Recent content by hulgy

OK.. 40-Hs=(47-Hs)z 34-Hs=(47-Hs)z^2 I square the first equation (40-Hs)^2=(47-Hs)^2*z^2 and divide by the second, so I get: (40-Hs)^2/(34-Hs) = (47-Hs) ...FFFFFUUUUUUUUUUUUUUUUUUUUU I was doing this all along, but I was writing some of my signs wrong... so yeah. Anyway thanks...

It doesn't work, it only leads to a polynomial that can't be factored. If it helps at all Newton's law of cooling is dH/dt = -k(H-Hs), the equation above is derived from it.

Homework Statement Suppose that the temperature of a pan of warm water obeys Newton's law of cooling. The water (47 degrees Celsius) was put in a room and 10 minutes later the water's temperature was 40 degrees Celsius. After another 10 minutes, the temperature of the water was 34 degrees...

I guess I'm going to get points off for notation on my AP calc test then. Well thanks for all the help, really cleared things up.

∫0t (50)dt for 1<t<3

Hmmm... what if t stood for a value like 1<t<3 which you would have to place into the 0 to t limit (this would technically eliminate the variable...right?). Would the integral as a whole still be wrong then?

Say the limit of the integral is from 0 to t and the integral is ended with a dt. Is this okay? Generally, all the integrals I see with a variable limit end with a d-letter that is not the same as the variable in the limit. ie: limit is from 0 to t, ends with du