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Math Help - questionable questions

  1. #1
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    questionable questions

    Greetings and salutations!

    I have some questions that I can't figure out.

    1
    the perimeter of a rectangular yard is 290 feet. If its length is 25 feet greater than its width, what are the dimensions of the yard?

    2
    What is the equation in standard form of the line through (4,1) and (-2,3)?

    3
    Write in simplest form without negative exponents.
    (2/3)^-2(4/5)^-1
    [I think i have an answer to this one:
    (2/3)^-2(4/5)^-1=(2^-2/3^-2)(5/4)=(2^-2*{1/3}^-2)(5/4)=({1/2}9)(5/4)=(9/2)(5/4)=48/8=6]

    Thanks
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  2. #2
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    Quote Originally Posted by NitroKnight View Post
    Greetings and salutations!

    I have some questions that I can't figure out.

    1
    the perimeter of a rectangular yard is 290 feet. If its length is 25 feet greater than its width, what are the dimensions of the yard?
    Set one side to be x and the other becomes x+25..
    2
    What is the equation in standard form of the line through (4,1) and (-2,3)?
    You want to find the inclination of the line. That is given by: \frac{\Delta y}{\Delta x}.
    \Delta y = 1-3
    \Delta x = 4-(-2) = 6
    now we have y = \frac{-2}{6}\times x = \frac{-1}{3}x
    Now you want to find the constant...ill leave that part to you
    3
    Write in simplest form without negative exponents.
    (2/3)^-2(4/5)^-1
    [I think i have an answer to this one:
    (2/3)^-2(4/5)^-1=(2^-2/3^-2)(5/4)=(2^-2*{1/3}^-2)(5/4)=({1/2}9)(5/4)=(9/2)(5/4)=48/8=6]
    Im not sure how this is menth to be cause I believe there are som bracetts missing.

    Thanks
    Hope it helps
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  3. #3
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    how do you make the math look like math and not gibberish like I did?
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  4. #4
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    Hold the

    Double click on the math text I have written and a window pops up showing how I have written.
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  5. #5
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    (2/3)^{-2}(4/5)^{-1}\rightarrow(2^{-2}/3^{-2})(5/4)\rightarrow(2^{-2}*{1/3}^{-2})(5/4)\rightarrow((1/2)9)(5/4) \rightarrow(9/2)(5/4)=48/8=6

    This is about how it looked in my head and on my working paper, am I correct in my working?
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  6. #6
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    Quote Originally Posted by NitroKnight View Post
    (2/3)^{-2}(4/5)^{-1}\rightarrow(2^{-2}/3^{-2})(5/4)\rightarrow(2^{-2}*{1/3}^{-2})(5/4)\rightarrow((1/2)9)(5/4) \rightarrow(9/2)(5/4)=48/8=6

    This is about how it looked in my head and on my working paper, am I correct in my working?
    (2/3)^2= 4/9 so (2/3)^{-2}= 9/4. (5/4)^{-1}= 4/5 so (5/2)^{-2}(5/4)^{-1}= (9/4)(4/5)= 9/5.
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  7. #7
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    Quote Originally Posted by NitroKnight View Post
    (2/3)^{-2}(4/5)^{-1}\rightarrow(2^{-2}/3^{-2})(5/4)\rightarrow(2^{-2}*{1/3}^{-2})(5/4)\rightarrow((1/2)9)(5/4) \rightarrow(9/2)(5/4)=48/8=6

    This is about how it looked in my head and on my working paper, am I correct in my working?
    That's still quite ambigious

    \left(\frac{2}{3}\right)^{-2} - 2\left(\frac{4}{5}\right)^{-1}

    With fractions raised to the negative power you can flip the fraction and change the sign of the exponent: \left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^n


    \left(\frac{2}{3}\right)^{-2} - 2\left(\frac{4}{5}\right)^{-1} = \left(\frac{3}{2}\right)^2 - 2\left(\frac{5}{4}\right)


    When raising fractions to powers you need to apply the exponent to both the numerator and the denominator: \left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}

    \frac{9}{4} - \frac{10}{4} = -\frac{1}{4}
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  8. #8
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    Quote Originally Posted by HallsofIvy View Post
    (2/3)^2= 4/9 so (2/3)^{-2}= 9/4. (5/4)^{-1}= 4/5 so (5/2)^{-2}(5/4)^{-1}= (9/4)(4/5)= 9/5.
    Quote Originally Posted by e^(i*pi) View Post
    That's still quite ambigious

    \left(\frac{2}{3}\right)^{-2} - 2\left(\frac{4}{5}\right)^{-1}

    With fractions raised to the negative power you can flip the fraction and change the sign of the exponent: \left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^n


    \left(\frac{2}{3}\right)^{-2} - 2\left(\frac{4}{5}\right)^{-1} = \left(\frac{3}{2}\right)^2 - 2\left(\frac{5}{4}\right)


    When raising fractions to powers you need to apply the exponent to both the numerator and the denominator: \left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}

    \frac{9}{4} - \frac{10}{4} = -\frac{1}{4}
    Now I'm confused.....
    Let me try to rewrite the original problem with my new found skills:

    Write in the simplest form without negative exponents:
    (\frac{2}{3})^{-2}(\frac{4}{5})^{-1}

    My working
    (\frac{2}{3})^{-2}(\frac{4}{5})^{-1}\rightarrow(\frac{2^{-2}}{3^{-2}})(\frac{5}{4})\rightarrow(2^{-2})(\frac{1}{3^2})(\frac{5}{4})\rightarrow (\frac{1}{2}*\frac{1}{9})(\frac{5}{4}) \rightarrow (\frac{1}{18})(\frac{5}{4})\rightarrow\frac{5}{72} ?
    I really think I went wrong here......What do I do?
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  9. #9
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    Quote Originally Posted by NitroKnight View Post
    Now I'm confused.....
    Let me try to rewrite the original problem with my new found skills:

    Write in the simplest form without negative exponents:
    (\frac{2}{3})^{-2}(\frac{4}{5})^{-1}

    My working
    (\frac{2}{3})^{-2}(\frac{4}{5})^{-1}\rightarrow(\frac{2^{-2}}{3^{-2}})(\frac{5}{4})\rightarrow(2^{-2})(\frac{1}{3^2})(\frac{5}{4})\rightarrow (\frac{1}{2}*\frac{1}{9})(\frac{5}{4}) \rightarrow (\frac{1}{18})(\frac{5}{4})\rightarrow\frac{5}{72} ?
    I really think I went wrong here......What do I do?
    You'd said that \frac{1}{3^{-2}} = \frac{1}{9}. That expression is equal to 9: \frac{1}{\frac{1}{9}} = 9

    You can flip the expression inside the brackets for any power as long as you change the sign:

    \left(\frac{2}{3}\right)^{-2} = \left(\frac{3}{2}\right)^2 = \frac{3^2}{2^2} = \frac{9}{4}
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  10. #10
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    Ok....my bad, so my answer should be:
    (\frac{2}{3})^{-2}(\frac{4}{5})^{-1}\rightarrow(\frac{3}{2})^2(\frac{5}{4})\rightarr  ow(\frac{9}{4})(\frac{5}{4})\rightarrow \frac{9*5}{4*4}\rightarrow\frac{45}{16} ?
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