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Artemon [7]
3 years ago
11

Eber ran 3 miles. He stopped to take a break every {3}{4) mile.How many times did Eber stop, including his final stop at the end

of the run?
Mathematics
2 answers:
Darina [25.2K]3 years ago
6 0

Answer:

The Anwser is UNKNOWN

Step-by-step explanation:

sveta [45]3 years ago
3 0

Answer:

4 times

Step-by-step explanation:

The total model is 3 miles, divided by 3/4 equals the multiplication problem 3/1 times 3/4 but use the reciprocal of 3/4 which is 4/3 so it would be 3/1 times 4/3 which is 12/3 which is equivalent to 4. :)

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77julia77 [94]

Answer:

The counterclockwise circulation is \frac{7}{12} and the outward flux is \frac{11}{15}

Step-by-step explanation:

We are given the field F(x,y) = (7xy,2y^2). A picture of the region and the path we are considering is attached. Recalll the following theorems.

Given a field of the form F(x,y)=(f(x,y),g(x,y) with f,g having continous partial derivates, C is a closed path counterclockwise oriented, R is the region enclosed by C and n is the normal vector pointing outwards of the path C. Then

\oint_C F\cdot dr =\iint_R \frac{\partial f}{dy}- \frac{\partial g}{dx} dA(this one calculates the counterclockwise circulation)

\oint_C F\cdot n ds =\iint_R (\frac{\partial f}{dx}+ \frac{\partial g}{dy} dA (This one calculates the outward flux)

Then, recall that in our case f(x,y) = 7xy, g(x,y)=2y^2[/tex]. Then

\frac{\partial f}{dx} = 7y,\frac{\partial f}{dy} = 7x

\frac{\partial g}{dx}=0, \frac{\partial g}{dy} = 4y.

Note that we just need to describe our region R. The region R lies between the parabola y=x^2 and the line y=x. Thus, one way to describe the region is as follows 0\leq x \leq 1, x^2\leq y \leq x. Then, using the previous results, we get that

\oint_C F\cdot dr =\int_{0}^{1}\int_{x^2}^{x}7x-0dydx = 7\int_{0}^1x(x-x^2)dx = 7 \left.(\frac{x^3}{3}-\frac{x^4}{4})\right|_{0}^1 = 7(\frac{1}{3}-\frac{1}{4}) = \frac{7}{12} (circulation)

\oint_C F\cdot n ds=\int_{0}^{1}\int_{x^2}^{x}7y+4ydydx = \frac{11}{2}\int_{0}^1\left.y^2\right_{x^2}^{x}dx = \frac{11}{2}\int_{0}^{1}x^2-x^4 dx = \frac{11}{2}\left(\frac{x^3}{3}-\frac{x^5}{5})\right|_{0}^{1}=\frac{11}{2}(\frac{1}{3}-\frac{1}{5})=\frac{11}{15}(flux)

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Answer:

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Answers:

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