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Math Help - 2 math problems 9th grade~~

  1. #1
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    2 math problems 9th grade~~

    1. suppose the population of a species decreases at a rate of 3.5 % each year. In the place you are studying, there are 80 of these animals.

    a) predict the number of animals that will remain after ten years

    b)at this rate, after how many years will the population first drop below 40 animals?

    2. Area=36in squared Perimeter= 36 in

    Write equation using one variable and find dimesions of the rectangle.


    I need help on how to solve this.

    Please show work how you did solve the two problems!

    Thank You Very Much!!!
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  2. #2
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    Smile

    Hint for (a) :

     <br />
\frac{3.5}{100}\times80=2.8<br />
for one year.
    for 10 years 2.8\times 10 = 28 (animals)
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  3. #3
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    Hello, tecktonikk!

    They're teaching exponentials and logarithms in 9th grade??


    1. Suppose the population of a species decreases at a rate of 3.5 % each year.
    In the place you are studying, there are 80 of these animals.

    a) Predict the number of animals that will remain after ten years
    Every year, the population is only 96.5% of the previous year's population.

    The formula we want is: . P(n) \:=\:800(0.965)^n . where n is the number of years.


    In ten years: . P(10) \:=\:800(0.965)^{10} \:=\:56.022... \:\approx\:56\text{ animals.}



    b) At this rate, after how many years will the population first drop below 40 animals?
    When will P(n) be less than 40?


    We have: . 80(0.965)^n \:<\:40 \quad\Rightarrow\quad (0.965)^n \:<\:0.5

    Take logs: . \log(0.965)^n \:<\:\log(0.5) \quad\Rightarrow\quad n\log(0.965) \:<\:0.5


    Divide by \log(0.965) . . . a negative quantity.

    . . n \;\;{\color{red}>} \:\;\frac{\log(0.5)}{\log(0.965)} \:=\:19.455...


    The population will drop before 40 in about 19\tfrac{1}{2} years.

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  4. #4
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    Quote Originally Posted by Soroban View Post
    Hello, tecktonikk!

    They're teaching exponentials and logarithms in 9th grade??

    Every year, the population is only 96.5% of the previous year's population.

    The formula we want is: . P(n) \:=\:800(0.965)^n . where n is the number of years.


    In ten years: . P(10) \:=\:800(0.965)^{10} \:=\:56.022... \:\approx\:56\text{ animals.}

    that makes my previous post wrong,does it ?
    (i thought it could be done by simple logic)
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  5. #5
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    Wait... but isn't it supposed to be

    y=80(1.035)^x??
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  6. #6
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    Quote Originally Posted by tecktonikk View Post
    Wait... but isn't it supposed to be

    y=80(1.035)^x??
    that would be for an increase in the animal population by 3.5%, not a decrease.
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  7. #7
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    Quote Originally Posted by tecktonikk View Post
    2. Area=36in squared Perimeter= 36 in

    Write equation using one variable and find dimesions of the rectangle.
    Perimeter of a rectangle = 2L + 2W = 36, so L + W = 18. If we let x equal to L, then W = 18 - L. Area of a rectangle = LW = 36, so x(18- x) = 36. Your equation becomes
    x^2 - 18x + 36 = 0
    so solve for x.


    01
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