1. ## 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

2. Originally Posted by rafaeli
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?
$\displaystyle a^{-m} = \frac1{a^m}$

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

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

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

Originally Posted by rafaeli
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.
$\displaystyle f\left(\sqrt3\right) = \left(\sqrt3\right)^4$, not $\displaystyle \sqrt{t^4}$. Do your substitutions correctly.

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

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

Like the hint says, you can factor the numerator.

Originally Posted by rafaeli
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, $\displaystyle y = a(x - h)^2 + k$, and the vertex will be at $\displaystyle (h,\;k)$.

$\displaystyle y=3x^2 - 5x + 1$

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

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

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

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

Now you should be able to find the vertex.

Originally Posted by rafaeli
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 $\displaystyle y$-axis) with vertex at $\displaystyle (h,\;k)$ will have an equation of the form $\displaystyle y = a(x - h)^2 + k$. So we have $\displaystyle y = a(x - 5)^2 + 12$. Now substitute the second point in and solve for $\displaystyle a$.

Originally Posted by rafaeli
Given that , , , , , and , what is
Beats me. We can't evaluate this without knowing $\displaystyle g(0)$.

3. He4llo, rafaeli!

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

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

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

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

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