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WARRIOR [948]
3 years ago
10

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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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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