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4 years ago

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Suppose that the relation G is defined as follows. G= (-1, -4), (8,4), (-4,4) Domain= Range=

1 answer:

il63 [147K]4 years ago

5 0

Answer:

Domain = {-4,-1,8}

range = {4,-4,4}

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3 years ago

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>

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3 years ago

Factor: 3.x2 + 7x + 2

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Answer:

x= -1/3 and -2

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3 years ago

Andrea put 2 quarts of water and 1 1/2 cups of lemon juice in a pitcher. How many total liquids are in the pitcher

gizmo_the_mogwai [7]

Key: 1 quart = 4 cups

So two quarts would equal 8 cups...So we add...

8 + 1 1/2 = 9 1/2

9 1/2 is the total. ^^

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4 years ago

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