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 = -3 and x = -3/2
Step-by-step explanation:
After writing down the polynomial, split it; put a line between 3x^2 and -18x. Look and 2x^3 + 3x^2 and -18x - 27 separately and factor them both:
p(x) = 2x^3 + 3x^2 <u>- 18x -27</u>
p(x) = x^2(2x+3) <u>-9(2x+3)</u>
Now notice how x^2 and -9 have the same factor (2x+3). That means x^2 and -9 can go together:
p(x) = (x^2 - 9)(2x+3)
Factor it once more because there's a difference of squares:
p(x) = (x+3)(x-3)(2x+3)
Now just plug in whatever makes the each bracket equal 0:
x = -3, x = 3, and x = -3/2
Those are your zeros.
In 2 hours,hope this helps :)
I’m pretty sure your counting by 3’s because It’s goes 9,3 . However it might be in the negatives
