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Math Help - Help with some problems on Schaum Outline

  1. #1
    Super Member 11rdc11's Avatar
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    Help with some problems on Schaum Outline

    Ok so I'm reviewing algebra since it was the only math class I didn't do well in and thought I would work out all the problems in the Schaum's Outline. I have ran across 3 problems I'm not sure how to do.

    a)Two numbers have the ratio 3:4. If 4 is added to each of the numbers the resluting ratio is 4:5. Find the two numbers.



    b) Determine whether the graph of each relation is symmetric with respect to the y axis, x axis, or origin.

    y = (x-3)^3

    I want to work this one out rather than look at the graph.

    y = (x-3)^3

    -y= (-x-3)^3

    (-x-3)^3 = (-x)^3 -3(-x)^2(-3) +3(-x)(-3)^2 -(-3)^3

    (-x-3)^3 = -x^3 + 9x^2 -27x +27

    -y= -x^3 + 9x^2 -27x +27

    y = x^3 - 9x^2 + 27x -27

    So it is symmetric with respect to the origin but is there a faster way of showing that. I tried going this route but got stuck

    -y = (-x-3)^3

    -y = (-1)^3(x+3)^3

    and then got stuck.

    c)Find the largest rectangle which can be inscribed in a right triangle whose legs are 6 and 8 inches repectively.

    I know A= xy

    and
     x^2 + y^2 = 100

    Thanks in advance
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  2. #2
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    Quote Originally Posted by 11rdc11 View Post
    a)Two numbers have the ratio 3:4. If 4 is added to each of the numbers the resluting ratio is 4:5. Find the two numbers.
    I would set this problem out as follows

    a:b = 3:4

    a+4:b+4 = 4:5
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  3. #3
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by pickslides View Post
    I would set this problem out as follows

    a:b = 3:4

    a+4:b+4 = 4:5
    I tried that and didn't get the correct answer. Let me try again and see if I made an error.
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  4. #4
    Super Member 11rdc11's Avatar
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    Lol yep made a careless error. Thanks pickslides
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    I'm not sure what the answer book says but I arrived at a=12 and b = 16
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    Quote Originally Posted by 11rdc11 View Post

    c)Find the largest rectangle which can be inscribed in a right triangle whose legs are 6 and 8 inches repectively.

    I know A= xy

    and
     x^2 + y^2 = 100

    Thanks in advance
    Largest rectangle implies find a maximum i.e \frac{dA}{dx} = 0

    where A= xy and y = \sqrt{100-x^2}

    gives A= x\times \sqrt{100-x^2} now find \frac{dA}{dx} = 0 and solve for x
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  7. #7
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by pickslides View Post
    Largest rectangle implies find a maximum i.e \frac{dA}{dx} = 0

    where A= xy and y = \sqrt{100-x^2}

    gives A= x\times \sqrt{100-x^2} now find \frac{dA}{dx} = 0 and solve for x
    Ok I did exactly the same thing you just showed but I and ended up with 5\sqrt{2}
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  8. #8
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    Quote Originally Posted by 11rdc11 View Post
    b) Determine whether the graph of each relation is symmetric with respect to the y axis, x axis, or origin.

    y = (x-3)^3
    The relation you posted is not symmetric to the origin.

    y = (x-3)^3

    -y= (-x-3)^3

    (-x-3)^3 = (-x)^3 -3(-x)^2(-3) +3(-x)(-3)^2 -(-3)^3????
    This is where you went wrong. For a lack of a better explanation, you "counted" the negatives twice. Look at the binomial cubic expansion in the general case:
    (a + b)^3 = a^3 + 3a^2 b + 3ab^2 + b^3
    (a \;{\color{red}-}\;b)^3 = a^3 \;{\color{red}-}\; 3a^2 b + 3ab^2 \;{\color{red}-}\; b^3

    If you're saying that a = -x and b = -3, then you follow the first rule:
    (-x-3)^3 = (-x+({\color{red}-3}))^3 = (-x)^3 \;{\color{red}+}\;3(-x)^2(-3) +3(-x)(-3)^2 {\color{red}+}\;(-3)^3

    But, if you're saying that a = -x and b = 3, then you follow the second rule:
    (-x-3)^3 = (-x)^3 -3(-x)^2({\color{red}3}) +3(-x)({\color{red}3})^2 -({\color{red}3})^3

    You were combining both rules.

    In either case,
    (-x-3)^3 = -x^3 - 9x^2 - 27x - 27.
    And since
    -y = -x^3 + 9x^2 - 27x + 27,
    the two polynomials are not equal, so this relation is not symmetric at the origin.

    In fact, this relation is not symmetric to the x-axis, y-axis, or the origin. You'll now have to show algebraically that the first two are not the case.


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    Last edited by yeongil; August 3rd 2009 at 10:49 PM.
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  9. #9
    Super Member 11rdc11's Avatar
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    Lol yes another careless error. I kept thinking this couldn't be symmetric with the origin but the back of the book said it was. Once again thanks for pointing out that I was using a negative 3 instead of a positive 3.
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  10. #10
    No one in Particular VonNemo19's Avatar
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    For the first problem.

    \frac{x}{y}=\frac{3}{4}

    \frac{x+4}{y+4}=\frac{4}{5}

    Solving for y in the first gives

    \frac{4}{3}x=y

    Substituting into the second

    \frac{x+4}{\frac{4}{3}x+4}=\frac{4}{5}

    x+4=\frac{16}{15}x+\frac{16}{5}

    \frac{1}{15}x=\frac{4}{5}

    x=12

    You can finish up.
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