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2 years ago

What is the greatest common factor of 3x^4, 15x^3", and 21x^2??

Mathematics

1 answer:

Umnica [9.8K]2 years ago

5 0

Answer:3x^2 each coefficient is divisible by 3 and and you can take x^2 from both of them

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What is 3,996 + 7899

ddd [48]

3996 + 7899 = 11895

:)

3 0

3 years ago

Read 2 more answers

Use the general slicing method to find the volume of the following solids. The solid whose base is the region bounded by the cur

Y_Kistochka [10]

Answer:

The volume is $V=\frac{64}{15}$

Step-by-step explanation:

The General Slicing Method is given by

<em>Suppose a solid object extends from x = a to x = b and the cross section of the solid perpendicular to the x-axis has an area given by a function A that is integrable on [a, b]. The volume of the solid is</em>

$V=\int\limits^b_a {A(x)} \, dx$

Because a typical cross section perpendicular to the x-axis is a square disk (according with the graph below), the area of a cross section is

The key observation is that the width is the distance between the upper bounding curve $y = 2 - x^2$ and the lower bounding curve $y = x^2$

The width of each square is given by

$w=(2-x^2)-x^2=2-2x^2$

This means that the area of the square cross section at the point x is

$A(x)=(2-2x^2)^2$

The intersection points of the two bounding curves satisfy $2 - x^2=x^2$ , which has solutions x = ±1.

$2-x^2=x^2\\-2x^2=-2\\\frac{-2x^2}{-2}=\frac{-2}{-2}\\x^2=1\\\\x=\sqrt{1},\:x=-\sqrt{1}$

Therefore, the cross sections lie between x = -1 and x = 1. Integrating the cross-sectional areas, the volume of the solid is

$V=\int\limits^{1}_{-1} {(2-2x^2)^2} \, dx\\\\V=\int _{-1}^14-8x^2+4x^4dx\\\\V=\int _{-1}^14dx-\int _{-1}^18x^2dx+\int _{-1}^14x^4dx\\\\V=\left[4x\right]^1_{-1}-8\left[\frac{x^3}{3}\right]^1_{-1}+4\left[\frac{x^5}{5}\right]^1_{-1}\\\\V=8-\frac{16}{3}+\frac{8}{5}\\\\V=\frac{64}{15}$

5 0

3 years ago

Find the slope of the line that contains these points.(5,1) (-3,1)

shtirl [24]

Slope equals change in y divided by change in x.

m=(y2-y1)/(x2-x1)

m=(1-1)/(5--3)

m=0/8

m=0

The slope is zero.

So this is a horizontal line of the form y=1

3 0

3 years ago

I need help with this question

Vesnalui [34]

That is not a question

3 0

2 years ago

Read 2 more answers

what is the equation of the line in point slope form that goes through (-2,-1) and is perpendicular to y=5x+3?

stich3 [128]

Point slope form is $y-y1=m(x-x1)$

Since we have to write the equation of a perpendicular line we will have to use the slope that is opposite and reciprocal.

$y - (-1) = - \frac{1}{5} (x - (-2))$
$y + 1 = - \frac{1}{5} (x + 2)$

Hope this helps :)

8 0

2 years ago