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sertanlavr [38]
2 years ago
12

Please help. I am not good with this

Mathematics
2 answers:
aniked [119]2 years ago
4 0

Answer:

X= 7

Z= 117

Step-by-step explanation:

Z would be 117 because on the bottom half of the equation it has to be equal to 180. It gave you 63 already and so I just took the 180 and subtracted the 63 to get 117.

kondaur [170]2 years ago
4 0

Answer:

z= 117

x= 7

Step-by-step explanation:

Since angle 63 and z are on the same axis, this creates a straight line. And a straight line = 180 degrees. So z can be found by:

180-63= 117

Then x. You can still use the 180 degrees method as well.

180= 117 + (10x-7) (subtract 117 on both sides)

63= 10x-7 (add 7 on both sides)

70= 10x (divide by 10 on each side)

7=x

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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:

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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

\int{ycos(xy)} \, dx = \int cos(u) \, du = sen(u) + C = sen(xy) + C(y,z)  

(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.

\int {-ye^{yz}} \, dy = \int {-e^{u} \, dy} = -e^u +K = -e^{yz} + K(z)

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)

The endpoints of the curve are r(0) = (0,0,1) and r(1) = (1,π/2,0). FOr the Fundamental Theorem for Line integrals, the integral of the gradient of f over C is f(c(1)) - f(c(0)) = f((0,0,1)) - f((1,π/2,0)) = (cos(0)-0e^(0))-(cos(π/2)-π/2e⁰) = 0-(-π/2) = π/2.

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