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3 years ago

Which of the following solution sets is all real numbers? Ixl >-2 Ixl <-2 Ixl = -2

Mathematics

2 answers:

Lisa [10]3 years ago

6 0

The first one, |x|>-2

Alekssandra [29.7K]3 years ago

3 0

Answer

Ixl >-2

Step-by-step explanation:

Ixl >-2 it would be this cause the unknown number was positive number. Positive was bigger than negative.

Ixl <-2 it can’t -2 cause the unknown number was positive

Ixl = -2 it can’t be equal cause the number was unknown and it wasn’t negative

I hope I can help you

I wish you a open path to your bright future.

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2 years ago

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3 years ago

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3 years ago

Read 2 more answers

G(x) = -2x^3 – 15x^2 + 36x

shusha [124]

Consider the function $G(x) = -2x^3 - 15x^2 + 36x.$ First, factor it:

$G(x) = -2x^3 - 15x^2 + 36x=-x(2x^2+15x-36)=\\ \\=-x\cdot 2\cdot \left(x-\dfrac{-15-\sqrt{513}}{4}\right)\cdot \left(x-\dfrac{-15+\sqrt{513}}{4}\right).$

The x-intercepts are at points $\left(\dfrac{-15-\sqrt{513} }{4},0\right),\ (0,0),\ \left(\dfrac{-15+\sqrt{513} }{4},0\right).$

1. From the attached graph you can see that

function is positive for $x\in \left(-\infrty, \dfrac{-15-\sqrt{513} }{4}\right)\cup \left(0,\dfrac{-15+\sqrt{513} }{4}\right);$
function is negative for $x\in \left(\dfrac{-15-\sqrt{513} }{4},0\right)\cup \left(\dfrac{-15+\sqrt{513} }{4},\infty\right).$

2. Since

$G(-x) = -2(-x)^3 - 15(-x)^2 + 36(-x)=2x^3-15x^2-36x\neq G(x)\ \text{and }\neq -G(x)$ the function is neither even nor odd.

3. The domain is $x\in (-\infty,\infty),$ the range is $y\in (-\infty,\infty).$

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3 years ago

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