# Thread: lots of math problems, and fast????

1. ## lots of math problems, and fast????

ok, I've got a couple and instead of splitting them into seperate threads, I thought I would post it all here, ok:

1: (simplify with only positive exponents)
sqrt(4x-16) / 4thsqrt (x-4)^3

and I can only make it 2(square root of (x-4))times(4th square root of (x-4) / x-4
is that right?

2: [(1/x^-2) + (4/x^-1*y^-1) + (1/y^-2)]^(-1/2)
I got square root of (x^2 + 4xy + y^2)/x^2 + 4xy + y^2
is that right?

3: find domain: y=log(2x-12)
is it all real numbers?
4: same deal, y=square root of tanx
all real numbers?
5: y=sqrt (square root) of (x-3) - sqrt(x+3)
all real numbers?

6: (solve albsolute value inequalities)
abs(x+1) < & = abs(x-3)
is is 2 <&= x <&= -4?

7: solve and sign chart:
2x^2 + 4x <&= 3
is it x= .581, 2.581?

8: use synthetic division to factor P(x) then solve P(x)=0
I've tried all numbers and can't get it to work
P(x) = x^3 - 6x^2 + 3x - 10

9: find vertical and horizontal asymptotes
y= (x+4)/(x^2 - 1)
and
y= (x^2 - x - 6) / (x^3 - x^2 + x - 6)

THANK YOU!
-meg-

(you don't have to know all of them either, just please I'll take anything, I'm kinda desperare at this point)
also, I work backwards, giving me the answer will do, i can figure it out myself, in fact, I'd probably learn it better if i figured it out by myself, I just need input

2. First read the LaTeX Tutorial

3) To get the domain of $\displaystyle y=\log(2x-12)$ you gotta set two things: $\displaystyle 2x-12$ can't be zero nor negative, then you have to set $\displaystyle 2x-12>0$

The rest are similar.

3. 8: use synthetic division to factor P(x) then solve P(x)=0
I've tried all numbers and can't get it to work
$\displaystyle P(x) = x^{3} - 6x^{2} + 3x - 10$
Are you sure that isn't $\displaystyle x^{3}+6x^{2}+3x-10$?.

The other one doesn't factor very nicely and has two non real solutions.

9: find vertical and horizontal asymptotes
$\displaystyle y= \frac{(x+4)}{(x^2 - 1)}$
and
$\displaystyle y= \frac{(x^2 - x - 6)}{(x^3 - x^2 + x - 6)}$
Asymptotes are easy if you just know some rules. To find the vertical asymptotes, find what x value(s) makes the denominator equal 0.

To find the horizontal asymptotes, if the power of the numerator is less than the power of the denominator the x-axis is the horizontal asymptote.

4. ^
That's exactly what I thought, it might be a mistake ... erk
and thanks for the asymptotes

5. Originally Posted by cassiopeia1289
ok, I've got a couple and instead of splitting them into seperate threads, I thought I would post it all here, ok:

1: (simplify with only positive exponents)
sqrt(4x-16) / 4thsqrt (x-4)^3

and I can only make it 2(square root of (x-4))times(4th square root of (x-4) / x-4
is that right?

...
Hello,

$\displaystyle \frac{\sqrt{4x-16}}{\sqrt[4]{(x-4)^3}} = 2 \cdot (x-4)^{\frac{1}{2}} \cdot (x-4)^{-\frac{3}{4}} = 2 \cdot (x-4)^{-\frac{1}{4}}=\frac{2}{(x-4)^{\frac{1}{4}}}$

6. Originally Posted by cassiopeia1289
ok, I've got a couple and instead of splitting them into seperate threads, I thought I would post it all here, ok:

...

2: [(1/x^-2) + (4/x^-1*y^-1) + (1/y^-2)]^(-1/2)
I got square root of (x^2 + 4xy + y^2)/x^2 + 4xy + y^2
is that right?

...
I assume that you mean:

$\displaystyle \left(\frac{1}{x^{-2}} + \frac{4}{x^{-1} \cdot y^{-1}} + \frac{1}{y^{-2}}\right)^{-\frac{1}{2}}$

If so your answer is right but I would add that x and y must be unequal zero and the complete term in bracket must be greater than zero.

7. Originally Posted by cassiopeia1289
ok, I've got a couple and instead of splitting them into seperate threads, I thought I would post it all here, ok:

...
4: same deal, y=square root of tanx
all real numbers?
5: y=sqrt (square root) of (x-3) - sqrt(x+3)
all real numbers?

...
to #4:

$\displaystyle y = \sqrt{\tan(x)}$. That means $\displaystyle \tan(x) \geq 0$ which is only possible if $\displaystyle D=\{x|x \in [n \cdot \pi, n \cdot \pi + \frac{\pi}{2}]\}$ with $\displaystyle n \in \mathbb{Z}$

to #5:

I assume that you mean:

$\displaystyle y=\sqrt{\sqrt{x-3} - \sqrt{x+3}}$

The domain of $\displaystyle \sqrt{x-3}$ is $\displaystyle D_1=\{x|x\geq 3\}$

The domain of $\displaystyle \sqrt{x+3}$ is $\displaystyle D_2=\{x|x\geq -3\}$

But because $\displaystyle \sqrt{x-3} < \sqrt{x+3}$ the radicand is allways negative that means the domain is the empty set.

= + = + = + = + = + = + = + = + = + = + = + = + = + = + = + = + = + = +

If you mean:

$\displaystyle y=\sqrt{x-3} - \sqrt{x+3}$ then the domain is: $\displaystyle D=\{x|x\geq 3\}$

8. Originally Posted by cassiopeia1289
ok, I've got a couple and instead of splitting them into seperate threads, I thought I would post it all here, ok:

...

6: (solve albsolute value inequalities)
abs(x+1) < & = abs(x-3)
is is 2 <&= x <&= -4?

...
Translate your inequality: "The graph of the left function should be below the graph of the right function." (see attachment)

$\displaystyle |x+1| \leq |x+3|~\Longleftrightarrow~|x+1| - |x+3|\leq 0$

Since
$\displaystyle |x+1|=\left\{\begin{array}{lr}-(x+1) & x < -1 \\x+1 & x\geq -1 \end{array}\right.$
and
$\displaystyle |x-3|=\left\{\begin{array}{lr}x-3 & x \geq 3 \\-(x-3) & x< 3 \end{array}\right.$

$\displaystyle \left\{\begin{array}{lr}-(x+1)-(-(x-3))\leq 0 & x <-1 \\ x+1-(-(x-3))\leq 0 & -1\leq x < 3 \\ x+1 - (x-3) \leq 0 &x \geq 3\end{array}\right.$
$\displaystyle x < -1 ~\wedge -1 \leq x < 1$. That means the solution is the set of x with x < 1.