The temperature is changing at a certain rate. What is the temperature when t=5?

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Homework Statement



A pizza, heated to a temperature of 350 degrees Fahrenheit, is taken out of an oven and placed in a 75°F room at time t=0 minutes. The temperature of the pizza is changed at a rate of -110e-0.4t degrees Fahrenheit per minute. TO the nearest degree, what is the temperature of the pizza at time t=5 minutes?

Answer: 112°F

Homework Equations



f'(t) = -ky?
All I can think of is that this problem is related to Newton's Law of Cooling.

The Attempt at a Solution



The rate of change or derivative is -110e-0.4t. I think what you're suppose to do is find how much the temperature dropped and subtract this number to the original temperature, 350°. I originally thought, if I plugged in t=5 into
-110e-0.4t and subtract this number from 350°, then I would have my answer. But, of course, this is wrong.
 
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lude1 said:
The rate of change or derivative is -110e-0.4t

So we have this meaning that

dT/dt = -110e-0.4t ⇒ T= ∫(dT/dt) dt

So what is T equal to?
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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