Which numbers are in the solution set of the inequality 3/4(2 + 2/3k) > -1/2(3k - 2)? Select all that apply. Multi-select Choices: (A)-1/4 (B)0 (C)1/2 (D)-1/2 (E)-6 (F)-3

Simplify Inequality: Simplify both sides of the inequality. 3/4(2 + 2/3k) > -1/2(3k - 2) = (3/4)(2 + 2/3k) > (-1/2)(3k - 2) = (3/4)(2 + 2/3k) > (-1/2)(3k - 2) Distribute Fractions: Distribute the fractions on both sides. = 3/4 * 2 + 3/4 * 2/3k > -1/2 * 3k + 1/2 * 2 = 3/2 + 1/2k > -3/2k + 1 Rearrange Terms: Get all terms involving k on one side and constant terms on the other. 1/2k + 3/2k > 1 - 3/2 = 4/2k > -1/2 = 2k > -1/2 Solve for k: Solve for k by dividing both sides by 2. 2k/2 > -1/2 / 2 = k > -1/4 Check Options: Check which options are greater than -1/4. (A) -1/4 is not greater than -1/4. (B) 0 is greater than -1/4. (C) 1/2 is greater than -1/4. (D) -1/2 is not greater than -1/4. (E) -6 is not greater than -1/4. (F) -3 is not greater than -1/4.

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Which numbers are in the solution set of the inequality \frac{3}{4}(2 + \frac{2}{3}k) > -\frac{1}{2}(3k - 2)? Select all that apply. $\newline$ Multi-select Choices: $\newline$ (A) $-\frac{1}{4}$ $\newline$ (B) $0$ $\newline$ (C) $\frac{1}{2}$ $\newline$ (D) $-\frac{1}{2}$ $\newline$ (E) $-6$ $\newline$ (F) $-3$

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Algebra 1

Simplify exponential expressions using the multiplication and division rules

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Q. Which numbers are in the solution set of the inequality

\frac{3}{4}(2 + \frac{2}{3}k) > -\frac{1}{2}(3k - 2)

? Select all that apply.

\newline

Multi-select Choices:

\newline

(A)

-\frac{1}{4}

\newline

(B)

0

\newline

(C)

\frac{1}{2}

\newline

(D)

-\frac{1}{2}

\newline

(E)

-6

\newline

(F)

-3

Simplify Inequality: Simplify both sides of the inequality. $\newline$ \frac{3}{4}(2 + \frac{2}{3}k) > -\frac{1}{2}(3k - 2) $\newline$ = \left(\frac{3}{4}\right)(2 + \frac{2}{3}k) > \left(-\frac{1}{2}\right)(3k - 2) $\newline$ = \left(\frac{3}{4}\right)(2 + \frac{2}{3}k) > \left(-\frac{1}{2}\right)(3k - 2)
Distribute Fractions: Distribute the fractions on both sides. $\newline$ = \frac{3}{4} \times 2 + \frac{3}{4} \times \frac{2}{3}k > -\frac{1}{2} \times 3k + \frac{1}{2} \times 2 $\newline$ = \frac{3}{2} + \frac{1}{2}k > -\frac{3}{2}k + 1
Rearrange Terms: Get all terms involving $k$ on one side and constant terms on the other.\frac{1}{2}k + \frac{3}{2}k > 1 - \frac{3}{2}= \frac{4}{2}k > -\frac{1}{2}= 2k > -\frac{1}{2}
Solve for k: Solve for k by dividing both sides by $2$ .\frac{2k}{2} > \frac{-1}{2} / 2= k > -\frac{1}{4}
Check Options: Check which options are greater than $-\frac{1}{4}$ .
(A) $-\frac{1}{4}$ is not greater than $-\frac{1}{4}$ .
(B) $0$ is greater than $-\frac{1}{4}$ .
(C) $\frac{1}{2}$ is greater than $-\frac{1}{4}$ .
(D) $-\frac{1}{2}$ is not greater than $-\frac{1}{4}$ .
(E) $-6$ is not greater than $-\frac{1}{4}$ .
(F) $-\frac{1}{4}$ $1$ is not greater than $-\frac{1}{4}$ .