Answer:
so the answer will be 1
Step-by-step explanation:
An oil tanker can be emptied by the main pump in 5 hours.
An auxilary pump can empty the tanker in 14 hours.
If the main pump is started at 7pm, when should the auxilary pump be started
so that the tanker is emptied by 11pm?
:
Let t = no. of hrs to run the Aux pump
:
The main pump will run 4 hr (7 - 11 pm)
:
Let the completed job = 1; (an empty Tanker)
:
The shared work equation
4%2F5 + t%2F14 = 1
Multiply equation by 70 to get rid of the equation, results:
14(4) + 5t = 70
56 + 5t = 70
5t = 70 - 56
5t = 14
t = 14%2F5
t = 2.8 hrs to run the aux pump
:
2.8 hr = 2 + .8(60) = 2 hrs 48 min
:
Subtract 2:48 from 11:00 = 8:12 PM start the aux pump
;
;
Check solution
4/5 + 2.8/14 =
.8 + .2 = 1
Multiplying each side by 3/4, we get (3/4)V=pir^3. Dividing both sides by pi, we then get r^3=(3/4)V/pi
For the second part, we get that 381.51=(4/3)πr^3, so plugging 381.51 into the equation above we get (3/4)(381.51)/pi=around 91, and r= the cube root of r^3, which would equal around 4.5 here
Answer:
B
Step-by-step explanation:
