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2 years ago

Pls help me! I'm confused

Mathematics

1 answer:

inna [77]2 years ago

5 0

Just put the numbers in order from greatest to least

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Find the value of x:

nadya68 [22]

Answer:

x = 3

Step-by-step explanation:

These angles are vertical angles. Vertical angles equals to on another but on the opposite side. Since they equal to another you do:

(10x + 1) = (12x - 5)

Add 5 on both sides:

10x + 6 = 12x

Subtract 10x on both sides:

6 = 2x

Divide 2 on both sides:

3 = x

6 0

3 years ago

Read 2 more answers

Three different whole numbers add up to 31.

lubasha [3.4K]

Let's find the least possibilities:

First number: 6
Second number: 8
Third number: 10

6 + 8 + 10 = 24

Now we can see that we are still missing 31 - 24 = 7.

7 can be gained by adding 3 and 4, so:

First number: 6 + 3 = 9
Second number: 8 + 4 = 12
Third number: 10

9 + 12 + 10 = 31

5 0

2 years ago

frank is buying a couch for his new apartment. the couch costs d dollars, and the sale tax is 8%. Select all of the following th

givi [52]

Answer:

Step-by-step explanation On the first day of Christmas my true love sent to me love me:

6 0

3 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

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