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The temperature of a certain cup of coffee 10 minutes after it was pou

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The temperature of a certain cup of coffee 10 minutes after it was pou  [#permalink]

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New post 28 Jul 2010, 09:11
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The temperature of a certain cup of coffee 10 minutes after it was poured was 120 degrees Fahrenheit. If the temperature F of the coffee t minutes after it was poured can be determined by the formula F = 120*2^(-at) + 60, where F is in degrees Fahrenheit and a is a constant. Then the temperature of the coffee 30 minutes after it was poured was how many degrees Fahrenheit?

A. 65
B. 75
C. 80
D. 85
E. 90
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Re: The temperature of a certain cup of coffee 10 minutes after it was pou  [#permalink]

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New post 29 Jul 2010, 04:03
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kilukilam wrote:
The tempreature of a certain cup of coffee 10 minutes after it was poured was 120 degree Faranite. If the temp F of the coffee t minues after it was poured can be determinde by the formula F=120 (2^-at)+60 , where F is degree Fahrenhite and a is constant, then the tempreature of the cofffee 30 minues after it was poured was how many degree fahrenhite

a. 65
b. 75
c. 80
d. 85
e. 90


We have the formula for calculating temperature: \(F=120*2^{-at}+60\) and the value for \(t=10\) --> \(F(t=10)=120=120*2^{-10a}+60\) --> \(\frac{1}{2^{10a}}=\frac{1}{2}\) --> \(10a=1\) --> \(a=0.1\). So formula is \(F=120*2^{-0.1t}+60\)

Now for \(t=30\) the temperature in Fahrenheits will be \(F=120*2^{-0.1*30}+60=\frac{120}{2^3}+60=75\).

Answer: B.
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Re: The temperature of a certain cup of coffee 10 minutes after it was pou  [#permalink]

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New post 12 Mar 2014, 05:11
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First, we have to find a.

We know that after t=10 minutes the temperature F=120 degrees.

Hence:
120 = 120*(2^-10a)+60
60 = 120*(2^-10a)
60/120 = 2^-10a
1/2 = 2^-10a
2^-1 = 2^-10a
-1 = -10a
1/10 = a

Now we need to find F after t=30 minutes:
F = 120*(2^-1/10*30)+60
F = 120* (2^-3) +60
F = 120* (1/2^3) +60
F = 120*1/8 +60
F = 15+60 = 75

Answer B!
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Re: The temperature of a certain cup of coffee 10 minutes after it was pou  [#permalink]

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New post 28 Jul 2010, 11:57
Hi,

Since F=120 (2^-at)+60,
replace F=120, t=10 to find a.
Once you find a, then to find the temperature 30 min after it was poured simply replace
t=30 and value of a in the equation.

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Re: The temperature of a certain cup of coffee 10 minutes after it was pou  [#permalink]

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New post 28 Jul 2010, 13:26
1
Is the question correctly written?? I am unable to find the solution.

\(F= 120* (\frac{1}{2^at}) + 60\)
Is this formula correct?
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Re: The temperature of a certain cup of coffee 10 minutes after it was pou  [#permalink]

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New post 17 Apr 2014, 01:36
Substitute the values f=120 and t=10 in given equation, this gives value of a=0.1

now equation is f=120(2^-0.1t)+60

substitute t=30 in above equation to find f

Answer B
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Re: The temperature of a certain cup of coffee 10 minutes after it was pou  [#permalink]

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New post 27 Mar 2015, 06:34
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Can someone explain me how
1/2 = 2^-10a becomes

2^-1 = 2^-10a
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Re: The temperature of a certain cup of coffee 10 minutes after it was pou  [#permalink]

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New post 27 Mar 2015, 06:59
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lzielen wrote:
Can someone explain me how
1/2 = 2^-10a becomes

2^-1 = 2^-10a


Hi lzielen!

\(\frac{1}{2}\) is \(2^{-1}\).

\(x^{-n} = 1/x^n\)
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Re: The temperature of a certain cup of coffee 10 minutes after it was pou  [#permalink]

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New post 22 Mar 2018, 09:27
Hussain15 wrote:
Is the question correctly written?? I am unable to find the solution.

\(F= 120* (\frac{1}{2^at}) + 60\)
Is this formula correct?


a*t is the power of 2 it is 2^(-at)
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Re: The temperature of a certain cup of coffee 10 minutes after it was pou  [#permalink]

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New post 13 Sep 2018, 05:47
Bunuel wrote:
kilukilam wrote:
The tempreature of a certain cup of coffee 10 minutes after it was poured was 120 degree Faranite. If the temp F of the coffee t minues after it was poured can be determinde by the formula F=120 (2^-at)+60 , where F is degree Fahrenhite and a is constant, then the tempreature of the cofffee 30 minues after it was poured was how many degree fahrenhite

a. 65
b. 75
c. 80
d. 85
e. 90


We have the formula for calculating temperature: \(F=120*2^{-at}+60\) and the value for \(t=10\) --> \(F(t=10)=120=120*2^{-10a}+60\) --> \(\frac{1}{2^{10a}}=\frac{1}{2}\) --> \(10a=1\) --> \(a=0.1\). So formula is \(F=120*2^{-0.1t}+60\)

Now for \(t=30\) the temperature in Fahrenheits will be \(F=120*2^{-0.1*30}+60=\frac{120}{2^3}+60=75\).

Answer: B.



You're solution is quite clear to me, I only have one question:


How do you conduct this step: \(\frac{1}{2^{10a}}=\frac{1}{2}\) --> \(10a=1\)

I would solve it like this:

\(\frac{1}{2^{10a}}=\frac{1}{2}\) --> 2^-1 = 2^-10a --> -1 lg(2) = -10a lg(2) --> 1 = 10a

Is there a easier possibility? Is it even right to solve the question by the use of logarithm?

Thanks in advance!
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Re: The temperature of a certain cup of coffee 10 minutes after it was pou  [#permalink]

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New post 13 Sep 2018, 05:50
DominikSterk wrote:
Bunuel wrote:
kilukilam wrote:
The tempreature of a certain cup of coffee 10 minutes after it was poured was 120 degree Faranite. If the temp F of the coffee t minues after it was poured can be determinde by the formula F=120 (2^-at)+60 , where F is degree Fahrenhite and a is constant, then the tempreature of the cofffee 30 minues after it was poured was how many degree fahrenhite

a. 65
b. 75
c. 80
d. 85
e. 90


We have the formula for calculating temperature: \(F=120*2^{-at}+60\) and the value for \(t=10\) --> \(F(t=10)=120=120*2^{-10a}+60\) --> \(\frac{1}{2^{10a}}=\frac{1}{2}\) --> \(10a=1\) --> \(a=0.1\). So formula is \(F=120*2^{-0.1t}+60\)

Now for \(t=30\) the temperature in Fahrenheits will be \(F=120*2^{-0.1*30}+60=\frac{120}{2^3}+60=75\).

Answer: B.



You're solution is quite clear to me, I only have one question:


How do you conduct this step: \(\frac{1}{2^{10a}}=\frac{1}{2}\) --> \(10a=1\)

I would solve it like this:

\(\frac{1}{2^{10a}}=\frac{1}{2}\) --> 2^-1 = 2^-10a --> -1 lg(2) = -10a lg(2) --> 1 = 10a

Is there a easier possibility? Is it even right to solve the question by the use of logarithm?

Thanks in advance!



\(\frac{1}{2^{10a}}=\frac{1}{2}\);

Cross-multiply: \(2=2^{10a}\)

Equate the powers: \(10a=1\)
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Re: The temperature of a certain cup of coffee 10 minutes after it was pou  [#permalink]

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New post 11 Dec 2018, 09:26
Alright, so I now understand how ...
1/2 = 2^-10a
becomes
2^-1 = 2^-10a

but how does it then turn into...
-1 = -10a

to me it should be
1=-10a

any help here?
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Re: The temperature of a certain cup of coffee 10 minutes after it was pou  [#permalink]

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New post 13 Feb 2019, 06:28
took me over 5mins to solve, is this high 600 level question ?
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Re: The temperature of a certain cup of coffee 10 minutes after it was pou  [#permalink]

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New post 13 Feb 2019, 06:48
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Re: The temperature of a certain cup of coffee 10 minutes after it was pou  [#permalink]

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New post 20 May 2019, 19:13
kilukilam wrote:
The temperature of a certain cup of coffee 10 minutes after it was poured was 120 degrees Fahrenheit. If the temperature F of the coffee t minutes after it was poured can be determined by the formula F = 120*2^(-at) + 60, where F is in degrees Fahrenheit and a is a constant. Then the temperature of the coffee 30 minutes after it was poured was how many degrees Fahrenheit?

A. 65
B. 75
C. 80
D. 85
E. 90


First we need to find the value of a using F = 120 and t = 10:

120 = 120 * 2^(-10a) + 60

60 = 120 * 2^(-10a)

1/2 = 2^(-10a)

2^(-1) = 2^(-10a)

With a common base, we can equate the exponents:

-1 = -10a

a = 1/10

Now we can find F using a = 1/10 and t = 30:

F = 120 * 2^(-(1/10)(30)) + 60

F = 120 * 2^(-3) + 60

F = 120 * 1/8 + 60

F = 15 + 60

F = 75

Answer: 2/B
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Re: The temperature of a certain cup of coffee 10 minutes after it was pou  [#permalink]

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New post 05 Jun 2019, 16:23
Can you elaborate this part more? I still have a hard time understanding how you find A.

How do you find it to be .1 from 2^(1/10a)?



1/2^{10a}[/fraction]=1/2[/m];

Cross-multiply: \(2=2^{10a}\)

Equate the powers: \(10a=1\)
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Re: The temperature of a certain cup of coffee 10 minutes after it was pou   [#permalink] 05 Jun 2019, 16:23
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