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3 years ago

What is the solution set to the inequality 5(x – 2)(x + 4) > 0?

Mathematics

2 answers:

Sidana [21]3 years ago

7 0

Answer:

(-∞, -4) U (2,∞)

Step-by-step explanation:

5(x – 2)(x + 4) > 0

First we solve for x, we replace the inequality sign by = sign

5(x – 2)(x + 4) = 0

Divide both sides by 5

(x – 2)(x + 4) = 0

Now we set each factor =0 and solve for x

x-2 =0 , so x= 2

x+4 =0, so x= -4

Now we use number line and make three intervals

First interval -infinity to -4

second interval -4 to 2

third interval 2 to infinity

Now we check each interval with our inequality

First interval -infinity to -4, pick a number in this interval and check with our inequality. lets pick -5

5(-5 – 2)(-5 + 4) > 0

35>0 is true

second interval -4 to 2, pick a number in this interval and check with our inequality. lets pick 0

5(0– 2)(0 + 4) > 0

-40>0 is false

Third interval 2 to infinity, pick a number in this interval and check with our inequality. lets pick 3

5(3 – 2)(3 + 4) > 0

35>0 is true

solution set are the intervals that make the inequalities true

(-∞, -4) U (2,∞)

olga nikolaevna [1]3 years ago

6 0

5(x-2)(x+4)>0 For this to be true, both parenthetical terms must be both positive or both negative.

x<-4 and x>2

x=(-oo,-4),(2,+oo)

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3 years ago

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If you're using the app, try seeing this answer through your browser: brainly.com/question/2787701

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$\mathsf{tan\,\theta=\dfrac{1}{4}\qquad\qquad(sin\,\theta\ \textgreater \ 0)}\\\\\\
\mathsf{\dfrac{sin\,\theta}{cos\,\theta}=\dfrac{1}{4}}\\\\\\
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Square both sides:

$\mathsf{(4\,sin\,\theta)^2=(cos\,\theta)^2}\\\\
\mathsf{4^2\,sin^2\,\theta=cos^2\,\theta}\\\\
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Since $\mathsf{sin\,\theta}$ is positive, you can discard the negative sign. So,

$\mathsf{sin\,\theta=\dfrac{1}{\sqrt{17}}\qquad\quad\checkmark}$

Substitute this value back into $\mathsf{(i)}$ to find $\mathsf{cos\,\theta:}$

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Tags: <em>trigonometric identity relation trig sine cosine tangent sin cos tan trigonometry precalculus</em>

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3 years ago

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3241004551 [841]

It would be 2(x+3)(x+1)=0

Explanation:
I used factor by grouping. You multiply the first term (2) by the last term (6). This gives you 12 then take the factors of 12 that add up to the middle term 8. You get 6 and 2.
It should look like 2x^2+6x+2x+6=0
when you do factor by grouping you factor the first two terms and then the last two terms separately. So you get (2x+2) and (x+3). (2x+2) could be factored into 2(x+1). Then you put everything together and get 2(x+3)(x+1)=0

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cluponka [151]

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