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ziro4ka [17]
3 years ago
11

Se

Mathematics
1 answer:
Musya8 [376]3 years ago
4 0

Answer:

No, this is not a right triangle.

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Tpy6a [65]

\bf \begin{array}{|cc|ll} \cline{1-2} x&y\\ \cline{1-2} 40&32\\ 28&16\\ 16&12\\ \cline{1-2} \end{array}~\hfill \stackrel{\textit{average rate of change}}{slope} \\\\[-0.35em] ~\dotfill\\\\ (\stackrel{x_1}{40}~,~\stackrel{y_1}{32})\qquad (\stackrel{x_2}{28}~,~\stackrel{y_2}{16}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{16-32}{28-40}\implies \cfrac{-16}{-12}\implies \boxed{\cfrac{4}{3}} \\\\[-0.35em] ~\dotfill

\bf (\stackrel{x_1}{28}~,~\stackrel{y_1}{16})\qquad (\stackrel{x_2}{16}~,~\stackrel{y_2}{12}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{12-16}{16-28}\implies \cfrac{-4}{-12}\implies \boxed{\cfrac{1}{3}}

8 0
3 years ago
True or False: No matter which factors you start with, you will always end up with the same numbers in the end using prime facto
seropon [69]

Answer:

True

Step-by-step explanation:

In Prime factorization, we are expected to obtain factors that are prime numbers that can multiply themselves to give the original number. So long as the first factor can divide the number without a remainder, other remaining factors can be multiplied together to give the original number.

Prime factorization of the number, 15 goes thus;

15/3=5

5/5=1

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Each week. You make $10.00 but you spend $3.00 and save the rest. How much would you save in 8 weeks?
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7 0
3 years ago
Read 2 more answers
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
What is a quotient of a number and 15?
Alenkinab [10]

Answer:450 i think


Step-by-step explanation:

Quotient' means you will divide the numbers.

Also, to be 'no greater' means to be 'less than or equal to'.

6 0
3 years ago
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