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3 years ago

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Mathematics

2 answers:

bazaltina [42]3 years ago

3 0

Answer:

Step-by-step explanation:

First cuz the y-intercept is 3 so not B or D

And the slope is -2/3

so the answer is C.

i'm not really sure about the slope part.

zimovet [89]3 years ago

3 0

A mañanansnsnsjS2927410

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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:

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

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

$M_{x}$ 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

Center mass coordinates for the lamina are (15.75,4.5)

3 0

3 years ago

4.Find the length of the diagonal of a rectangle having dimensions 16 cm and 12 cm.

nignag [31]

Answer:

20 cm

Step-by-step explanation:

So from the problem we get that the two sides of the rctangle are 16 cm 12 cm which are base perpendicular of the right angle triangle. H = 20 cm. So the diagonal of the rectangle = 20 cm.

7 0

3 years ago

Read 2 more answers

6. If n (E) = 40, n (A) = 22, n (ANB) = 8 and n ((AUB)') = 6, determine n(B)

Dimas [21]

Answer:

Step-by-step explanation:

First n(AUB)=n(E) - n((AUB)')

So n(AUB) = 40 - 6 = 34

Now all u hv to do is use the formulae,

n(AUB) = n(A) + n(B) - n(ANB)

So when u substitute above values,

34=22 + n(B) - 8

So,

n(B) = 4

4 0

3 years ago

Which statement is true about the expression 12- 7 + 3?

Mrrafil [7]

Answer:

I don't see any statements here.

Anyway, the answer is 8.

$12 - 7 + 3 = 8$

3 0

3 years ago

10) Show that in a group of 10 people (where any two people are either friends or enemies), there are either three mutual friend

ddd [48]

Answer with explanation:

Number of People in the group =10 People

The combination between two people is that, they can be either friends or enemies.

Total number of Possible Relation

=18 +16+14+12+10+8+6+4+2

= 90 Relations in all.

Out of 90 relations , 45 will be friends and 45 will be enemies.

⇒Now, we have to prove that, between 10 people, there are either three mutual friends or four mutual enemies, and there are either three mutual enemies or four mutual friends.

→→If there are three mutual friends, total number of people in the group =3 ×2=6 people in the group

And, four mutual enemies in a group means there are 2 people in each group.

So, total number of people if we combine the two groups in which there are either three mutual friends or four mutual enemies

=6 +4

=10

Hence proved.

→→→→Second part is ,in this group there can be either three mutual enemies or four mutual friends.

⇒If there are three mutual enemies, total number of people in the group =3 ×2=6 people in the group

And, four mutual friends in a group means there are 2 people in each group.

So, total number of people if we combine the two groups in which there are either three mutual enemies or four mutual enemies

=6 +4

=10

Hence proved.

3 0

3 years ago