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3 years ago

A, B, and C are polynomials, where:

Mathematics

1 answer:

viktelen [127]3 years ago

7 0

Answer:

2x square -25x + 7

Step-by-step explanation:

a square - (b+c)

(3x-4)square - {(x+7)+(x square+2)}

3x square-2.3x.4+4 square - (x+7+x square+2)

3x square - 24x + 16 -x - 7 - x square - 2

3x square - x square -24x - x + 16-7-2

2x square-25x+7

or,

x(2x - 25)+7

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Simplify 2x + x<br> o 2x2<br> 3x<br> O 3x²

DochEvi [55]

Answer:

Step-by-step explanation:

Step-1 : Multiply the coefficient of the first term by the constant 2 • 2 = 4

Step-2 : Find two factors of 4 whose sum equals the coefficient of the middle term, which is 3 .

-4 + -1 = -5

-2 + -2 = -4

-1 + -4 = -5

1 + 4 = 5

2 + 2 = 4

4 + 1 = 5

Observation : No two such factors can be found !!

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3 years ago

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saveliy_v [14]

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3 years ago

Which explains whether sophia is correct

IgorLugansk [536]

Answer:

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2 years ago

Find a parametric representation for the part of the cylinder y2 + z2 = 49 that lies between the planes x = 0 and x = 1. x = u y

sweet-ann [11.9K]

Answer:

The equation for z for the parametric representation is $z = 7 \sin (v)$ and the interval for u is $0\le u\le 1$ .

Step-by-step explanation:

You have the full question but due lack of spacing it looks incomplete, thus the full question with spacing is:

Find a parametric representation for the part of the cylinder $y^2+z^2 = 49$ , that lies between the places x = 0 and x = 1.

$x=u\\ y= 7 \cos(v)\\z=? \\ 0\le v\le 2\pi \\ ?\le u\le ?$

Thus the goal of the exercise is to complete the parameterization and find the equation for z and complete the interval for u

Interval for u

Since x goes from 0 to 1, and if x = u, we can write the interval as

$0\le u\le 1$

Equation for z.

Replacing the given equation for the parameterization $y = 7 \cos(v)$ on the given equation for the cylinder give us

$(7 \cos(v))^2 +z^2 = 49 \\ 49 \cos^2 (v)+z^2 = 49$

Solving for z, by moving $49 \cos^2 (v)$ to the other side

$z^2 = 49-49 \cos^2 (v)$

Factoring

$z^2 = 49(1- \cos^2 (v))$

So then we can apply Pythagorean Theorem:

$\sin^2(v)+\cos^2(v) =1$

And solving for sine from the theorem.

$\sin^2(v) = 1-\cos^2(v)$

Thus replacing on the exercise we get

$z^2 = 49\sin^2 (v)$

So we can take the square root of both sides and we get

$z = 7 \sin (v)$

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3 years ago

I don't know how to help my sister with this please help

Orlov [11]

It’s 3/12 you just need to find common multiples

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3 years ago