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3 years ago

Cho A= [ -4;4] B= ( 1;7] Xác định A B

Mathematics

1 answer:

antiseptic1488 [7]3 years ago

4 0

To find the intersection of A and B figure who has the bigger lower bounds and smaller upper bounds out of A and B.

The lower bound of A is -4 while the lower bound of B is (but not including) 1. So the lower bound will be (1.

The upper bound of A is 4 while the upper bound of B is 7. So the resulting upper bound will be 4].

Put these together and get (1,4].

Hope this helps :)

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Read 2 more answers

Find the counterclockwise circulation and outward flux of the field Fequals7 xy Bold i plus 2 y squared Bold j around and over t

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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3 years ago

Ike and Brian live in a tall apartment building. Ike’s bedroom window is 52 feet above the ground. Brian's bedroom window is 79

Alenkasestr [34]

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Step-by-step explanation:

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3 years ago

What are the variables if PT= 2x, TR= y+4, QT= x+2, TS=y

Brilliant_brown [7]

The missing aspect of this question is shown in the attached image, which is that the letters correspond to vertices of a parallelogram.

A key feature of the parallelogram is that the diagonals bisect each other, therefore:

PT = TR
QT = TS

With this information we can now plug in the equations and solve for the variables x and y.

PT = TR
2x = y+4
y = 2x - 4

QT = TS
x + 2 = y

We now have two equations for the variable y. With this we can solve for x.

2x - 4 = x + 2
x = 6

y = x + 2
y = 6 + 2
y = 8

Once we solved for the variable x, we simply placed that value back into one of the previous equations and solved for y. The results are x = 6, y = 8.

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3 years ago