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yarga [219]
4 years ago
14

Which of the following polynomials corresponds to the subtraction of the multivariate polynomials 19x^3+44x^2y+17 and y^3-11xy^2

+2x^2y-13x^3
A. y^3-6x^3+33x^2y+2x^2y+17

B. 20x^3-y^3+33x^2y+2x^2y+17

C. 31x^3-6x^3+44x^2y+11x^2y+17

D. 32x^3-y^3+42x^2y+11x^2y+17
Mathematics
1 answer:
VARVARA [1.3K]4 years ago
7 0
Name the two given polynomials as
f(x) = 19x³ + 44x²y + 17
g(x) = y³ - 11xy² + 2x²y - 13x³

Create the subtraction f(x) - g(x).
f(x)-g(x) = 19x³ + 44x²y + 17 - (y³ - 11xy² + 2x²y - 13x³)
             = (19 + 13)x³ + (44 - 2)x²y + 11xy² - y³ + 17
             = 32x³ - y³ + 42x²y + 11xy² + 17

Answer: D
Note that answer D has a typo. 11xy² is correct, but 11x²y is not.

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If sin f° = 8/9 and the measure of YW is 24 units, what is the measure of YX?

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Drag the tiles to the correct boxes to complete the pairs. Find the distance between each pair of points. 5 units 4 units 2 unit
Olin [163]

Answer:

a) Distance between points A (5, 4) and B( 5, -2) is 6 units

b) Distance between points E (-2, -1) and F( -2, -5) is 4 units

c) Distance between points C (-4, 1) and D( 1, 1) is 5 units

d) Distance between points G(3, -5) and H(6, -5) is 3 units

Step-by-step explanation:

We need to find the distance between each pair

a) A (5, 4) and B( 5, -2)

the distance formula is:

d(A,B) = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}

Putting values: x₁ = 5, x₂=5 and y₁= 4 and y₂= -2

d(A,B) = \sqrt{(5-5)^2+(-2-4)^2}

d(A,B) = \sqrt{(0)^2+(-6)^2}

d(A,B) = \sqrt{36}

d(A,B) = 6

Distance between points A (5, 4) and B( 5, -2) is 6 units

b)  E (-2, -1) and F( -2, -5)

the distance formula is:

d(E,F) = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}

Putting values: x₁ = -2, x₂=-2 and y₁= -1 and y₂= -5

d(E,F) = \sqrt{(-2-(-2))^2+(-5-(-1))^2}

d(E,F) = \sqrt{(0)^2+(-4)^2}

d(E,F) = \sqrt{16}

d(E,F) = 4

Distance between points E (-2, -1) and F( -2, -5) is 4 units

c)  C (-4, 1) and D( 1, 1)

the distance formula is:

d(C,D) = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}

Putting values: x₁ = -4, x₂=1 and y₁= 1 and y₂= 1

d(C,D)= \sqrt{(1-(-4))^2+(1-(1))^2}

d(C,D) = \sqrt{(5)^2+(0)^2}

d(C,D) = \sqrt{25}

d(C,D) = 5

Distance between points C (-4, 1) and D( 1, 1) is 5 units

d) G(3, -5) and H(6, -5)

the distance formula is:

d(G,H) = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}

Putting values: x₁ = 3, x₂=6 and y₁= -5 and y₂= -5

d(G,H)= \sqrt{(6-(3))^2+(-5-(-5))^2}

d(G,H) = \sqrt{(3)^2+(0)^2}

d(G,H) = \sqrt{9}

d(G,H) = 3

Distance between points G(3, -5) and H(6, -5) is 3 units

8 0
3 years ago
Line g is shown on the graph.
alisha [4.7K]

The linear equation that defines the line G can be written as:

y = -(1/3)*x + 5

<h3>How to get the equation of the line G?</h3>

A general linear equation written in the slope-intercept form is:

y = a*x + b

Where a and b are real numbers, a is the slope and b is the y-intercept.

If we know that the line passes through two points (x₁, y₁) and (x₂, y₂), then the slope is:

a = (y₂ - y₁)/(x₂ - x₁)

In this case we know that the line passes through (-3, 6) and (0, 5), then the slope is:

a = (5 - 6)/(0 - (-3)) = -1/3

The linear equation is something like:

y = (-1/3)*x + b

We know that it passes through (0, 5), then:

5 = (-1/3)*0 + b

5 = b

We conclude that the linear equation can be written as:

y = -(1/3)*x + 5

If you want to learn more about linear equations:

brainly.com/question/1884491

#SPJ1

7 0
1 year ago
How to solve <br> -3(4R-8)-36
natima [27]

Answer: -12r-12

Step-by-step explanation: There is no equals sign so you can't solve it but you can simplify it. You start out with -3(4r-8)-36. Then, you use PEMDAS. (-3×4r)+(-3×-8)-36. Then, it turns into -12r×+24-36. You just continue to simplify until you can't anymore. You end up with -12r-12.

7 0
4 years ago
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