3.072 seconds, (38.4/12.5)
The initial value of 100 that doubles over each interval.
without the answer choices, I can only describe it and give you an example of the graph.
I'm assuming the function is 100*(2)^x because if it is as listed it would be a quadratic function with a vertical stretch of 100.
So, f[x] = 1/4x^2 - 1/2Ln(x)
<span>thus f'[x] = 1/4*2x - 1/2*(1/x) = x/2 - 1/2x </span>
<span>thus f'[x]^2 = (x^2)/4 - 2*(x/2)*(1/2x) + 1/(4x^2) = (x^2)/4 - 1/2 + 1/(4x^2) </span>
<span>thus f'[x]^2 + 1 = (x^2)/4 + 1/2 + 1/(4x^2) = (x/2 + 1/2x)^2 </span>
<span>thus Sqrt[...] = (x/2 + 1/2x) </span>
At this rate, it should take 7,500 years to dissolve 15m thiccccc
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