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Hi goosey. First you simplify 28/68 which is about .41176. Then you must take that to the (1/14)th power, not to the 14th power. That's because opposites cancel each other out in a way.The problem is 28/68=b^14 and I need to solve for b. I get .41176^14 which is not the right answer(which is .93859) What am I doing wrong?
f(t) represents a temperature after time "t", but we don't know what temperature we're looking for. We need to have a temperature before we can find at what time it reaches that temperature.got it now. Ok, so the last part of this question my professor said do not find the solution as the book says by graphing. The final part of the problem is take f(t)=22+68(.93859)^t. He said to do it the way we learned through exponential and the answer is 21. Again, I must be off since I am not getting it right. The whole problem is taking newtons law to find different temps. at different room temperatures.
Hi goosey00,93 to 60 after 13 min in a room of 25 degrees. how long before it gets to 40 degrees. Although this is a different problem I am stuck at the same point. so far I have .49295 (1/13) which I got .94704. Now what is it do I do now w/out graphing to get it?
And how did you obtain, \(.49295^\frac{1}{13}\) ?93 to 60 after 13 min in a room of 25 degrees.