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Math Help - Explain 4 questions + 1 random question

  1. #1
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    Explain 4 questions + 1 random question

    The first 4 attached questions do not make sense to me. The 'correct' answers are dotted but I'd like to see the proof for it. (They are user-generated problems from the Sparknotes website so I anticipated some errors in them. I can't tell whether or not I'm wrong or the problem had wrong answer choices)

    Status problem 1:Solved - Thanks!
    Status problem 2:Unconfirmed: I checked for m when g(m) = x^2 + 4 and when g(m) = 2x + 4; 3 does not fit.
    Status problem 3: [Same as problem 2]
    Status problem 4:Unsolved




    Random question:
    Also, the forumula \frac {p}{q} is used to solve polynomials correct? Polynomials and or Binomials?
    Attached Thumbnails Attached Thumbnails Explain 4 questions + 1 random question-wrong-1.jpg   Explain 4 questions + 1 random question-untitled.jpg   Explain 4 questions + 1 random question-wrong-2.jpg   Explain 4 questions + 1 random question-wrong-3.jpg  
    Last edited by Masterthief1324; February 10th 2010 at 02:51 PM.
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Masterthief1324 View Post
    The first 4 attached questions do not make sense to me. The 'correct' answers are dotted but I'd like to see the proof for it. (They are user-generated problems from the Sparknotes website so I anticipated some errors in them. I can't tell whether or not I'm wrong or the problem had wrong answer choices)

    Status problem 1:Unsolved
    Status problem 2:Unsolved
    Status problem 3:Unsolved
    Status problem 4:Unsolved




    Random question:
    Also, the forumula \frac {p}{q} is used to solve polynomials correct? Polynomials and or Binomials?
    For 1

    3p(3h)=3\Rightarrow{p}(3h)=1\Rightarrow\frac{3h}{2  }-5=1
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  3. #3
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    Quote Originally Posted by VonNemo19 View Post
    For 1

    3p(3h)=3\Rightarrow{p}(3h)=1\Rightarrow\frac{3h}{2  }-5=1
    It's asking for the value of 'h'.
    My reasoning:
    In order for the second equation to be true, ( 3p(3h)=3), the function p(3h) has to equal to 1.

    The function p(x) = \frac {x}{2-5} or p(x) = \frac {x}{-3}. Thus when x = 3h, the function p(x) becomes:

    p(3h) = \frac {3h}{-3}

    Since p(3h) HAS to equal to 1, 'h' equals -1 because:
    1 = \frac {3(-1) or (-3)}{-3}
    Last edited by Masterthief1324; February 10th 2010 at 01:51 PM.
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  4. #4
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Masterthief1324 View Post
    It's asking for the value of 'h'.
    My reasoning:
    In order for the second equation to be true, ( 3p(3h)=3), the function p(3h) has to equal to 1.

    The function p(x) = \frac {x}{2-5} or p(x) = \frac {x}{-3}. Thus when x = 3h, the function p(x) becomes:

    p(3h) = \frac {3h}{-3}

    Since p(3h) HAS to equal to 1, 'h' equals -1 because:
    1 = \frac {3(-1) or (-3)}{-3}


    So 4 is the wrong answer even though it was given as an answer choice. They asked for h and no answer choices are -1 so all answer choices are wrong! Take that sparknotes!
    No.

    Like I said before...

    3p(3h)=3 means that p(3h)=1.

    Which means that

    \frac{3h}{2}-5=1.

    Solve for h.

    By the way, association is very important. you were given p(x)=\frac{x}{2}-5, not p(x)=\frac{x}{2-5}.

    Oh, and four is CORRECT.
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  5. #5
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    1, 3, 3, 3, 5, 5, 5, 5, 5, 5, 7, 7, 7, 7, 7, 7, 7, 9, 9, 9, 9, 9, 9, 9, 9, 9, 11. Just count!

    Or:
    1 occurs 1 time
    3 occurs 3 times
    5 occurs 5 times
    7 occurs 7 times
    9 occurs 9 times

    so 11 occurs after 1+ 3+ 5+ 7+ 9 other numbers.

    "Let the function g be defined by g(x)= x^2+ 4. If p is a positive number such that g(2m)= g(m)+ 22 which of the following could be a value of m?"

    There's an obvious typo in this problem. They must mean "If m is a positive number...".

    g(2m)= (2m)^2+ 4= 4m^2+ 4. g(2m)= g(m)+ 22 becomes 4m^2+ 4= (m^2+ 4)+ 22. Solve that equation for m.

    Your other problem appears to be just a copy of that.

    As for your random question, "Also, the forumula \frac{p}{q} is used to solve polynomials correct? Polynomials and or Binomials?", I can make no sense out of it. First, \frac{p}{q} is a not a "formula", it is just a fraction or, more generally, an "expression". Second, you don't "solve polynomials", you solve polynomial equations. Most importantly, since you have not said what "p" and "q" are, or how they are connected to the equation, I don't know how you would intend to use \frac{p}{q} to solve the equation.
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  6. #6
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    [quote=Masterthief1324;451914]It's asking for the value of 'h'.
    My reasoning:
    In order for the second equation to be true, ( 3p(3h)=3), the function p(3h) has to equal to 1.

    The function p(x) = \frac {x}{2-5} or p(x) = \frac {x}{-3}. Thus when x = 3h, the function p(x) becomes:

    p(3h) = \frac {3h}{-3}

    Since p(3h) HAS to equal to 1, 'h' equals -1 because:
    1 = \frac {3(-1) or (-3)}{-3}
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  7. #7
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    Quote Originally Posted by HallsofIvy View Post
    1, 3, 3, 3, 5, 5, 5, 5, 5, 5, 7, 7, 7, 7, 7, 7, 7, 9, 9, 9, 9, 9, 9, 9, 9, 9, 11. Just count!

    Or:
    1 occurs 1 time
    3 occurs 3 times
    5 occurs 5 times
    7 occurs 7 times
    9 occurs 9 times

    so 11 occurs after 1+ 3+ 5+ 7+ 9 other numbers.

    "Let the function g be defined by g(x)= x^2+ 4. If p is a positive number such that g(2m)= g(m)+ 22 which of the following could be a value of m?"

    There's an obvious typo in this problem. They must mean "If m is a positive number...".

    g(2m)= (2m)^2+ 4= 4m^2+ 4. g(2m)= g(m)+ 22 becomes 4m^2+ 4= (m^2+ 4)+ 22. Solve that equation for m.

    Your other problem appears to be just a copy of that.

    As for your random question, "Also, the forumula \frac{p}{q} is used to solve polynomials correct? Polynomials and or Binomials?", I can make no sense out of it. First, \frac{p}{q} is a not a "formula", it is just a fraction or, more generally, an "expression". Second, you don't "solve polynomials", you solve polynomial equations. Most importantly, since you have not said what "p" and "q" are, or how they are connected to the equation, I don't know how you would intend to use \frac{p}{q} to solve the equation.
    For the question concerning g(2m), I solved and had a decimal answer:

     <br />
4m^2 + 4 = (m^2 + 4) + 22<br />

    m = 2.7...; Therefore the answer is not 3 correct?
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