All categories

2 years ago

Explain how Han Fei Tzu's beliefs differed from the beliefs of Confucius. Be sure to cite specific examples from your lesson.

1 answer:

4 0

Answer: Han Fei Tzu was the creator of Legalism. Han Fei Tzu believed that people needed harsh rules and punishments just for people to do their everyday jobs. Confucius was the founder of Confucianism. Confucius believed that the duties to the family are important.

Step-by-step explanation: This is right bro.

You might be interested in

What digit can be in the ones place of a number that has 5 as a factor

saveliy_v [14]

Only numbers possible would be five and zero

8 0

3 years ago

2x+1> <img src="https://tex.z-dn.net/?f=2x%20%2B%201%20%5Cgeqslant%205%20-%20x" id="TexFormula1" title="2x + 1 \geqslant 5

melisa1 [442]

What are we supposed solve for? Be for specific please

4 0

3 years ago

Candy is at the balloon shop and sees that 10 balloons cost 0.15 he needs 50 balloons to decorate his room how much will 50 ball

Kryger [21]

50/10 = 5
5 * 0.15 = 0.75
it will cost 75 cents

8 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

Due tomorrow please help :)

coldgirl [10]

Okay, so your first one, number six. You have two shapes there, a cone and a circle. Now you have a full cone and a half of a circle. Let's start with the cone. The formula for the volume of a cone is πr²h/3. You have your height, that is 12, so get your paper out and plug that in. H = 12. Now, your radius is where it gets a bit tricky. Have you ever heard of the Pythagorean theorem? It's a²+b²=c². Look at the cone, but look at it like it's a two-dimensional shape. You see the triangle in the middle, ignore the rest and pretend it's only that triangle. You now have a right triangle, You know your height is 12, and by looking at the triangle adjacent to it, and know it's equal, you know that your hypotenuse (the longest side of your triangle) is 15. So plug that into the Pythagorean theorem. Now you have 12²+b²=15². Use algebra to work that out, and you're left with 9. So your radius is 9. Now plug that into your volume of a cone equation. The volume of the cone is about 1017.88. Now find the circle or sphere. The formula is 4/3πr³. Just plug in your radius and get your answer, which is 3053.63. Now cut that in half, as you've only got half a circle. Now you've got 1526.82. Now add that to the volume of your cone, and you've got your final answer. 2544.7, or about that. I used π on a calc, and not 3.14, so it might be a little different.

Do you think you can do the next one? Just look at your triangle in there. You've already got one side and a hypotenuse, so plug it into the Pythagorean theorem and get your last side. Then look at the little marks on there and you can see if places are the same. If you need help with it don't hesitate to ask. I'm staying up all night anyway. Homeschooled and behind, so I'm not going anywhere.

7 0

2 years ago

Explain how Han Fei Tzu's beliefs differed from the beliefs of Confucius. Be sure to cite specific examples from your lesson.​

Explain how Han Fei Tzu's beliefs differed from the beliefs of Confucius. Be sure to cite specific examples from your lesson.