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3 years ago

Solve the inequality:(x+2)/(x^2-3x)>0

Mathematics

1 answer:

makkiz [27]3 years ago

7 0

Answer:

The inequality is equivalent to x(x+2)(x−3)>0 , with the additional conditions that x≠0 and x≠3 .

Since x(x+2)(x−3) only changes signs when crossing −2 , 0 and 3 , from the fact that the evaluating the polynomial at 4 yields 24 , we see that the polynomial is

positive over (3,∞)

negative over (0,3)

positive over (−2,0)

negative over (−∞,−2)

Thus the solution set for your inequality is (−2,0)∪(3,∞) .

Step-by-step explanation:

hi Rakesh here is your answer :)

#shadow

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Answer:

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Step-by-step explanation:

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$5^{x-2}-7^{x-3}=7^{x-5}+11\cdot5^{x-4}\\\\5^{x-2}-7^{x-2-1}=7^{x-2-3}+11\cdot5^{x-2-2}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\5^{x-2}-\dfrac{7^{x-2}}{7^1}=\dfrac{7^{x-2}}{7^3}+11\cdot\dfrac{5^{x-2}}{5^2}\\\\5^{x-2}-\dfrac{1}{7}\cdot7^{x-2}=\dfrac{1}{343}\cdot7^{x-2}+\dfrac{11}{25}\cdot5^{x-2}\\\\-\dfrac{1}{7}\cdot7^{x-2}-\dfrac{1}{343}\cdot7^{x-2}=\dfrac{11}{25}\cdot5^{x-2}-5^{x-2}\\\\\left(-\dfrac{1}{7}-\dfrac{1}{343}\right)\cdot7^{x-2}=\left(\dfrac{11}{25}-1\right)\cdot5^{x-2}$

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3 years ago

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I need help!!<br> I don’t get it!!

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Answer:

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Step-by-step explanation:

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3 years ago

Solve the inequality:(x+2)/(x^2-3x)>0​

Solve the inequality:(x+2)/(x^2-3x)>0