The first pump empties half the pond in 3 hours, so in 1/6 that time (1/2 hour), it empties (1/6)·(1/2) = 1/12 of the pond.
The second pump empties the other 5/12 of the pond in that half hour, so has a pumping rate of (1/2 h)/(5/12 pond) = (6/5 h)/pond.
The second pump could do the entire job alone in 1 hour and 12 minutes.
Answer:
D its D
Step-by-step explanation:
Answer:

Step-by-step explanation:
The rate of infection is jointly proportional to the number of infected troopers and the number of non-infected ones. It can be expressed as follows:

Rearranging and integrating

At the initial breakout (t=0) there was one trooper infected (I=1)

In two days (t=2) there were 5 troopers infected

Rearranging, we can model the number of infected troops (I) as

You have to keep order of operations in mind.
1. Take care of the parentheses; distribute.
-3+6x-3=-20-8x
Simplified: 6x-6=-20-8x
2. Isolate the x variable.
Add 8x to both sides and add 6 to both sides.
14x=-14
3. Solve for x.
Divide by 14 on both sides.
x=-1