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:
a) 8x+6a
b)5q+7q
Step-by-step explanation:
a) we need to combine the like(same) terms(with same variable) together
same terms: 3x and 5x
2a and 4a
so
3x+5x=8x
2a+4a=6a
8x+6a
b)a) we need to combine the like(same) terms together
same terms: 7p-2p=5q
6q+q=7q
Answer:
x= 70
Step-by-step explanation:
In this question you have to use BODMAS. This indicates that you have to do the equation in the brackets first. Therefore, 3x15= 45. Then you add 25, which will give you 70.
