All categories

3 years ago

Please answer asap!!!!!!!!

Mathematics

2 answers:

pentagon [3]3 years ago

4 0

Jeff’s box holds more popcorn.

bazaltina [42]3 years ago

3 0

I think Jeff’s because it’s a deeper bag which can hold more

You might be interested in

How to put 17/4 into a mixed number

KiRa [710]

Divide the top by bottom. So 17 divided by 4. Which would be around 4.25

4 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):

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

Hi guys can you answer my math question

Nikitich [7]

Answer:

Hindi KO po Alam yan

Step-by-step explanation:

pagaralan mo nang mabuti

8 0

3 years ago

Read 2 more answers

I need the answer fast pls

ludmilkaskok [199]

A. C. D. E. F.

These all have variability.

6 0

2 years ago

Read 2 more answers

Please help me with this.A rectangular field is

Karolina [17]

First, let's get the perimeter of the rectangle:
$P=2W+2L$
$P=130m+210m$
$P=340m$
Then, let's get the area of the bigger one:
$A=WL$
$A=65m*105m$
$A=6825m^2$

Then let's try using a rectangle with a smaller ratio:
$P=100m+240m$
$P=340m$
Then:
$A=50m*120m$
$A=6000m^2$

If you used a square:
$P=170+170$
$P=340$
$A=WL$
$A=85^2$
$A=7225$

There you have it. A rectangle with a smaller area with the same perimeter.
What does it show? The smaller the difference you get from width and length, the larger the area is.

5 0

3 years ago

Read 2 more answers