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3 years ago

Please answer ASAP!

Mathematics

1 answer:

romanna [79]3 years ago

8 0

Answer:

volume is the product of the 3 dimensions. so devide the volume by the height and by the length to get just the width

1176 / 14 / 8

= 10.5

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In circle O, RT and SU are diameters. If m = m, what is m? 47° 52° 64° 87°

vlada-n [284]

From the given figure, it can be seen that 13x = 15x - 8 because they are vertical angles and thus are equal.

13x = 15x - 8
15x - 13x = 8
2x = 8
x = 8/2 = 4

Thus, 15x - 8 = 15(4) - 8 = 60 - 8 = 52.

RT is a diameter, which means that mRT = 180

mRV + mVU + 52 = 180

mRV + mVU = 180 - 52 = 128

Now, given that mRV = mVU,

Thus, 2mVU = 128

Therefore, mVU = 128 / 2 = 64°

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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):

Points (0,0) and (2,1):

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

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

Points (2,1) and (0,3):

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

Mass of the lamina that occupies region D is 4.

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:

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