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3 years ago

A large 2 topping pizza l costs $2 more than a medium 3 topping pizza m what is the dependent and independent variable

Mathematics

1 answer:

denis-greek [22]3 years ago

6 0

Independent - Size/Type of pizza

Dependent - The cost of pizza

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What is another name for CA

abruzzese [7]

Answer:

i think its b

Step-by-step explanation:

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2 years ago

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Find a factorization of x² + 2x³ + 7x² - 6x + 44, given that −2+i√√7 and 1 - i√/3 are roots.

Levart [38]

A factorization of $x^4+2x^3+7x^2-6x+44$ is $(x^2+4x+11)(x^2-2x+4)$ .

<h3>What are the properties of roots of a polynomial?</h3>

The maximum number of roots of a polynomial of degree $n$ is $n$ .
For a polynomial with real coefficients, the roots can be real or complex.
The complex roots of a polynomial with real coefficients always exist in a pair of conjugate numbers i.e., if $a+ib$ is a root, then $a-ib$ is also a root.

If the roots of the polynomial $p(x)=ax^4+bx^3+cx^2+dx+e$ are $r_1,r_2,r_3,r_4$ , then it can be factorized as $p(x)=(x-r_1)(x-r_2)(x-r_3)(x-r_4)$ .

Here, we are to find a factorization of $p(x)=x^4+2x^3+7x^2-6x+44$ . Also, given that $-2+i\sqrt{7}$ and $1-i\sqrt{3}$ are roots of the polynomial.

Since $p(x)=x^4+2x^3+7x^2-6x+44$ is a polynomial with real coefficients, so each complex root exists in a pair of conjugates.

Hence, $-2-i\sqrt{7}$ and $1+i\sqrt{3}$ are also roots of the given polynomial.

Thus, all the four roots of the polynomial $p(x)=x^4+2x^3+7x^2-6x+44$ , are: $r_1=-2+i\sqrt{7}, r_2=-2-i\sqrt{7}, r_3=1-i\sqrt{3}, r_4=1+i\sqrt{3}$ .

So, the polynomial $p(x)=x^4+2x^3+7x^2-6x+44$ can be factorized as follows:

$\{x-(-2+i\sqrt{7})\}\{x-(-2-i\sqrt{7})\}\{x-(1-i\sqrt{3})\}\{x-(1+i\sqrt{3})\}\\=(x+2-i\sqrt{7})(x+2+i\sqrt{7})(x-1+i\sqrt{3})(x-1-i\sqrt{3})\\=\{(x+2)^2+7\}\{(x-1)^2+3\}\hspace{1cm} [\because (a+b)(a-b)=a^2-b^2]\\=(x^2+4x+4+7)(x^2-2x+1+3)\\=(x^2+4x+11)(x^2-2x+4)$

Therefore, a factorization of $x^4+2x^3+7x^2-6x+44$ is $(x^2+4x+11)(x^2-2x+4)$ .

To know more about factorization, refer: brainly.com/question/25829061

#SPJ9

3 0

1 year ago

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HELP!!!! BRAINIEST ANSWER + POINTS!!! PLEASE BE SERIOUS (I NEED ALL 5 POINTS + EXPLANATION)

skelet666 [1.2K]

Answer:

slope=3

y-intercept=-1

Step-by-step explanation:

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2 years ago

Find the area of the shaded region

o-na [289]

so hmmm let's get the area of the whole hexagon, and then get the area of the circle inside it, then subtract the area of the circle from that of the hexagon's, what's leftover is what we didn't subtract, namely the shaded part.

$\textit{area of a regular polygon}\\\\ A=\cfrac{1}{4}ns^2\cot\stackrel{\stackrel{degrees}{\downarrow }}{\left( \frac{180}{n} \right)}~ \begin{cases} n=\textit{number of sides}\\ s=\textit{length of a side}\\[-0.5em] \hrulefill\\ n=\stackrel{hexagon}{6}\\ s=\frac{9}{2} \end{cases}\implies A=\cfrac{1}{4}(6)\left( \cfrac{9}{2} \right)^2 \cot\left( \cfrac{180}{6} \right)$

$A=\cfrac{1}{4}(6)\cfrac{9^2}{2^2} \cot(30^o)\implies A=\cfrac{243}{8}\cot(30^o)\implies A=\cfrac{243\sqrt{3}}{8} \\\\[-0.35em] ~\dotfill\\\\ \textit{area of circle}\\\\ A=\pi r^2~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=\frac{4}{5} \end{cases}\implies A=\pi \left( \cfrac{4}{5} \right)^2\implies A=\cfrac{16\pi }{25} \\\\[-0.35em] ~\dotfill$

$\stackrel{\textit{area of the hexagon}}{\cfrac{243\sqrt{3}}{8}}~~ - ~~\stackrel{\textit{area of the circle}}{\cfrac{16\pi }{25}}\implies \cfrac{6075\sqrt{3}-128\pi }{200}$