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hammer [34]
3 years ago
13

Do the rectangular prisms over lap?

Mathematics
2 answers:
mojhsa [17]3 years ago
8 0

Answer:

The prism that starts at (0, 2, 5)and has a size of (9, 20, 5)is ending at (9, 22, 10). To check for overlapping prisms (A and B), use the start and end points of these. The two prisms have to overlap in all dimensions. To check overlapping in X dimension

Step-by-step explanation:

so basically its a yes hope it helps? :))

NeTakaya3 years ago
4 0
Yes it does







yes it does
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Use the Fundamental Theorem for Line Integrals to find Z C y cos(xy)dx + (x cos(xy) − zeyz)dy − yeyzdz, where C is the curve giv
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Answer:

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Step-by-step explanation:

We have to find a funtion f such that its gradient is (ycos(xy), x(cos(xy)-ze^(yz), -ye^(yz)). In other words:

f_x = ycos(xy)

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we can find the value of f using integration over each separate, variable. For example, if we integrate ycos(x,y) over the x variable (assuming y and z as constants), we should obtain any function like f plus a function h(y,z). We will use the substitution method. We call u(x) = xy. The derivate of u (in respect to x) is y, hence

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(Remember that c is treated like a constant just for the x-variable).

This means that f(x,y,z) = sen(x,y)+C(y,z). The derivate of f respect to the y-variable is xcos(xy) + d/dy (C(y,z)) = xcos(x,y) - ye^{yz}. Then, the derivate of C respect to y is -ze^{yz}. To obtain C, we can integrate that expression over the y-variable using again the substitution method, this time calling u(y) = yz, and du = zdy.

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Where, again, the constant of integration depends on Z.

As a result,

f(x,y,z) = cos(xy) - e^{yz} + K(z)

if we derivate f over z, we obtain

f_z(x,y,z) = -ye^{yz} + d/dz K(z)

That should be equal to -ye^(yz), hence the derivate of K(z) is 0 and, as a consecuence, K can be any constant. We can take K = 0. We obtain, therefore, that f(x,y,z) = cos(xy) - e^(yz)

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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