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vazorg [7]
1 year ago
5

Question 2(Multiple Choice Worth 1 points) (01.04 LC) Solve for x: −2x − 4 > 8

Mathematics
1 answer:
frutty [35]1 year ago
8 0

The required solution of the inequality is x  < - 6.


Given that,
To solve the equation −2x − 4 > 8.

<h3>What is inequality?</h3>

Inequality can be described as the consideration of the equation including the symbol of ( ≤, ≥, <, >) instead of the equal sign in an equation.


Here,
The given equation
−2x − 4 > 8
adding 4 on both sides
-2x > 12
multiply by -1
when multiplying the negative -1 it will change the behavior of the inequality, so.
2x < - 12
Divide by 2 into both sides
x < - 6

Thus, the required solution of the inequality is x  < - 6.

Learn more about inequality here:

brainly.com/question/14098842

#SPJ1

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Perhaps you meant to write that x\to-2 instead? In that case, you would have

\displaystyle\lim_{x\to-2}\frac{x^2-x+6}{x+2}=\lim_{x\to-2}\frac{(x+2)(x-3)}{x+2}=\lim_{x\to-2}(x-3)=-2-3=-5
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3 years ago
Prove that :( 1 + 1/<img src="https://tex.z-dn.net/?f=tan%5E%7B2%7DA" id="TexFormula1" title="tan^{2}A" alt="tan^{2}A" align="ab
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Answer:

See explanation

Step-by-step explanation:

Simplify left and right parts separately.

<u>Left part:</u>

\left(1+\dfrac{1}{\tan^2A}\right)\left(1+\dfrac{1}{\cot ^2A}\right)\\ \\=\left(1+\dfrac{1}{\frac{\sin^2A}{\cos^2A}}\right)\left(1+\dfrac{1}{\frac{\cos^2A}{\sin^2A}}\right)\\ \\=\left(1+\dfrac{\cos^2A}{\sin^2A}\right)\left(1+\dfrac{\sin^2A}{\cos^2A}\right)\\ \\=\dfrac{\sin^2A+\cos^2A}{\sin^2A}\cdot \dfrac{\cos^2A+\sin^A}{\cos^2A}\\ \\=\dfrac{1}{\sin^2A}\cdot \dfrac{1}{\cos^2A}\\ \\=\dfrac{1}{\sin^2A\cos^2A}

<u>Right part:</u>

\dfrac{1}{\sin^2A-\sin^4A}\\ \\=\dfrac{1}{\sin^2A(1-\sin^2A)}\\ \\=\dfrac{1}{\sin^2A\cos^2A}

Since simplified left and right parts are the same, then the equality is true.

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Step-by-step explanation:

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