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Fiesta28 [93]
3 years ago
5

What is the quarter of 100?

Mathematics
2 answers:
Ghella [55]3 years ago
8 0

A quarter or 1/4 of 100 is 25

You can determine this by multiplying 25 by 4 which gives you 100

Hope this helps you

-AaronWiseIsBae

aalyn [17]3 years ago
6 0
the answer is 25 I think
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Find the surface area of each figure. Round your answer to the nearest tenth if necessary.​
dedylja [7]

Answer:

556

Step-by-step explanation:

Surface area of the cuboid=2*(lb+bh+lh)

Surface area=2*(110+80+88)=556

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Mekhanik [1.2K]
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This is to confusing​
erastova [34]

Answer:

<u>40%</u>  

Step-by-step explanation:

because there is 10 parts and if the 10 multipied by 10 (to find the percent) is 100 then the 4 is 40 (the unshaded part would be 60%)

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3 years ago
I need help with this pleasee
Sauron [17]

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120

Step-by-step explanation:

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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

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
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