20% if it will determine the pay of how much it is.
Answer: c did on a p e x
Step-by-step explanation:
Answer:
The amount of water must be added to this task = 60 
Step-by-step explanation:
Let amount of water added = x 
Then from the given conditions
A chemist would like to dilute a 90-cc solution that is 5% acid to one that is 3% acid. So
90 (0.05) = 0.03 ( 90 + x )
4.5 = 2.7 + 0.03 x
0.03 x = 1.8
x = 60 
Therefore the amount of water must be added to this task = 60 
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

