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White raven [17]
2 years ago
15

This is for people who just want to talk

Mathematics
2 answers:
stiks02 [169]2 years ago
3 0

Answer:

hello

Step-by-step explanation:

hello is a greeting word for people that wants to be friendly to other people and the sigh language for hello is just waving your have for people you want to greet and another word for hello is hi, hi is a word for greeting also the sigh language of hi is waving your hand for example

Hello:

Hi:

34kurt2 years ago
3 0
Hey lol I’m doing paper work atm so I’m here to help me with it
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In a right triangle, the acute angles measure 2x+50 and 3x degrees
xenn [34]
2x+50 + 3x= 90 degrees
5x +50 = 90
5x =40
x=8

Angle1= 2(8) +50= 66
Angle2= 3(8)=24

66+24+90= 180
6 0
2 years ago
Read 2 more answers
You want to make a picture frame for a
docker41 [41]

Answer:

i think you'll need 1 foot 6 inches of wood

Step-by-step explanation:

7 0
3 years ago
What is 4/9 • t when t=1/3
Reptile [31]
\frac{4}{9}* \frac{1}{3}= \frac{4}{27}
8 0
3 years ago
Read 2 more answers
Use Stokes' Theorem to evaluate C F · dr where C is oriented counterclockwise as viewed from above. F(x, y, z) = yzi + 4xzj + ex
natima [27]

Answer:

The result of the integral is 81π

Step-by-step explanation:

We can use Stoke's Theorem to evaluate the given integral, thus we can write first the theorem:

\displaystyle \int\limits_C \vec F \cdot d\vec r = \int \int_S curl \vec F \cdot d\vec S

Finding the curl of F.

Given F(x,y,z) = < yz, 4xz, e^{xy} > we have:

curl \vec F =\left|\begin{array}{ccc} \hat i &\hat j&\hat k\\ \cfrac{\partial}{\partial x}& \cfrac{\partial}{\partial y}&\cfrac{\partial}{\partial z}\\yz&4xz&e^{xy}\end{array}\right|

Working with the determinant we get

curl \vec F = \left( \cfrac{\partial}{\partial y}e^{xy}-\cfrac{\partial}{\partial z}4xz\right) \hat i -\left(\cfrac{\partial}{\partial x}e^{xy}-\cfrac{\partial}{\partial z}yz \right) \hat j + \left(\cfrac{\partial}{\partial x} 4xz-\cfrac{\partial}{\partial y}yz \right) \hat k

Working with the partial derivatives

curl \vec F = \left(xe^{xy}-4x\right) \hat i -\left(ye^{xy}-y\right) \hat j + \left(4z-z\right) \hat k\\curl \vec F = \left(xe^{xy}-4x\right) \hat i -\left(ye^{xy}-y\right) \hat j + \left(3z\right) \hat k

Integrating using Stokes' Theorem

Now that we have the curl we can proceed integrating

\displaystyle \int\limits_C \vec F \cdot d\vec r = \int \int_S curl \vec F \cdot d\vec S

\displaystyle \int\limits_C \vec F \cdot d\vec r = \int \int_S curl \vec F \cdot \hat n dS

where the normal to the circle is just \hat n= \hat k since the normal is perpendicular to it, so we get

\displaystyle \int\limits_C \vec F \cdot d\vec r = \int \int_S \left(\left(xe^{xy}-4x\right) \hat i -\left(ye^{xy}-y\right) \hat j + \left(3z\right) \hat k\right) \cdot \hat k dS

Only the z-component will not be 0 after that dot product we get

\displaystyle \int\limits_C \vec F \cdot d\vec r = \int \int_S 3z dS

Since the circle is at z = 3 we can just write

\displaystyle \int\limits_C \vec F \cdot d\vec r = \int \int_S 3(3) dS\\\displaystyle \int\limits_C \vec F \cdot d\vec r = 9\int \int_S dS

Thus the integral represents the area of a circle, the given circle x^2+y^2 = 9 has a radius r = 3, so its area is A = \pi r^2 = 9\pi, so we get

\displaystyle \int\limits_C \vec F \cdot d\vec r = 9(9\pi)\\\displaystyle \int\limits_C \vec F \cdot d\vec r = 81 \pi

Thus the result of the integral is 81π

5 0
3 years ago
The percentage of people given an antirheumatoid who suffer severe, moderate, or minor side effects are 11, 20 and 69%, respecti
Snowcat [4.5K]

Answer:2.9235

Step-by-step explanation:

Since 20 people are administered the medicine and;

11% of the 20 people suffer severe i.e 11% of 20 = 2.2 people

20% of them suffer moderate i.e

20% of 20 = 4 people and;

69% of the 20 people suffer minor side i.e

69% of 20 = 13.8 people

Therefore the probability 2, 4, and 14 people will suffer severe, moderate, or minor side effects, respectively will be;

2/2.2 + 4/4 + 14/13.8 (i.e possible outcome/total outcome)

This will give us 2.9235

6 0
3 years ago
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