Aiko jump roped for 20 minutes before stopping at 8:05. Just count 20 minutes back in order to see when she started.
8:05 - 20 minutes = 7:45
The answer to the question
The given equations are

(1)

(2)
When t=0, obtain

Obtain derivatives of (1) and find x'(0).
x' (t+1) + x - 4√x - 4t*[(1/2)*1/√x = 0
x' (t+1) + x - 4√x -27/√x = 0
When t=0, obtain
x'(0) + x(0) - 4√x(0) = 0
x'(0) + 9 - 4*3 = 0
x'(0) = 3
Here, x' means

.
Obtain the derivative of (2) and find y'(0).
2y' + 4*(3/2)*(√y)*(y') = 3t² + 1
When t=0, obtain
2y'(0) +6√y(0) * y'(0) = 1
2y'(0) = 1
y'(0) = 1/2.
Here, y' means

.
Because

, obtain

Answer:
The slope of the curve at t=0 is 1/6.
You will need this formula:
Years = ln (Total / Principal) / rate
(where "ln" means natural logarithm)
and we'll use $100 and $200 for beginning and ending amount
Years = ln (200 / 100) / rate
Years = 0.69314718056 / .052
Years =
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<span>
<span>
13.3297534723
</span>
</span>
</span>
Rounding to the nearest tenth of a year:
Years =
<span>
<span>
<span>
13.3
Source:
http://www.1728.org/rate2.htm
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