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3 years ago

8

there is a total of 10 pages of homework. sophia worked on some pages on monday if she has 3 pages left how many pages did sophi

a work on monday?

1 answer:

Oxana [17]3 years ago

3 0

Answer:

7

Step-by-step explanation:

Since, Sophia had three pages of homework and she only has three left you can set up an equation to solve. The equation would be 10=x-3, then you just solve the equation which would be 7, hope that helped.

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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]

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

Answer:

As

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

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

As

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

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

Step-by-step explanation:

As we know that for any polynomial let say<em> </em> $p(x)$ <em>, </em> $c$ 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, $p(x)=x^4-9x^2-4x+12$ <em>, </em>we must have to evaluate <em> </em> $p(x)$ <em> </em>for each of these values to determine if the output of the function gets zero.

So,

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

<em> </em> $p(x)=x^4-9x^2-4x+12$

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

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

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

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

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

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

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

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

$=16+20-36$

$=0$

Thus,

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

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

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

<em> </em> $p(x)=x^4-9x^2-4x+12$

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

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

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

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

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

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

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

Thus,

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

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

Keywords: polynomial, root

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