70 - 55
y = 15
input value of y into equ1
x + 15 = 55
x = 55 - 15
x = 40
Therefore, The Watson family sprinkler is 40 L/hr while Thompson family sprinkler is 15 L/hr
He actually borrowed P=21349-3000=18349 (present value)
Assume the monthly interest is i.
then future value due to loan:
F1=P(1+i)^n=18349(1+i)^(5*12)=18349(1+i)^60
future value from monthly payment of A=352
F2=A((1+i)^n-1)/i=352((1+i)^60-1)/i
Since F1=F2 for the same loan, we have
18349(1+i)^60=352((1+i)^60-1)/i
Simplify notation by defining R=1+i, then
18349(R^60)-352(R^60-1)/(R-1)=0
Simplify further by multiplication by (R-1)
f(R)=18349*R^60*(R-1)-352(R^60-1)=0
Solve for R by trial and error, or by iteration to get R=1.004732
The APR is therefore
12*(1.004732-1)=0.056784, or 5.678% approx.
The answer is d because 4+8=12 and they are 4 units away from eachother:)
=7x^4+15x^3+4y^2 I hope this helps!!
