Answer:
what do you want us to solve
Step-by-step explanation:
i dont see anything to find a awnser
Answer: Option C)1 over 15 minus 1 over x equals 1 over 20
Explanation:
Since, Micah can fill a box with books in 15 minutes.
Therefore, the work done by Micah in one minute= 1/15
Also, Sydney takes the books out puts them on a shelf.
And the times taken by Micah when Sydney is also taking the books outside from the self= 20 minutes
Therefore, the work done by Micah in one minute when Sydney taking books out of the box= 1/20
Let Sydney alone takes x minutes to take books outsides the shelf.
Then, work done by Sydney in one minute=1/x
Thus, the work done by Sydney( by taking books out of the box)= the work done by Micah - work done by Micah and Sydney simultaneously= 1/15-1/20
⇒1/x=1/15-1/20
⇒1/15-1/20=1/x
⇒1/15-1/x=1/20 is the required expression.
Therefore, Option C is correct.
Answer:

Step-by-step explanation:
We are given that

R=150 ohm
L=5 H
V(t)=10 V





I(0)=0
Substitute t=0


Substitute the values


Answer:25
Step-by-step explanation: