Answer:
1
Step-by-step explanation:
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:
end of third year = 1,200*(1.02)^3 = 1,273.45 plus
end of second year = 1,200*(1.02)^2 = 1,248.48 plus
end of first year = 1,200*(1.02) = 1,224.00
Total = 3,745.9
Step-by-step explanation:
You are multiplying by 4 or dividing by 4 8 16 20 28
2 4 5 7
<span>Let's find out the ball's acceleration
V^2 = U^2 + 2as
Where the distance S = 6 feet and U = 45.
Suppose he just took off so V = 0 and we are left with U^2 =2as
(45)^2 = 2a(6)
a = 2025/12 = 168.75 ft/s^2
Also S = ut + 1/2 at^2
Here S = 5ft and U= 0; a = 168.75. So we have 5 = (1/2) * 168.75 * t^2
So we have 5 * 2 = 168.75 t^2
T^2 = âš 10/ 168.75 = 0.00596 = âš0.006
T = 0.0077s</span>
