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Math Help - algebraic problem

  1. #1
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    algebraic problem

    1 The current price of a pair of basketball shoes is $30. The original price had been reduced by 25%. that reduced price was then lowered by 50% to arrive at the current price. What was the original price?



    In this last word problem it has a sign in it that i have no idea what it means and i also cant type it here for you to see but i'll try to explain it.

    It looks like a (cp) but just no space at all between the c and the p.

    Here is what the word problem is askin g me to do.

    2. If (m cp n) = 1/m + 1/n + 1/m^2 + 1/n^2 for any values of m nd n, what is the value of ( 2 cp 4)? express your answer as a common fraction.
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  2. #2
    A riddle wrapped in an enigma
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    Quote Originally Posted by xsriel View Post
    1 The current price of a pair of basketball shoes is $30. The original price had been reduced by 25%. that reduced price was then lowered by 50% to arrive at the current price. What was the original price?
    Hello xsriel,

    Find the original price (op)

    {\color{red}op}-\underbrace{.25{\color{red}op}}_{\text{1st discount}}-\underbrace{.5({\color{red}op}-.25{\color{red}op})}_{\text{2nd discount}}=30

    {\color{red}op}-.25{\color{red}op}-.5{\color{red}op}+.125{\color{red}op}=30

    .375{\color{red}op}=30

    {\color{red}op}=80


    Quote Originally Posted by xsriel View Post
    In this last word problem it has a sign in it that i have no idea what it means and i also cant type it here for you to see but i'll try to explain it.

    It looks like a (cp) but just no space at all between the c and the p.

    Here is what the word problem is askin g me to do.

    2. If (m cp n) = 1/m + 1/n + 1/m^2 + 1/n^2 for any values of m nd n, what is the value of ( 2 cp 4)? express your answer as a common fraction.
    That cp thingy is just a symbol they created to define a specific operation. I'll use \spadesuit instead.


    If (m\; \spadesuit \;n)=\frac{1}{m}+\frac{1}{n}+\frac{1}{m^2}+\frac{1  }{n^2}, then

     (2\; \spadesuit \;4)=\frac{1}{2}+\frac{1}{4}+\frac{1}{2^2}+\frac{1  }{4^2}

     (2\; \spadesuit \;4)=\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{  16}=\boxed{\frac{17}{16}}
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