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Lelechka [254]
2 years ago
13

Tamara and clyde got different answers when dividing 2x4 7x3 – 18x2 11x – 2 by 2x2 – 3x 1. analyze their individual work.

Mathematics
1 answer:
Harman [31]2 years ago
5 0

The process to divide the polynomial is discussed below:

What is Polynomial Division?

Polynomial long division is an algorithm for dividing a polynomial by another polynomial of the same or lower degree, a generalized version of the familiar arithmetic technique called long division.

Given:

( 2x^{4}+7x^{3}-18x^{2}+11x-2 )/ 2x^{2}-3x+1

Steps to follow:

Divide the leading term of the dividend by the leading term of the divisor: 2x^{4}/2x²=x².

Multiply it by the divisor: x²(2x²−3x+1)=2x^{4}−3x³+x².

Subtract the dividend from the obtained result:

( 2x^{4}+7x^{3}-18x^{2}+11x-2 )- ( x^{4}−3x³+x². ) =10x³−19x²+11x−2.

Divide the leading term of the obtained remainder by the leading term of the divisor: 10x³/2x²=5x.

Multiply it by the divisor: 5x(2x²−3x+1)=10x³−15x²+5x.

Subtract the remainder from the obtained result: (10x³−19x²+11x−2)−(10x³−15x²+5x)=−4x²+6x−2.

Divide the leading term of the obtained remainder by the leading term of the divisor: −4x²/2x²=−2.

Multiply it by the divisor: −2(2x²−3x+1)=−4x²+6x−2.

Subtract the remainder from the obtained result:

(−4x²+6x−2)−(−4x²+6x−2)=0.

Learn more about long division here:

brainly.com/question/14780388

#SPJ1

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Write the standard equation of a circle that passes through (−2, 10) with center (−2, 6).
Liono4ka [1.6K]
Standard form of a circle" (x-h)²+(y-k)²=r², (h,k) being the center, r being the radius.
in this case, h=-2, k=6, (x+2)²+(y-6)²=r²
use the point (-2,10) to find r: (-2+2)²+(10-6)²=r², r=4
so the equation of the circle is: (x+2)²+(y-6)²=4²
6 0
3 years ago
Read 2 more answers
Graph ∆KLM with vertices K(2,2), L(0,- 1), and M(-3,2). Graph ∆K'L'M' with a scale factor of 2. What are the coordinates of the
olasank [31]

Answer:

The coordinates of the new vertices are:

K'(4,4), L'(0,- 2), and M'(-6,4).

Step-by-step explanation:

In order to dilate a shape by 2, you can simply multiply the coordinates of the orignal triangle each by 2 to find the new coordinates of the new vertices.

K(2,2), L(0,- 1), and M(-3,2)

K'(2*2,2*2), L'(0*2,- 1*2), and M'(-3*2,2*2).

Now become...

K'(4,4), L'(0,- 2), and M'(-6,4).

You can now graph ∆K'L'M' using these coordinates.

5 0
2 years ago
There were 10 bunches of roses and tulips. Each bunch had 5 of the same flowers. There were 2 more bunches of tulips than roses.
____ [38]

Answer:

7

Step-by-step explanation:

6 0
3 years ago
*brainliest for best answer thank you* Solve for b.
vekshin1

Hello from MrBillDoesMath!

Answer:

The fourth choice,  b = +\- sqrt( sg + a^2)

Discussion:

s = (b^2 - a^2)/g     =>              multiply both sides by "g"

sg = b^2 - a^2        =>               add a^2 to both sides

sg + a^2 = b^2       =>                take the square root of each side

b = +\- sqrt( sg + a^2)


which is the fourth choice.


Thank you,

MrB

7 0
3 years ago
Giving BRAINLIEST to the CORRECT answer.
olga nikolaevna [1]

Answer:a

Step-by-step explanation:

it became wider by decreasing two on the graph.

7 0
3 years ago
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