00:01
In this problem, we're going to use newton's lower cooling.
00:02
And as you can see, i have our function listed below.
00:05
And what is going to do is tell us the temperature, which is capital t, of the soup after t, lowercase t, number of minutes.
00:13
So let's go to part a.
00:15
Part a says, what is the initial temperature of the soup? well, that would mean our time would equal to zero minutes.
00:21
So we're going to substitute lowercase t in place with zero.
00:25
So that means we're trying to find t of zero.
00:27
So this would equal to 62 plus 100.
00:30
149 e to the negative 0 .05 times zero power.
00:35
Well, anything times zero is equal to zero.
00:38
And e raised to the zero power is simply just one.
00:41
So really, we're just left with 62 plus 149.
00:45
Well, 62 plus 149 is equal to 211.
00:49
So initially, our soup would be 211 degrees fahrenheit.
00:54
All right.
00:55
So now let's take a look at part b.
00:57
In part b, they want us to find the temperature after 10 minutes.
01:00
Well, that would mean that t, or lowercase t, would equal to 10.
01:04
So we'll just substitute 10 in place of t.
01:07
So we'd have t of 10 equal to 62 plus 149e raised to the negative 0 .05 times 10 power.
01:17
Well, negative 0 .05 times 10 is equal to negative 0 .5.
01:22
So i have 62 plus 149 e to the negative 0 .5 power.
01:27
And from here, you can go right to your calculator.
01:29
So to type in 62 plus 149, the constant e raised in a negative 0 .5 power.
01:37
And it looks like your directions want you to round to one decimal place...