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solong [7]
2 years ago
12

8xy^3*xy ^8 plsssssssssssssssssssss help

Mathematics
1 answer:
Sindrei [870]2 years ago
4 0

Answer:

x • (x^3 - 3x^2y + 3xy^2 + 7y^3)

Step-by-step explanation:

(8x • (y^3)) +  x • (x - y)^3

2^3xy^3 +  x • (x - y)^3

Evaluate :  (x-y)^3   =   x^3-3x^2y+3xy^2-y^3

Pull out like factors :

x^4 - 3x^3y + 3x^2y^2 + 7xy^3  =

x • (x^3 - 3x^2y + 3xy^2 + 7y^3)

x^3 - 3x^2y + 3xy^2 + 7y^3 is not a perfect cube

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Write an equation in slope-intercept form that describes the line with a slope of -2 and (5,4) is on the line
marin [14]

Answer:

The equation in slope-intercept form will be:

  • y = -2x + 14

Step-by-step explanation:

Given

  • slope = m = -2
  • point (5, 4)

We know that the slope-intercept form of the line equation is

y = mx+b

where m is the slope of the line and b is the y-intercept

substituting m = -2 and the point (5, 4) in the slope-intercept form to get the y-intercept 'b'

y = mx+b

4 = -2(5) + b

4 = -10 + b

b = 4+10

b = 14

now substituting m = -2 and b = 14  in the slope-intercept form to get the equation in slope-intercept

y = mx+b

y = -2x + 14

Therefore, the equation in slope-intercept form will be:

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3 0
3 years ago
Any got an answer for this math question?
Kamila [148]

Answer:the answer  this is negative 3

Step-by-step explanation:

3 0
3 years ago
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30 POINTS AVAILABLE
kherson [118]

Answer:

\large\boxed{(x-2)^2+(y-1)^2=34}

Step-by-step explanation:

The equation of a circle in standard form:

(x-h)^2+(y-k)^2=r^2

(h, k) - center

r - radius

We have the endpoints of the diameter: (-1, 6) and (5, -4).

Midpoint of diameter is a center of a circle.

The formula of a midpoint:

\left(\dfrac{x_1+x_2}{2};\ \dfrac{y_1+y_2}{2}\right)

Substitute:

h=\dfrac{-1+5}{2}=\dfrac{4}{2}=2\\\\k=\dfrac{6+(-4)}{2}=\dfrac{2}{2}=1

The center is in (2, 1).

The radius length is equal to the distance between the center of the circle and the endpoint of the diameter.

The formula of a distance between two points:

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Substitute the coordinates of the points (2, 1) and (5, -4):

r=\sqrt{(5-2)^2+(-4-1)^2}=\sqrt{3^2+(-5)^2}=\sqrt{9+25}=\sqrt{34}

Finally we have:

(x-2)^2+(y-1)^2=(\sqrt{34})^2

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