I've been stuck on this question for a while now, I need help!

Mathematics

1 answer:

MaRussiya [10]3 years ago

6 0

In order to solve it you must make each component negative. Multiply them by negative one!

M < 2

Now which numbers would satisfy it? M is less than two... so, m could be one, or any negative number, yeah?

Multiply by negative one. - M and -2 are what you have. If it were one... negative one is less than negative two. Turns out M is less than two.

M < 6

M is less than six. It could be 1, 2, 3, 4, or 5.

-M and -6

Is negative 3 less than -6? All of the numbers are! So, m is actually less than 6

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∠ ABD = 5(2X+1)
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In the Figure below is shown the graph of this function. We have the following function:

$f(x)=x^3+2x^2-x-2$

The $y-intercept$ occurs when $x=0$ , so:

$f(0)=(0)^3+2(0)^2-(0)-2=-2$

Therefore, the $y-intercept$ is the given by the point:

$\boxed{(0,-2)}$

From the figure we have three $x-intercepts$ :

$\boxed{P_{1}(-2,0)} \\ \boxed{P_{2}(-1,0)} \\ \boxed{P_{3}(1,0)}$

So, the $x-intercepts$ occur when $y=0$ . Thus, proving this:

$f(x)=x^3+2x^2-x-2 \\ \\ For \ P_{1}:\\ If \ x=-2, \ y=(-2)^3+2(-2)^2-(-2)-2=0 \\ \\ For \ P_{2}:\\ If \ x=-1, \ y=(-1)^3+2(-1)^2-(-1)-2=0 \\ \\ For \ P_{3}:\\ If \ x=1, \ y=(1)^3+2(1)^2-(1)-2=0$