What is (6^3)/(12n^2)(2)<= 0.00001 ?

The solution to (6^3)/(12n^2)(2) = sqrt((36)/(0.00001))

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(6^3)/(12n^2)(2)<= 0.00001

\frac{6 ^{3}}{12 n ^{2}} (2) \leq 0.00001

n \leq - \frac{36}{0.00001} or n \geq \frac{36}{0.00001}

Interval Notation

(- \infty, - \frac{36}{0.00001}] \cup [\frac{36}{0.00001}, \infty)

Decimal Notation

n \leq - 1897.36659 \dots or n \geq 1897.36659 \dots

Solution steps

\frac{6 ^{3}}{12 n ^{2}} \cdot 2 \leq 0.00001

Rewrite in standard form

\frac{432}{12 n ^{2}} - 0.00001 \leq 0

Identify the intervals

n \leq - \frac{36}{0.00001} or n \geq \frac{36}{0.00001}

\frac{w}{2} + 9 + 7 = - 10

\frac{15}{2 6}

8 x + 5 (- 4 x) = 24

- 9 = - 11 + \frac{b}{8}

5 x^{2} - 6 \geq 14

What is (6^3)/(12n^2)(2)<= 0.00001 ?
The solution to (6^3)/(12n^2)(2)<= 0.00001 is n<=-sqrt((36)/(0.00001))\lor n>= sqrt((36)/(0.00001))