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:
She was wrong because if you were to find the quotient it would be 2 because if it were to be 3 it'll look like 6 divided by 18= 3. so what she has to do is find what times 6 equals 2 and that'll be 6 divided by 12= 2
Step-by-step explanation:
1. find volume
V=LWH
V=25*13*14
V=4550
each covers 1000
how many will cover the whole thing?
we can't put 4 becuase that leaves 550 uncovered
add another
5 is answer (it is too much , but at leas not too little) answer is D
2.
V=LWH
V=6*4*10
V=240
answer is D
3. count how many are there, don't forget the hidden ones
13 cubes
volume=legnth tiems width times height
cubes so
v=side^3
side=3
3^3=27
13 cubes
13*27=351 in^3
challenge (this is meant to challenge you, not for you to ask people, but I will solve anyway, robbin you of experience)
V=LWH
V=5880
H=30
5880=LW30
divide both sides by 30
196=LW
they have square bases so L=W
196=L^2
sqrt both sides
14=L=W
the side legnth is 14in
Answer:
<h2>y=60+25h+0.3r</h2>
Step-by-step explanation:
let y be the total charges
given that the hour is h
and the cost for raw material is r
therefore
1. The first hour of service call = $60
2. For each additional hour= 25h
3. 30% over the cost of the materials.= 30/100*r=0.3r
total charges
y=60+25h+0.3r
therefore the expression is
y=60+25h+0.3r
