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Math Help - A mixed bag of problems to solve

  1. #1
    Member MathBlaster47's Avatar
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    A mixed bag of problems to solve

    I have a fair variety of problems that I am having some issues with--most I have worked out but need confirmation on, others I have no real idea what to do with.

    The ones I need to double check are:

    (1)
    Assuming that none of the solutions are undefined,simplify each of the following:

    (a) \frac{a^2 b^2 c}{abc}

    My answer: Either abc or ab(I'm not too sure which, though I am leaning towards "ab")

    (b) \frac{a^{2m} b^{m+1}}{a^m b}

    My answer: a^{2m-m}b^m or a^mb^m
    __________________________________________________ ______________
    (2)
    Simplify:
    (a) \frac{x-y}{x+y} + \frac{x+y}{x-y}

    My answer: \frac{x-y}{x+y} + \frac{x+y}{x-y} \rightarrow \frac{(x-y)(x-y)}{(x+y)(x-y)} + \frac{(x+y)(x+y)}{(x+y)(x-y)} \rightarrow \frac{(x-y)^2}{(x+y)(x-y)} + \frac{(x+y)^2}{(x+y)(x-y)} \rightarrow \frac{(x-y)^2+(x+y)^2}{(x+y)(x-y)} \rightarrow (x-y)+(x-y)
    __________________________________________________ ______________

    (3)
    If you are h feet above the earth's surface, the distance, d, that you can see is approximately d (in miles)=
    \sqrt{\frac{3h}{2}}.
    Suppose you are on the roof of a building 900 ft. tall. how far can you see?(calculate to two decimal places)

    My answer: h=900 so \sqrt{\frac{3h}{2}}=d \rightarrow \sqrt{\frac{3*900}{2}}=d \rightarrow \sqrt{\frac{2700}{2}}=d \rightarrow \sqrt{1250}=d \  \therefore \  d=36.74 Thus, I can see for 36.74 Mi.
    __________________________________________________ ______________

    (4)
    According to Descartes' Rule of Signs, (a) how many positive real roots does each of the following have? (b) how many negative roots?

    (1)
    f(a)=a^5-4a^2-7
    My answer:
    (a):1
    (b):0

    (2)
    f(x)=3x^3+9x^2+8x
    My answer:
    (a):0
    (b):1
    -------------------------------------------------------
    Now for the ones I have no idea about.

    (1)
    Solve the equation
    x^3+9x^2+x+2=0, if -2 is a root.
    (I have tried to solve this one but when I do synthetic division on it I get the equation:
    x^2+x=0 which I don't think will work)

    (2)

    Find the dimensions of a rectangle a
    with the greatest area whose perimeter is 30 feet.
    (I got nothing--I'm totally confused as to what to do with this one.)

    Thank you!

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  2. #2
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    Hi there,

    In future please post one problem per thread, makes life easier for all concerned.

    Quote Originally Posted by MathBlaster47 View Post

    The ones I need to double check are:

    (1)
    Assuming that none of the solutions are undefined,simplify each of the following:

    (a) \frac{a^2 b^2 c}{abc}

    My answer: Either abc or ab(I'm not too sure which, though I am leaning towards "ab")


    \frac{a\times a \times b\times b \times c}{a\times b \times c}


    After some cancelation you get ab
    Quote Originally Posted by MathBlaster47 View Post



    (b) \frac{a^{2m} b^{m+1}}{a^m b}

    My answer: a^{2m-m}b^m or a^mb^m

    You need to do it like this

    a^{2m-m}b^{m+1-1} =a^mb^m
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  3. #3
    Member MathBlaster47's Avatar
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    Quote Originally Posted by pickslides View Post
    Hi there,

    In future please post one problem per thread, makes life easier for all concerned.
    Ok, I'll be sure to do that. I posted all of my questions in one place because I didn't want it to look like I was spamming the board.
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  4. #4
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    Quote Originally Posted by MathBlaster47 View Post
    Ok, I'll be sure to do that. I posted all of my questions in one place because I didn't want it to look like I was spamming the board.
    I understand your concern, thanks for that.

    Maybe post 1 or 2 quesitons per thread, get the required help, solve them and come back later to post more.
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  5. #5
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    How did you arrive at the last expression?

    Quote Originally Posted by MathBlaster47 View Post

    __________________________________________________ ______________
    (2)
    Simplify:
    (a) \frac{x-y}{x+y} + \frac{x+y}{x-y}


    ......

    My answer: \frac{(x-y)^2+(x+y)^2}{(x+y)(x-y)} \rightarrow (x-y)+(x-y)
    also if correct (x-y)+(x-y) = 2 (x-y)
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  6. #6
    Member MathBlaster47's Avatar
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    Quote Originally Posted by pickslides View Post
    How did you arrive at the last expression?



    also if correct (x-y)+(x-y) = 2 (x-y)
    I arrived at the last expression by expanding to \frac{[(x-y)(x-y)]+[(x+y)(x+y)]}{(x-y)(x+y)} then I canceled (x-y) and (x+y) from the expression: [(x-y)(x-y)]+[(x+y)(x+y)]/(x-y)(x+y) became [(x-y)(x-y)]+[(x+y)(x+y)]/(x-y)(x+y) .

    I guess I'm not quite getting it.
    Where am I going wrong?
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  7. #7
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    Quote Originally Posted by MathBlaster47 View Post

    (2)

    Find the dimensions of a rectangle a
    with the greatest area whose perimeter is 30 feet.
    (I got nothing--I'm totally confused as to what to do with this one.)

    Thank you!


    This can be done in two ways.

    The algebraic approach is as follows,
    Let x and y be the sides of the rectangle in feet.

    Therefore, the perimeter of the rectangle P = 2(x+y) = 30

    Now, the area of the rectangle A = xy


    Now, since the Arithmetic mean of two numbers>= Their geometric mean,

    (x+y)/2 >= \sqrt{xy}

    Thus 0.25P >= 2\sqrt{A}

    Hope you can proceed from here.
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  8. #8
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    Quote Originally Posted by MathBlaster47 View Post
    Solve the equation x^3+9x^2+x+2=0, if -2 is a root.
    (I have tried to solve this one but when I do synthetic division on it I get the equation:
    x^2+x=0 which I don't think will work)
    x(x + 1) = 0
    x = 0 or x + 1 = 0
    OK?
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  9. #9
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    Quote Originally Posted by MathBlaster47 View Post
    I arrived at the last expression by expanding to \frac{[(x-y)(x-y)]+[(x+y)(x+y)]}{(x-y)(x+y)} then I canceled (x-y) and (x+y) from the expression.
    1. Expand the brackets in the numerator:

    \frac{[(x-y)(x-y)]+[(x+y)(x+y)]}{(x-y)(x+y)} = \dfrac{x^2-2xy+y^2+x^2+2xy+y^2}{(x-y)(x+y)} =  \dfrac{2x^2+2y^2}{(x-y)(x+y)} =\dfrac{2(x^2+y^2)}{(x-y)(x+y)}

    2. You can't factor a sum of squares ( in \mathbb{R})

    3. In the dark ages of math you would have been boiled in oil for your "canceling"
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  10. #10
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    Quote Originally Posted by bandedkrait View Post
    This can be done in two ways.

    The algebraic approach is as follows,
    Let x and y be the sides of the rectangle in feet.

    Therefore, the perimeter of the rectangle P = 2(x+y) = 30

    Now, the area of the rectangle A = xy


    Now, since the Arithmetic mean of two numbers>= Their geometric mean,

    (x+y)/2 >= \sqrt{xy}

    Thus 0.25P >= 2\sqrt{A}

    Hope you can proceed from here.
    Thank you!
    Just so I'm clear: 15\geq \sqrt{A} because (x+y)/2 = (\frac{1}{4}) 30= 15.
    So: since 2x+2y=30 and xy=A then if x=5 and y=10, 2*5+2*10=30 and 5*10=50.
    How do I make sure that the area I found is the greatest possible?

    Quote Originally Posted by Wilmer View Post
    x(x + 1) = 0
    x = 0 or x + 1 = 0
    OK?
    Thank you muchly!


    Quote Originally Posted by earboth View Post
    1. Expand the brackets in the numerator:

    \frac{[(x-y)(x-y)]+[(x+y)(x+y)]}{(x-y)(x+y)} = \dfrac{x^2-2xy+y^2+x^2+2xy+y^2}{(x-y)(x+y)} =  \dfrac{2x^2+2y^2}{(x-y)(x+y)} =\dfrac{2(x^2+y^2)}{(x-y)(x+y)}

    2. You can't factor a sum of squares ( in \mathbb{R})

    3. In the dark ages of math you would have been boiled in oil for your "canceling"
    1.Thank You!! I felt like something was off.....

    2.Ah..... .

    3.More oil? I haven't even gotten over the deep fryer I'm in right now......
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  11. #11
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    Because the maximum value of the Geometric mean, \sqrt{ab} is 15/2 = 7.5

    This will happen for the situation x=y, i.e. when the rectangle is a square.

    so using this, you get the maximum value of x=15

    Hence find the maximum area, which is 225.
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