All categories

3 years ago

How do you stop your friend from committing suici??

Mathematics

1 answer:

insens350 [35]3 years ago

3 0

Answer:

call the hotline, or tell him to talk to a therapist

Step-by-step explanation:

You might be interested in

Evaluate the function for the given values to determine if the value is a root. p(−2) = p(2) = The value is a root of p(x).

bija089 [108]

Note: Since you missed to mention the the expression of the function $p(x)$ . After a little research, I was able to find the complete question. So, I am assuming the expression as $p(x)=x^4-9x^2-4x+12$ and will solve the question based on this assumption expression of $p(x)$ , which anyways would solve your query.

Answer:

$p\left(-2\right)=0$

Therefore, $x=-2$ is a root of the polynomial $p(x)=x^4-9x^2-4x+12$

$p\left(2\right)=-16$

Therefore, $x=2$ is not a root of the polynomial $p(x)=x^4-9x^2-4x+12$

Step-by-step explanation:

As we know that for any polynomial let say $p(x)$ , $c$ is the root of the polynomial if $p(c)=0$ .

In order to find which of the given values will be a root of the polynomial, $p(x)=x^4-9x^2-4x+12$ , we must have to evaluate $p(x)$ for each of these values to determine if the output of the function gets zero.

So,

Solving for $p\left(-2\right)$

$p(x)=x^4-9x^2-4x+12$

$p\left(-2\right)=\left(-2\right)^4-9\left(-2\right)^2-4\left(-2\right)+12$

$\mathrm{Simplify\:}\left(-2\right)^4-9\left(-2\right)^2-4\left(-2\right)+12:\quad 0$

$\left(-2\right)^4-9\left(-2\right)^2-4\left(-2\right)+12$

$\mathrm{Apply\:rule}\:-\left(-a\right)=a$

$=\left(-2\right)^4-9\left(-2\right)^2+4\cdot \:2+12$

$\mathrm{Apply\:exponent\:rule}:\quad \left(-a\right)^n=a^n,\:\mathrm{if\:}n\mathrm{\:is\:even}$

$=2^4-2^2\cdot \:9+8+12$

$=2^4+20-2^2\cdot \:9$

$=16+20-36$

$=0$

Thus,

$p\left(-2\right)=0$

Therefore, $x=-2$ is a root of the polynomial $p(x)=x^4-9x^2-4x+12$ .

Now, solving for $p\left(2\right)$

$p(x)=x^4-9x^2-4x+12$

$p\left(2\right)=\left(2\right)^4-9\left(2\right)^2-4\left(2\right)+12$

$\mathrm{Remove\:parentheses}:\quad \left(a\right)=a$

$p\left(2\right)=2^4-9\cdot \:2^2-4\cdot \:2+12$

$p\left(2\right)=2^4-2^2\cdot \:9-8+12$

$p\left(2\right)=2^4+4-2^2\cdot \:9$

$p\left(2\right)=16+4-36$

$p\left(2\right)=-16$

Thus,

$p\left(2\right)=-16$

Therefore, $x=2$ is not a root of the polynomial $p(x)=x^4-9x^2-4x+12$ .

Keywords: polynomial, root

Learn more about polynomial and root from brainly.com/question/8777476

#learnwithBrainly

7 0

3 years ago

Read 2 more answers

9/15 + 7/15= ? Pls help I’m being dumb :0

Mashutka [201]

I think the answer is 1.06666666667

7 0

3 years ago

Rewrite the expression as an equivalent expression that does not contain powers of trigonometric functions greater than 1.

stich3 [128]

We are asked to express cos^ 3 x in terms of powers of trigonometric functions not greater than 1

cos ^ 3 x is equal to cos x * cos^2 x.
cos^2 x = 1 - sin^2 x.
sin^2 x = (1 - cos 2x) /2

the answer is cos ^ 3 x = cos x * (1 - (1 - cos 2x) /2 )

5 0

3 years ago

Please help me! Please don't get it wrong ;-;

polet [3.4K]

Answer:

Its 40 degrees

Step-by-step explanation:

60+80=140

180-140=40 (a straight line is always 180 degrees)

Hope it helped

3 0

3 years ago