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3 years ago

8

Which expression represents the correct form for the quotient and remainder, written as partial fractions, of StartFraction 6 x

cubed + 15 x squared + 20 x + 36 Over 3 x squared + 7 EndFraction? 2 x + StartFraction A Over 3 x squared + 7 EndFraction 2 x + StartFraction A x + B Over 3 x squared + 7 EndFraction 2 x + 5 + StartFraction A Over 3 x squared + 7 EndFraction 2 x + 5 + StartFraction A x + B Over 3 x squared + 7 EndFraction

2 answers:

diamong [38]3 years ago

8 0

Answer:

The answer is "quotient =2x+5 and remainder =6x+1".

Step-by-step explanation:

Some of the data is missing that's why defined this question as follows:

Given:

$\to \frac{6x^3+15x^2+20x+36} {3x^2+7}$

when we divide the above value it will give the quotient =2x+5 and remainder =6x+1.

N76 [4]3 years ago

4 0

Answer:

Its D

Step-by-step explanation:

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Find the mass and center of mass of the lamina that occupies the region D and has the given density function rho. D is the trian

Alla [95]

Answer: mass (m) = 4 kg

center of mass coordinate: (15.75,4.5)

Step-by-step explanation: As a surface, a lamina has 2 dimensions (x,y) and a density function.

The region D is shown in the attachment.

From the image of the triangle, lamina is limited at x-axis: 0≤x≤2

At y-axis, it is limited by the lines formed between (0,0) and (2,1) and (2,1) and (0.3):

<u>Points (0,0) and (2,1):</u>

y = $\frac{1-0}{2-0}(x-0)$

y = $\frac{x}{2}$

<u>Points (2,1) and (0,3):</u>

y = $\frac{3-1}{0-2}(x-0) + 3$

y = -x + 3

Now, find total mass, which is given by the formula:

$m = \int\limits^a_b {\int\limits^a_b {\rho(x,y)} \, dA }$

Calculating for the limits above:

$m = \int\limits^2_0 {\int\limits^a_\frac{x}{2} {2(x+y)} \, dy \, dx }$

where a = -x+3

$m = 2.\int\limits^2_0 {\int\limits^a_\frac{x}{2} {(xy+\frac{y^{2}}{2} )} \, dx }$

$m = 2.\int\limits^2_0 {(-x^{2}-\frac{x^{2}}{2}+3x )} \, dx }$

$m = 2.\int\limits^2_0 {(\frac{-3x^{2}}{2}+3x)} \, dx }$

$m = 2.(\frac{-3.2^{2}}{2}+3.2-0)$

m = 2(-4+6)

m = 4

<u>Mass of the lamina that occupies region D is 4.</u>

<u />

Center of mass is the point of gravity of an object if it is in an uniform gravitational field. For the lamina, or any other 2 dimensional object, center of mass is calculated by:

$M_{x} = \int\limits^a_b {\int\limits^a_b {y.\rho(x,y)} \, dA }$

$M_{y} = \int\limits^a_b {\int\limits^a_b {x.\rho(x,y)} \, dA }$

$M_{x}$ and $M_{y}$ are moments of the lamina about x-axis and y-axis, respectively.

Calculating moments:

For moment about x-axis:

$M_{x} = \int\limits^a_b {\int\limits^a_b {y.\rho(x,y)} \, dA }$

$M_{x} = \int\limits^2_0 {\int\limits^a_\frac{x}{2} {2.y.(x+y)} \, dy\, dx }$

$M_{x} = 2\int\limits^2_0 {\int\limits^a_\frac{x}{2} {y.x+y^{2}} \, dy\, dx }$

$M_{x} = 2\int\limits^2_0 { ({\frac{y^{2}x}{2}+\frac{y^{3}}{3})}\, dx }$

$M_{x} = 2\int\limits^2_0 { ({\frac{x(-x+3)^{2}}{2}+\frac{(-x+3)^{3}}{3} -\frac{x^{3}}{8}-\frac{x^{3}}{24} )}\, dx }$

$M_{x} = 2.(\frac{-9.x^{2}}{4}+9x)$

$M_{x} = 2.(\frac{-9.2^{2}}{4}+9.2)$

$M_{x} = 18$

Now to find the x-coordinate:

x = $\frac{M_{y}}{m}$

x = $\frac{63}{4}$

x = 15.75

For moment about the y-axis:

$M_{y} = \int\limits^2_0 {\int\limits^a_\frac{x}{2} {2x.(x+y))} \, dy\,dx }$

$M_{y} = 2.\int\limits^2_0 {\int\limits^a_\frac{x}{2} {x^{2}+yx} \, dy\,dx }$

$M_{y} = 2.\int\limits^2_0 {y.x^{2}+x.{\frac{y^{2}}{2} } } \,dx }$

$M_{y} = 2.\int\limits^2_0 {x^{2}.(-x+3)+\frac{x.(-x+3)^{2}}{2} - {\frac{x^{3}}{2}-\frac{x^{3}}{8} } } \,dx }$

$M_{y} = 2.\int\limits^2_0 {\frac{-9x^3}{8}+\frac{9x}{2} } \,dx }$

$M_{y} = 2.({\frac{-9x^4}{32}+9x^{2})$

$M_{y} = 2.({\frac{-9.2^4}{32}+9.2^{2}-0)$

$M{y} =$ 63

To find y-coordinate:

y = $\frac{M_{x}}{m}$

y = $\frac{18}{4}$

y = 4.5

<u>Center mass coordinates for the lamina are (15.75,4.5)</u>

3 0

3 years ago

Distance between two points (−5,−5) and (-9, -2)

sergey [27]

Answer: 5

Step-by-step explanation: d = √(x^²-x^₁)^2+(y^²-y^₁)^2 d = √ (-9--5)^2+ (-2 - -5)^2 d = √(-4)^2 + (3)^2 d = √16 + 9 d = √25 d = 5

8 0

2 years ago

Which function passes through the points (2, 15) and (3, 26)?

igor_vitrenko [27]

<span>Which function passes through the points (2, 15) and (3, 26)?
</span>
Its B

7 0

3 years ago

Read 2 more answers

Michael’s weight can be represented by the expression 72x^5. Al’s weight can be represented by the expression 9x^7.

podryga [215]

Answer:

a) $\frac{72x^5}{9x^7}$

b) $=\frac{8}{x^2}$

c), Yes; $8x^2$

Step-by-step explanation:

Michael's Weight: $72x^5$

Al's weight: $9x^7$

a) Ratio of Michael's weight to Al's weight: $\frac{72x^5}{9x^7}$

b) This ratio simplifies to: $\frac{8\times 9x^5}{9x^5\timesx^2}$

$=\frac{8}{x^2}$

c) Yes, If the exponent in each expression were negative, then we have:

Ratio of Michael's weight to Al's weight: $\frac{72x^{-5}}{9x^{-7}}$

This ratio simplifies to: $8x^{-5--7}=8x^2$

The two ratios are not the same.

3 0

3 years ago

The diameter of a circle is 75 feet. What is

yan [13]

Answer:

option 5

Step-by-step explanation:

we have diameter=75 feet

we need to find the radius

radius=diameter/2

so we write

radius=75 feet/2

finally

we have

radius=37.5 feet

4 0

3 years ago