Ahot bowl of soup cools according to newton’s law of cooling. its temperature (degrees f) at time t is given by t(t)=76+105e^(-0.125t), where t is given in minutes. what was the initial temperature of the soup? what is the temperature of the soup after 30 minutes?

a) T=181°F

b) T=78.469°F

Step-by-step explanation:

In this case big T is temperature and little t is time in minutes, in physics this is usually the case.

Now for the initial temperature time is zero minutes so:

Let t=0

And so the initial temperature of the soup is 181°F.

The temperature after 30 minutes is:

Let t=30

So the temperature after 30 minutes is 78.469°F.  If you have a graphing calculator replace T(t) with y and t with x and look at when x=0 and x=30 your y will represent temperature.

a

step-by-step explanation:

show us the graphs dude

step-by-step explanation: