24>-3r>=-9

Solution

24 > - 3 r \geq - 9

- 8 < r \leq 3

Interval Notation

(- 8, 3]

Solution steps

24 > - 3 r \geq - 9

a > u \geq b

then

a > u and u \geq b

24 > - 3 r and - 3 r \geq - 9

24 > - 3 r : r > - 8

- 3 r \geq - 9 : r \leq 3

Combine the intervals

r > - 8 and r \leq 3

Merge Overlapping Intervals

- 8 < r \leq 3

x^{2} = - 13

\frac{40}{360} \cdot π (12)^{2}

∣ x - 2 ∣ \geq 8

- 4112 - 28

\frac{2 ^{n} \cdot 9 ^{2 n + 1}}{6 ^{n - 2}}