What is (a^2-81)/(a^2-9a)<= 0 ?

The solution to (a^2-81)/(a^2-9a)<= 0 is -9<= a<0

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(a^2-81)/(a^2-9a)<= 0

\frac{a ^{2} - 81}{a ^{2} - 9 a} \leq 0

- 9 \leq a < 0

Interval Notation

[- 9, 0)

Solution steps

\frac{a ^{2} - 81}{a ^{2} - 9 a} \leq 0

Factor

\frac{a ^{2} - 81}{a ^{2} - 9 a} : \frac{( a + 9 ) ( a - 9 )}{a ( a - 9 )}

\frac{( a + 9 ) ( a - 9 )}{a ( a - 9 )} \leq 0

Identify the intervals

- 9 \leq a < 0

0. 5^{10}

\frac{1}{2} - \frac{4}{9}

- 20 x^{2} y^{3} z + 32 y^{2} z^{2}

\frac{2}{3} (x - 7) = - 2

64 x - x^{4}

What is (a^2-81)/(a^2-9a)<= 0 ?
The solution to (a^2-81)/(a^2-9a)<= 0 is -9<= a<0