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Math Help - help me plz!

  1. #1
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    help me plz!

    hi again, i need more help plz! i'm about to cry.

    ok here the -3 is throwing me off, and do you also subtract the numbers, for example 3a-1a, 2a, i would think so right? could someone please show me the exact steps to do this?

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    Find the inverse of , yep this is embarassing to ask, but i really forgot how to do this.

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    Given that for all , find the ratio
    Hint, the numerator factors.
    i don't get this one, like i did t^4 - radical (t^4) all over t - radical 3
    thats where im stuck.
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    State the coordinate of the vertex of the parabola ,
    do u use the quadratic equation??
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    Find the equation of the parabola whose vertex is at , and that passes through the point
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    Given that , , , , , and , what is
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  2. #2
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    Quote Originally Posted by rafaeli View Post
    hi again, i need more help plz! i'm about to cry.

    ok here the -3 is throwing me off, and do you also subtract the numbers, for example 3a-1a, 2a, i would think so right? could someone please show me the exact steps to do this?
    a^{-m} = \frac1{a^m}

    As for subtracting the coefficients, what you are basically asking is whether \frac31 = 2; of course not. You subtract exponents, and then only when the base is the same.


    Quote Originally Posted by rafaeli View Post
    Find the inverse of , yep this is embarassing to ask, but i really forgot how to do this.
    Swap the dependent and independent variables:

    f(x) = \frac{5x}{1 + x}, so x = \frac{5f^{-1}(x)}{1 + f^{-1}(x)} (assuming the inverse exists). Now solve for f^{-1}(x).

    Quote Originally Posted by rafaeli View Post
    Given that for all , find the ratio
    Hint, the numerator factors.
    i don't get this one, like i did t^4 - radical (t^4) all over t - radical 3
    thats where im stuck.
    f\left(\sqrt3\right) = \left(\sqrt3\right)^4, not \sqrt{t^4}. Do your substitutions correctly.

    \left\lvert\frac{f(t) - f\left(\sqrt3\right)}{t - \sqrt3}\right\rvert

    =\left\lvert\frac{t^4 - \sqrt3^4}{t - \sqrt3}\right\rvert

    Like the hint says, you can factor the numerator.

    Quote Originally Posted by rafaeli View Post
    State the coordinate of the vertex of the parabola ,
    do u use the quadratic equation??
    No. Complete the square and put the equation in standard form, y = a(x - h)^2 + k, and the vertex will be at (h,\;k).

    y=3x^2 - 5x + 1

    \Rightarrow y = 3\left(x^2 - \frac53x\right) + 1

    \Rightarrow y = 3\left[x^2 - \frac53x + \left(\frac56\right)^2 - \left(\frac56\right)^2\right] + 1

    \Rightarrow y = 3\left(x^2 - \frac53x + \frac{25}{36}\right) - 3\left(\frac56\right)^2 + 1

    \Rightarrow y = 3\left(x - \frac56\right)^2 - \frac{13}{12}

    Now you should be able to find the vertex.

    Quote Originally Posted by rafaeli View Post
    Find the equation of the parabola whose vertex is at , and that passes through the point
    As I said, a parabola (with axis parallel to the y-axis) with vertex at (h,\;k) will have an equation of the form y = a(x - h)^2 + k. So we have y = a(x - 5)^2 + 12. Now substitute the second point in and solve for a.

    Quote Originally Posted by rafaeli View Post
    Given that , , , , , and , what is
    Beats me. We can't evaluate this without knowing g(0).
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  3. #3
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    He4llo, rafaeli!

    There are many ways to simplfy the first one . . .


    \left[\frac{3a^{\frac{1}{2}}b^{-\frac{1}{3}}c}{a^{-\frac{1}{2}}b^{\frac{1}{3}}c^{\frac{7}{2}}}\right]^{-3}
    I would simplify the "inside" first . . . "do" the division.

    . . \bigg[ 3\cdot a^{\frac{1}{2}-(-\frac{1}{2})} \cdot b^{-\frac{1}{3}-\frac{1}{3}} \cdot c^{1-\frac{7}{2}}\bigg]^{-3} \;=\;\bigg[3a^1\,b^{-\frac{2}{3}}\,c^{-\frac{5}{2}}\bigg]^{-3}

    Then: . (3)^{-3}(a^1)^{-3}(b^{-\frac{2}{3}})^{-3}(c^{-\frac{5}{2}})^{-3} \;=\; 3^{-3}a^{-3}b^2c^{\frac{15}{2}} \;=\;\frac{1}{3^3a^3}\cdot b^2c^{\frac{15}{2}}

    Therefore: . \frac{b^2c^{\frac{15}{2}}}{27a^3}

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