Can someone help me please?

Mathematics

1 answer:

Molodets [167]3 years ago

5 0

-2 should be correct. It is an input for that given graph.

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The opposite angles in a parallelogram have the same measure true or false

Snowcat [4.5K]

Diagonal angles have the same measure - not sure what u mean by oposite.

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4 years ago

1. What is the slope of the line that passes through the given points? <br> (2, 12) and (6, 11)

Y_Kistochka [10]

Answer:

-1/4x

Step-by-step explanation:

11-12/6-2 = -1/4

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If f(x) = 1-x^2+x^3, then f(-1)=

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Answer: -1

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3 years ago

I need help with this question!<br> Experts only please <br><br> Will mark brainliest

Naily [24]

3x + 2y(100)
yes he was correct
2 cups are added each time so 2 times 100 is 200 add the 3 you have 203.
i know i’m not an expert but i wanted to help :)

4 0

3 years ago

Find a polynomial P(x) with real coefficients having a degree 4, leading coefficient 4, and zeros 3-i and 4i.

Alisiya [41]

now, this polynomial has roots of 3-i and 4i, namely 3 - i and 0 + 4i.

let's bear in mind that a complex root never comes all by her lonesome, her sibling is always with her, the conjugate, so if 3 - i is there, 3 + i is also coming along, likewise if 0 + 4i is there, her sibling 0 - 4i is also there.

$\bf \begin{cases} x=3-i\implies &x-3+i=0\\ x=3+i\implies &x-3-i=0\\ x=4i\implies &x-4i=0\\ x=-4i\implies &x+4i=0 \end{cases}\\\\[-0.35em] ~\dotfill\\\\ (x-3+i)(x-3-i)(x-4i)(x+4i)=\stackrel{y}{0} \\[2em] \underset{\textit{difference of squares}}{[(x-3)+i][(x-3)-i]}\underset{\textit{difference of squares}}{[x-4i][x+4i]}=0$

$\bf [(x-3)^2-i^2][x^2-(4i)^2]=y\implies [(x-3)^2-(-1)][x^2-(4^2i^2)]=0 \\[2em] [(x-3)^2-(-1)][x^2-(16(-1))]=0\implies [(x-3)^2+1][x^2+16]=0 \\[2em] [(x^2-6x+9)+1][x^2+16]=y\implies (x^2-6x+10)(x^2+16)=0 \\\\\\ x^4-6x^3+10x^2+16x^2-96x+160=0 \\\\\\ x^4-6x^3+26x^2-96x+160=0 \\\\\\ \stackrel{\textit{multiplying both sides by 4}}{4(x^4-6x^3+26x^2-96x+160)=4(0)} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill 4x^4-24x^3+104x^2-384x+640=y~\hfill$

3 0

3 years ago