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anonymous

  • one year ago

The function f(x) = 396(3)x represents the growth of a dragonfly population every year in a remote swamp. Pierre wants to manipulate the formula to an equivalent form that calculates every month, not every year. Which function is correct for Pierre's purposes?

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  1. anonymous
    • one year ago
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    A. f(x)= 33(3)^x b. F(x)=396(3^12)x/12 C. f(x)=396(3^1/12)^12x D. f(x)=4752(3)^x

  2. anonymous
    • one year ago
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    So, I got C?

  3. IrishBoy123
    • one year ago
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    **absent typos**: \(\large 396(3^{\frac{1}{12}})^{12x} = 396(3^{\frac{1}{12} \times 12x}) = 396(3)^{x}\) ie back to where you started. plug a number in and see. none of these look right https://gyazo.com/5b74d0c9b3753f89f9467f1bc3f2858c

  4. IrishBoy123
    • one year ago
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    screenshot?

  5. anonymous
    • one year ago
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    @campbell_st

  6. anonymous
    • one year ago
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    f(x) = 33(3)x f(x) = 396(312)the x over 12 power f(x) = 396(3 to the one twelfth power)12x f(x) = 4752(3)x

  7. campbell_st
    • one year ago
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    this is the 2nd time I have seen one of these questions and they don't make much sense to me... I'd expect for this exponential model if the time changes to months the value of x needs to be divided by 12 so you can test by looking at the model \[f(1) = 396(3)^{\frac{1}{12}}\] then find f(2) and f(3) to see if there is growth hope it helps

  8. anonymous
    • one year ago
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    That"s not an answer choice:( @campbell_st

  9. campbell_st
    • one year ago
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    well remember the index law for power of a power, multiply the powers \[(x^a)^b = x^{a \times b}\] so you have \[3^{\frac{x}{12}} = (3^?)^x\]

  10. anonymous
    • one year ago
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    12?

  11. anonymous
    • one year ago
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    So would I be right when I said, C?

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