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13 Feb 2008, 12:07
Kudos
Bookmarks
Any tips on how to solve this one? Thanks!
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13 Feb 2008, 12:25
Kudos
Bookmarks
Fortunately this one is much simpler than it looks. Let's set up the initial equation:
subtract 60 from each side
simplify
plug in our 10 min for t (time)
divide each side by 120
a must be equal to
Now that we have our constant let's plug in our new information
and VOILA. There's your answer of 75.
It looks like a mess if you do it step by step, but if you know how the exponents work you can do this fairly easily without many calculations
subtract 60 from each side
simplify
plug in our 10 min for t (time)
divide each side by 120
a must be equal to
Now that we have our constant let's plug in our new information
and VOILA. There's your answer of 75.
It looks like a mess if you do it step by step, but if you know how the exponents work you can do this fairly easily without many calculations
13 Feb 2008, 12:31
Kudos
Bookmarks
Use the information from the first part of the problem to plug into their equation.
You should get this:
120 = 120*[2^(-10a)] + 60
120 - 120*[2^(-10a)] = 60
120*{1-[2^(-10a)]} = 60
1-[2^(-10a)] = 1/2
-2^(-10a) = -1/2
-2^(-10a) = -2^(-1)
-10a = -1
a = 1/10
Now that you have the constant 'a', you can use it to solve with the information in the second part.
You should get this:
Temp in Farenheit (Answer) = 120*[2^(-30*1/10)] + 60
120 *[2^(-3)] + 60
120 * (1/8) + 60
15 + 60
75
You should get this:
120 = 120*[2^(-10a)] + 60
120 - 120*[2^(-10a)] = 60
120*{1-[2^(-10a)]} = 60
1-[2^(-10a)] = 1/2
-2^(-10a) = -1/2
-2^(-10a) = -2^(-1)
-10a = -1
a = 1/10
Now that you have the constant 'a', you can use it to solve with the information in the second part.
You should get this:
Temp in Farenheit (Answer) = 120*[2^(-30*1/10)] + 60
120 *[2^(-3)] + 60
120 * (1/8) + 60
15 + 60
75
