Answer:
Step-by-step explanation:
The posssible factors are
x^2 + 4x - 32
(x+8)(x-4)
x^2+ 14x - 32
(x+16)(x-2)
(x+32)(x-1)
x^2+31x-32
Suppose the larger pump alone can empty the tank in L hours, and the smaller pump can finish the job in S hours, then each hour the large pump empties 1/L portion of the tank, and the small pump empties 1/S per hour
Working together for three hours, they empty the whole tank, which is 100% of it, so 3/L+3/S=100%=1
Larger pump can empty the tank in 4 hours less than the smaller one, so L=S-4
replace L: 3/(S-4)+3/S=1
Make the denominator the same to solve for:
3S/[S(S-4)] +3(S-4)/[S(S-4)]=1
(3S+3S-12)/[S(S-4)]=1
(3S+3S-12)=[S(S-4)]
S^2-10s+12=0
use the quadratic formula to solve for S
S is about 8.6
The answer is not whole hour.
Answer:
x = 5
Step-by-step explanation:
20(20-x)=300
1. Divide both sides by 20.
20(20-x)/20 = 300/20
2. Simplify.
20-x=15
3. Subtract 20 from both sides.
20-x-29=15-20
4. Simplify.
-x=-5
5. Divide both sides by -1.
-x/-1 = -5/-1
6. Simplify.
x=5
