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3 years ago

13

Factor the polynomial below.

1 answer:

Zolol [24]3 years ago

4 0

Answer:

Answer is Option D. $6x^{2} y^{2}z(xy+2z^{2})$

Step-by-step explanation:

In the expression given we will do the factors of two polynomials first and then we will take the common factors out of the bracket.

First we do the factors of $6x^{3} y^{3}z$ =6× $x^{2}$ ×x× $y^{2}$ ×y×z

Now we do the factors of $12x^{2} y^{2} z^{3}$ =2×6× $x^{2}$ $y^{2}$ ×z× $z^{2}$

Now take the common factors of both the polynomials out of the bracket

⇒ $3×2x^{2} y^{2}z(xy+2z^{2})$

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Is there a relationship between area or perimeter of a rectangle

Arturiano [62]

Hello,

For a given perimeter (P) there are an infinity of Area (A)
Let's say x the length, and y the wide of the rectangle

P=2(x+y)
A=xy
k=x-y >=0

As (x+y)²-4xy=(x-y)²: A²-4P=k² or P=(A²-k²)/4
In primus, you will find a graph (abacus) giving P for a A and k given.
Negative Area or P are excluded.(just remind the first quadrant, A>=0 and P>=0)

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3 years ago

Choose all expressions that are equivalent to 0.8

Nataly [62]

0.80, 0.800, 8.0. They are all equil to 0.8(but 8.0 might not be, sorry)

3 0

3 years ago

Read 2 more answers

You kept track of the numbr of minutes you spent on homework for one week. 18, 20, 22, 11, 19, 18, 18 What is range of minutes y

frosja888 [35]

Answer:

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Step-by-step explanation:

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4 0

3 years ago

Evaluate the double integral.

Fynjy0 [20]

Answer:

$\iint_D 8y^2 \ dA = \dfrac{88}{3}$

Step-by-step explanation:

The equation of the line through the point $(x_o,y_o)$ & $(x_1,y_1)$ can be represented by:

$y-y_o = m(x - x_o)$

Making m the subject;

$m = \dfrac{y_1 - y_0}{x_1-x_0}$

∴

we need to carry out the equation of the line through (0,1) and (1,2)

i.e

y - 1 = m(x - 0)

y - 1 = mx

where;

$m= \dfrac{2-1}{1-0}$

m = 1

Thus;

y - 1 = (1)x

y - 1 = x ---- (1)

The equation of the line through (1,2) & (4,1) is:

y -2 = m (x - 1)

where;

$m = \dfrac{1-2}{4-1}$

$m = \dfrac{-1}{3}$

∴

$y-2 = -\dfrac{1}{3}(x-1)$

-3(y-2) = x - 1

-3y + 6 = x - 1

x = -3y + 7

Thus: for equation of two lines

x = y - 1

x = -3y + 7

i.e.

y - 1 = -3y + 7

y + 3y = 1 + 7

4y = 8

y = 2

Now, y ranges from 1 → 2 & x ranges from y - 1 to -3y + 7

∴

$\iint_D 8y^2 \ dA = \int^2_1 \int ^{-3y+7}_{y-1} \ 8y^2 \ dxdy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \int ^{-3y+7}_{y-1} \ y^2 \ dxdy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( \int^{-3y+7}_{y-1} \ dx \bigg) dy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( [xy^2]^{-3y+7}_{y-1} \bigg ) \ dy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( [y^2(-3y+7-y+1)]\bigg ) \ dy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ([y^2(-4y+8)] \bigg ) \ dy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( -4y^3+8y^2 \bigg ) \ dy$

$\iint_D 8y^2 \ dA =8 \bigg [\dfrac{ -4y^4}{4}+\dfrac{8y^3}{3} \bigg ]^2_1$

$\iint_D 8y^2 \ dA =8 \bigg [ -y^4+\dfrac{8y^3}{3} \bigg ]^2_1$

$\iint_D 8y^2 \ dA =8 \bigg [ -2^4+\dfrac{8(2)^3}{3} + 1^4- \dfrac{8\times (1)^3}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ -16+\dfrac{64}{3} + 1- \dfrac{8}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ -15+ \dfrac{64-8}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ -15+ \dfrac{56}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ \dfrac{-45+56}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ \dfrac{11}{3}\bigg]$

$\iint_D 8y^2 \ dA = \dfrac{88}{3}$

4 0

3 years ago

Find the midpoint of the segment with the following endpoints.<br> (-4, 5) and (-1, 1)

Solnce55 [7]

Answer:

(-5/2 , 2)

Step-by-step explanation:

add the two x coordinates of the endpoints and divide by 2. Do the same for the y coordinates.

6 0

3 years ago