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Math Help - Calculus Question

  1. #1
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    Calculus Question

    I do not know how to find part b of this problem can anyone help me.

    When a cold drink is taken from a refrigerator, its temperature is 5C. After 25 minutes in a 20C room its temperature has increased to 10C. (Round the answers to one decimal place.)
    (a) What is the temperature of the drink after 35 minutes?
    C

    (b) When will its temperature be 16C?
    minutes
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  2. #2
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    Quote Originally Posted by skeltonjoe View Post
    I do not know how to find part b of this problem can anyone help me.

    When a cold drink is taken from a refrigerator, its temperature is 5C. After 25 minutes in a 20C room its temperature has increased to 10C. (Round the answers to one decimal place.)
    (a) What is the temperature of the drink after 35 minutes?
    C

    (b) When will its temperature be 16C?
    minutes
    I assume the green check means you got part (a) correct ... how did you figure that value?
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  3. #3
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    i did the problem out, then for part b I know your suppose to take like 16=20 - 16 * 2/3 raised to the t/25 and solve it for t which I get 85.5 but apparently its wrong.
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  4. #4
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    Quote Originally Posted by skeltonjoe View Post
    i did the problem out, then for part b I know your suppose to take like 16=20 - 16 * 2/3 raised to the t/25 and solve it for t which I get 85.5 but apparently its wrong.
    perhaps I wasn't clear in my request ...

    what equation did you use to solve part (a) , and how did you go about getting that equation?
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  5. #5
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    y(t) = -15 * 2/3 ^ t/25

    I keep getting 85.5 I just redid it 3 times
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  6. #6
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    Quote Originally Posted by skeltonjoe View Post
    y(t) = -15 * 2/3 ^ t/25

    I keep getting 85.5 I just redid it 3 times
    your equation is incorrect. if y is the temperature, then ...

    y(0) should = 5

    y(25) should = 10

    your equation does not yield those values for the given times.

    you should be deriving the temperature function from Newton's law of cooling/heating ...

    \frac{dy}{dt} = k(20 - y) , using the initial conditions y(0) = 5 and y(25) = 10
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    Quote Originally Posted by skeltonjoe View Post
    y(t) = -15 * 2/3 ^ t/25

    I keep getting 85.5 I just redid it 3 times
    Did you mean y(t)=20-15(\frac{2}{3})^{\frac{t}{25}} ?
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  8. #8
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    yes thats what I meant, can anyone please answer this question its due in 2 hours
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  9. #9
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    i got it the answer was 81.5
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