Solve for x. -9x+2 > 18quad" OR "quad13 x+15 <= -4 Choose 1 answer: A) x <= -(19)/(13) (B) x < -(16)/(9) (c) -(16)/(9) < x < -(19)/(13) (D) There are no solutions (E) All values of x are solutions

Solve inequality -9x + 2 > 18: First, let's solve the inequality -9x + 2 > 18. Subtract 2 from both sides to isolate the term with x. -9x + 2 - 2 > 18 - 2 -9x > 16 Now, divide both sides by -9. Remember that dividing by a negative number reverses the inequality sign. x < -16/9Isolate the term with x: Next, let's solve the inequality 13x + 15 <= -4. Subtract 15 from both sides to isolate the term with x. 13x + 15 - 15 <= -4 - 15 13x <= -19 Now, divide both sides by 13 to solve for x. x <= -19/13Divide both sides by -9: Now we have two inequalities to consider: x < -16/9 and x <= -19/13. We need to find the intersection of these two sets to find the solution for x. The number -16/9 is less than -19/13, so the solution set for x is the values that are less than -16/9 and also less than or equal to -19/13. This means that the solution set is x < -16/9, as it is the more restrictive condition.

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Solve for
x.

-9x+2 > 18quad" OR "quad13 x+15 <= -4
Choose 1 answer:
A)
x <= -(19)/(13)
(B)
x < -(16)/(9)
(c)
-(16)/(9) < x < -(19)/(13)
(D) There are no solutions
(E) All values of
x are solutions

Solve for $x$ . $\newline$ -9 x+2>18 \quad \text { OR } \quad 13 x+15 \leq-4 $\newline$ Choose $1$ answer: $\newline$ A) $x \leq-\frac{19}{13}$ $\newline$ (B) x<-\frac{16}{9} $\newline$ (c) \( -\frac{16}{9}

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Math Problems

Algebra 1

Domain and range of quadratic functions: equations

Full solution

Q. Solve for

x

\newline

-9 x+2>18 \quad \text { OR } \quad 13 x+15 \leq-4

\newline

Choose

1

answer:

\newline

x \leq-\frac{19}{13}

\newline

(B)

x<-\frac{16}{9}

\newline

(c)

-\frac{16}{9}<x<-\frac{19}{13}

\newline

(D) There are no solutions

\newline

(E) All values of

x

are solutions

Solve inequality -9x + 2 > 18: First, let's solve the inequality -9x + 2 > 18. $\newline$ Subtract $2$ from both sides to isolate the term with $x$ . $\newline$ -9x + 2 - 2 > 18 - 2 $\newline$ -9x > 16 $\newline$ Now, divide both sides by $-9$ . Remember that dividing by a negative number reverses the inequality sign. $\newline$ x < -\frac{16}{9}
Isolate the term with x: Next, let's solve the inequality $13x + 15 \leq -4$ . $\newline$ Subtract $15$ from both sides to isolate the term with x. $\newline$ $13x + 15 - 15 \leq -4 - 15$ $\newline$ $13x \leq -19$ $\newline$ Now, divide both sides by $13$ to solve for x. $\newline$ $x \leq -\frac{19}{13}$
Divide both sides by $-9$ : Now we have two inequalities to consider: $\newline$ x < -\frac{16}{9} and $x \leq -\frac{19}{13}$ . $\newline$ We need to find the intersection of these two sets to find the solution for $x$ . $\newline$ The number $-\frac{16}{9}$ is less than $-\frac{19}{13}$ , so the solution set for $x$ is the values that are less than $-\frac{16}{9}$ and also less than or equal to $-\frac{19}{13}$ . $\newline$ This means that the solution set is x < -\frac{16}{9}, as it is the more restrictive condition.