Question

@satellite73

Answer 1

@satellite73

Answer 2

i can do it, but not with the method you want

Answer 3

as a matter of fact we can do this in two minutes
one really

Answer 4

what do you mean? its calculus

Answer 5

i know what it is
it is really very easy and very basic
the temperature decays (decrease) to the ambient temp, which means the difference in the temperatures goes to zero proportionally
work only with the differences in termpratures '
initial difference in temp is
\[75-38=37\] after 30 minutes the difference is
\[60-38=22\] do it decreases by a factor of \(\frac{22}{38}\) every 30 minutes

Answer 6

oops i meant
\[\frac{22}{37}\]
to find the new temp in another 30 minutes compute
\[22\times \frac{22}{37}\]

Answer 7

uhh.. i don think my teacher wants it this way haha im dealing with dT/dt=k(T-38) and then integrating it but thanks

Answer 8

you get 13.08 add that to 38 and get 51 approx

Answer 9

yeah cant do it that way

Answer 10

ok then how about this
it is going to be
\[T=T_0e^{kt}\] and we need to solve for \(k\)

Answer 11

this is the way it is usually done, especially in pre - calc
you get the same answer, just more work