We can solve for x using a technique called successive approximations. step 1: if we assume that x is very small compared to 0.100 (such that 0.100 x ≈ 0.100) then our first approximation of x (let\'s call it x1) can be calculated as

Supposed that we are given the equation:

7.20 x 10^-5 = x(0.100+x)^2 

So assuming that x is very small so that:

0.100 + x ~ 0.100

and calling it x1, so:

7.20 x 10^-5 = x1 (0.100)^2 

Then we simply solve for x1:

x1 = 7.20 x 10^-5 / (0.100)^2 

x1 = 7.20 x 10^-3

step-by-step explanation:

12: 00 p.m. to 7: 00 p.m. is a total of 7 hours.

4: 00 p.m. to 7: 00 p.m. is a total of 3 hours.

this means that from 4: 00 p.m to 7: 00 p.m. this represents 3/7 of what was lost from 12: 00 p.m. to 7: 00 p.m.

after knowing this, after doing 11,600 - 6,800 = 4,800

4,800 represents 3/7 of what was lost for 7 hours.

we then divide 4,800 by 3 which gets 1,600.

because we know that 1,600 represents 1/7 of how much water was lost, we just have to multiply by 7, which gets 11,200.   after adding 11,200 to how much water was left at 7: 00 p.m. which was 6,800. 11,200 + 6,800 = 18,000

this means that the pool originally had 18,000 gallons in the pool.

you can also check your work by doing 18,000 - 4(1600), which becomes           1800 - 6400 which gets you 11,600.