Ask question

Ask question

All categories

3 years ago

6

Compute the slope of the line intersecting the points (1, 2) and (3, 8). *

1 answer:

DiKsa [7]3 years ago

6 0

Answer:

slope = 3

Step-by-step explanation:

Calculate the slope m using the slope formula

m = $\frac{y_{2}-y_{1} }{x_{2}-x_{1} }$

with (x₁, y₁ ) = (1, 2) and (x₂, y₂ ) = (3, 8)

m = $\frac{8-2}{3-1}$ = $\frac{6}{2}$ = 3

You might be interested in

Kareem ran three times last week. Here are the distances he rall

EleoNora [17]

Simply add the three decimals to find Kareem's total distance.

1.6 + 11.5 + 15 = 28.1

(0.6 + 0.5 = 1.1, 1 + 11 + 15 = 27, 27 + 1.1 = 28.1)

<h2>Answer:</h2>

<u>Kareem ran </u><u>28.1 km</u><u> on those three days.</u>

I hope this helps :)

8 0

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>

3 0

3 years ago

Solve the inequality. 22−2x<38 Which answer represents the solution set?

algol [13]

Answer:

x >-8

Step-by-step explanation:

22−2x<38

Subtract 22 from each side

22-22−2x<38-22

-2x < 16

Divide each side by -2, remembering to flip the inequality

-2x/-2 > 16/-2

x >-8

3 0

3 years ago

A newspaper’s cover page is 3/8 text, and photographs fill the rest. If 2/5 of the text is an article about endangered species,

jolli1 [7]

Answer:

$\frac{3}{20}$

Step-by-step explanation:

Let x be the cover page

We are given that a newspaper 's cover page is $\frac{3}{8}$ text and photograph fill the rest.

We have to find that how much part of cover's page is the article about endangered species

Text = $\frac{3}{8}x$

Remaining part = $x-\frac{3}{8} x$

Remaining part = $\frac{8x-3x}{8}=\frac{5}{8} x$

Therefore, a newspaper;s cover page fill with Photograph = $\frac{5}{8}$

If the article about endangered species= $\frac{2}{5}$ of text

Therefore, the article about endangered species= $\frac{2}{5}\times\frac{3x}{8}=\frac{3}{20}x$

Hence, the article about endangered species is $\frac{3}{20}$ of the cover page.

7 0

3 years ago

Read 2 more answers

Find the value of x please!! ASAP

cricket20 [7]

Answer:

x = 10

Step-by-step explanation:

By looking at the picture, I can tell the angle of the 3x is also 30 degrees.

This means 3x = 30.

When you solve for x by dividing each side by 3, you result with the answer of x = 10.

To check you work, plug in 10 for x to recieve 30 degrees again.

3 0

3 years ago