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Lemur [1.5K]
3 years ago
15

store one is selling three notebooks for $5.13 store to is selling 5 notebooks for $8.40 store 3 7/8 notebooks for $14 in which

store are the notebooks the cheapest​
Mathematics
1 answer:
sdas [7]3 years ago
5 0

Answer:

store 2

Step-by-step explanation:

This is because store one sells notebook for 5.13. Store 2 sells one notebook for 1.68, and the last stores sells one notebook for 2 or 1.75. Meaning that store 2 sells it for cheap.

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36 - 12%
x - 100%

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Given the flag ABCDE, determine the single rule (x,y) that creates the
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Step-by-step explanation:

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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

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

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3 years ago
At the arcade, nick spent 203 tokens on videogames in seven days. each day he spent five more tokens than the previous day. how
Dafna11 [192]

Suppose nick spent x number of tokens on day 1

So he spent x + 5 on day 2

He spent x + 10 on day 3

He spent x + 15 on day 4

He spent x + 20 on day 5

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And on day 7 he spent x + 30

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7x + 105 = 203

Subtracting 105 from both sides

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7x = 98

Dividing by 7 on both sides

x = 14

So on third day he spent x + 10

= 14 + 10

= 24 tokens

Tokens spent on day 3 = 24

8 0
3 years ago
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