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Math Help - Problem

  1. #1
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    athens
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    Problem

    how can i solve this exercise? ( The value of an investment varies according to the formula V=A^t/12,where V is the value of the investment in pounds, A is the constant to be found and t is the time in years after the investment was made. i)If the investment was worth 8000 pounds after three years,how much the value of A to the nearest pound is? ii)How much the value of the investment after 10 years will be?








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  2. #2
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    Re: Problem

    From
    V=A^{\frac{t}{12}}

    Raise both sides to the power of \frac{12}{t}

    V^{\frac{12}{t}}=A^{{\frac{t}{12}}^{\frac{12}{t}}}

    Remember the rules of indices. the indices {\frac{t}{12}} and {\frac{12}{t}} cancel to be just 1

    V^{\frac{12}{t}}=A^1

    You can now find A.
    After you find the constant A you can do the second part easily by using the original equation at t=10
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  3. #3
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    athens
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    Re: Problem

    shakari,,tell me what will be the answers with the data what i have written above..
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  4. #4
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    Re: Problem

    You'll learn more if you do it yourself
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  5. #5
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    athens
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    Re: Problem

    well if v=8000,then A=8000^4,in the first question, this is impossible..so plz tell me the answers
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