What's 400 divided by 5 show your work

Mathematics

1 answer:

vlada-n [284]2 years ago

6 0

5 goes into 40 8 times than add a zero

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Solving systems of equations with substitution: y= 2x-4 y= -x+5 help please!

Furkat [3]

y= 2x-4
y= -x+5
so

2x-4 = -x+5
2x +x = 5 +4
3x = 9
x =3

y = 2x-4
y =2(3) - 4
y = 6-4
y = 2

answer: (3,2)

4 0

2 years ago

One number is 16 more than a second number. The sum of the first number and twice the second number is 139. What is the answer ?

Nezavi [6.7K]

Answer first number = 57 second number = 41
x= 16 + y
x + 2y = 139

16+ y + 2y =139
16 + 3y =139
16-16 +3y = 139-16
3y = 123
3/3y = 123/3
y= 41

x= 16+y
x = 16+41
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2 years ago

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

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