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2 years ago

I will give BRAINLIEST

Mathematics

1 answer:

BabaBlast [244]2 years ago

5 0

Answer:

Step-by-step explanation:

6 + 4 is 10

Half of 10 is 5

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What is an expressions that are equivalent to 7(3c).

ra1l [238]

Well, there are lots, but if you get rid of the parenthesis...

you would get 21c and that equals to 7(3c).

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4 years ago

Juan builds a square dog pen that has an area of 100 square feet. He then builds a second pen that has an area of 49 square feet

tiny-mole [99]

Answer:

its 21.0

Step-by-step explanation:

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3 years ago

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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:

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

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

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

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

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

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7 0

3 years ago

Read 2 more answers

Please HELP! I DON'T KNOW HOW CAN I SOLVE THIS

Wewaii [24]

Answer:

Step-by-step explanation:

The Pythagorean Theorem:

$leg^2+leg^2=hypotenuse^2\to a^2+b^2=c^2$

$41.\\\\a=1,\ b=1,\ c=x\\\\x^2=1^2+1^2\\\\x^2=1+1\\\\x^2=2\to x=\sqrt2$

$42.\\a=1, b=2,\ c=x\\\\x^2=1^2+2^2\\\\x^2=1+4\\\\x^2=5\to x=\sqrt5$

$43.\\a=3,\ b=x,\ c=5\\\\3^2+x^2=5^2\\\\9+x^2=25\qquad\text{subtract 9 from both sides}\\\\x^2=16\to x=\sqrt{16}\to x=4$

$44.\\\\a=4,\ b=x,\ c=8\\\\4^2+x^2=8^2\\\\16+x^2=64\qquad\text{subtract 16 from both sides}\\\\x^2=48\to x=\sqrt{48}\\\\x=\sqrt{16\cdot3}\qquad\text{use}\ \sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\\\\x=\sqrt{16}\cdot\sqrt3\\\\x=4\sqrt3$

7 0

3 years ago

Read 2 more answers

Coach Taylor wants to know the total practice time if he has the players warm up for 15 min

mrs_skeptik [129]

18 mins it is ! I think your missing the question tho

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3 years ago