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
Answer:
If you look over the steps you can see that until 4x + x + 3 = 18, evertything is dandy. But the step after that 4x + x =21 seems a bit fishy.
Think about it they subtract 3 from both sides so the first side is correct
4x + x, but they added 3 to the other side:
not
4x+x = 21
Then we solve for 4x + x = 15
To solve for y we use :
y = x+3
y = 3+3 = 6
so (3,6) is the right answer
Answer:
Step-by-step explanation:
(x-5)2+8 = 92
2x-10+8 = 92
2x = 94
x = 47
The answer would be log base 2 of 6 over log base 2 of 3
