1. ## Inequalities

1.Graph the following on a number line.

a. {x]-3<=to x < 1, xER}

b. {s] s^2 > 9, sER}

2.Solve and graph each solution set. (belong to the set of real numbers)

a. 3a-2<= to 7

b. 2(s +3) - 3(s-5) >= to 12

3. Solve. (set of real numbers)

a. 5-4b divided by 3 <5

4. A function is g(x)= 2(x-3)^2+ 4

a. determine the range of g when x > 4

b. determine the range of g when o <= to x <= to 6

c. determine the domain of g when f(x) <= to 12.

It would be appreciated since im lost in this.

2. Originally Posted by Blast18
1.Graph the following on a number line.

a. {x]-3< or equal to x < 1, xER}

b. {s] s squared > 9, sER}

2.Solve and graph each solution set. (belong to the set of real numbers)

a. 3a-2< or equal to 7

b. 2(s +3) - 3(s-5) > or equal to 12

3. Solve. (set of real numbers)

a. 5-4b divided by 3 <5

4. A function is g(x)= 2(x-3) square + 4

a. determine the range of g when x > 4

b. determine the range of g when o < or equal to x < or equal to 6

c. determine the domain of g when f(x) < or equal to 12.

It would be appreciated since im lost in this.
i'm not sure i understand some of your questions. please retype them.

use:

<= for $\leq$
>= for $\geq$
^ for powers. that is, to say s squared, type s^2

3. fixed it

4. Hello, Blast18!

I figured out what the first one says . . .

1. Graph the following on a number line.

. . $a)\;\;\{x\,|\,-3 \leq x < 1,\;x \in R\}$

It says:. The set of all number $x$ such that
. . $x$ is between -3 and 1 (including -3)
. . and $x$ is a real number.

You can visualize this set of numbers, can't you? .Graph it.

. . . $- - \bullet=========\circ - -$
. . . . . $\text{-}3$ . . . . . . . . . . $1$

Use a "solid dot" at -3 to indicate that it is included.
Use an "empty dot" at +1 to indicate that is not included.

$b)\;\;\{s\,|\,s^2 > 9,\; s\in R\}$

We want values of $s$ such that: $s^2 \,>\,9$
. . We find that: . $s \,<\,-3$ or $s \,>\,3$

. . $= = = \circ - - - - \circ = = =$
. . . . . $\text{-}3$ . . . . . $3$

2. Solve and graph each solution set (real numbers).

$a)\;\;3a - 2\:\leq\:7$

You're expected to know how to solve an inequality.

Add 2 to both sides: . $3a \:\leq \:9$

Divide both sides by 3: . $a \:\leq \:3$

. . $=======\bullet - - -$
. - . - . . . . . . $3$

$b)\;\;2(s +3) - 3(s-5)\:\geq\:12$

We have: . $2s + 6 - 3s + 15 \:\geq \:12$

Then: . $-s + 21 \:\geq \:12$

Subtract 21 from both sides: . $-s \:\geq\:-9$

Multiply both sides by -1: . $s \:\leq \:9$
.
and reverse the inequality!

. . $=======\bullet--$
. - . - . . . . . . $9$

5. thanks hopefully someone can help me with the other questions