Help with pre-calculus questions

[watchthesky30]

Hi. I need some help with some of these pre-calculus questions. If you can help me it would be greatly appreciated. Thanks!

1) x(2-3x)?0
would you solve this like x?0 and 2-3x?0
or would you just do it like 2x-3x²?

2) 3-|2x+4|?1
For this i got the answers x?-1 and x?-3. i'm not sure if that's right.
[-1,-3] would that be the interval notation?

3) The manager of a weekend flea market knows from past experience that if she charges x dollars for a rental space at the flea market, then the number y of spaces she can rent is given by the equation y=200-4x.
a) Sketch a graph of this linear equation. (Remember that the rental charge per space and the number of spaces rented must both be nonnegative quantities.)
- So i know i have to make a table and plug in the numbers but would the graph be increasing or decreasing? because i got it to decrease, because it said that they both have to be nonnegative quantities (so the numbers that i put for x were all positive) but that doesn't make sense =S.

4) A set of real numbers is graphed. Find an inequality involving an absolute value that describes the set.
- For the record, my teacher never even taught us this at all.
so there's an arrow going to the left of -3 and an arrow going to the right of -1.

 

[BigGlenntheHeavy]

$\displaystyle 4) \ \ Let \ -2 \ < \ x \ <18 \ \implies \ |x-8| \ < \ 10.$

$\displaystyle Note: \ It \ is \ easy \ to \ find \ the \ value \ when \ given \ the \ absolute \ value, \ for \ example:$

$\displaystyle |x-10| \ <8, \ \implies \ \ -8 \ < \ x-10 \ <8, \ hence \ 2 \ < \ x \ <18$

$\displaystyle Now, \ find \ the \ absolute \ value \ of \ -13 \ < \ x \ < \ 27.$

$\displaystyle Answer \ for \ 4) \ -3 \ < \ x \ < \ -1 \ \implies \ |x+2| \ < \ 1.$

 

[BigGlenntheHeavy]

2)

$\displaystyle Solve \ 3-|2x+4| \ \le \ 1$

$\displaystyle -|2x+4| \ \le \ -2$

$\displaystyle |2x+4| \ \ge \ 2, \ (dividing \ by \ a \ negative, \ change \ the \ sense).$

$\displaystyle |2(x+2)| \ \ge \ 2$

$\displaystyle |2||x+2| \ \ge \ 2$

$\displaystyle |x+2| \ \ge \ 1$

$\displaystyle Hence, \ x+2 \ \ge \ 1 \ \implies \ x \ \ge \ -1 \ and \ x+2 \ \le \ -1 \ \implies \ x \ \le \ -3$

$\displaystyle Ergo, \ domain \ is \ (-\infty,-3] \ U \ [-1,\infty)$

 

[BigGlenntheHeavy]

1)

$\displaystyle x(2-3x) \ \le \ 0 \ \implies \ x \ \le \ 0 \ and \ (2-3x) \ \le \ 0 \ \implies \ x \ \ge \ \frac{2}{3}$

$\displaystyle Ergo, \ domain \ is \ (-\infty,0] \ U \ \bigg[\frac{2}{3},\infty\bigg).$