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3 years ago

Help me please explain do not need work just explain THANK YOU I NEED HELP

Mathematics

1 answer:

Mnenie [13.5K]3 years ago

6 0

Alright,

you will need to solve for each variable to find if they got the equation correctly or not..

Let's start with h
isolate h
(1/2)h = A/(b1+b2)
Now we need to get rid of 1/2 we may do so by multiplying both sides by 2

h = 2(A/(b1+b2))

And that's not how they did it for h so Option D doesn't apply

Let's see C
Also doesn't apply since they multiplied 2 only by A rather than A/h

Let's see B
Also not correct they made the same mistake as with option C

Let's see A
again same mistake
when multiplying both sides by 2
the 2 should be as following
$2\frac{A}{h}$

I believe you should contact your instructor and explain that non of the options is right.

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Answer:

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Step-by-step explanation:

I don't know what you mean by sketch pad but here is the working out

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3 years ago

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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):

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:

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