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ludmilkaskok [199]
3 years ago
7

Consider the vector field ????(x,y,z)=(5z+y)????+(4z+x)????+(4y+5x)????.

Mathematics
1 answer:
Fynjy0 [20]3 years ago
8 0

Answer:

a) 5xz + xy + 4yz

b) 10

Step-by-step explanation:

a) Here F(x,y,z)=(5z+y)i+(4z+x)j+(4y+5x)k

Since the case F  =  ∇f holds, then

∇f = f_xi+f_yj+f_zk = (5z+y)i+(4z+x)j+(4y+5x)k

So, f_x = 5z + y

If we integrate f_x with respect to x, we will get an integration constant C which is also a function that depends to y and z.

Hence,

f = \int f_xdx = 5xz + xy + g(y,z)

Now we need to find g(y,z).

So first let's take the derivative of g(y,z) with respect to y.

f_y = x + g_y(y,z) = 4z + x

Hence, g_y(y,z) = 4z

So now, if we integrate g_y with respect to y to find g(y,z)

g = \int g_ydy = 4yz + C

Thus,

f = 5xz + xy + g(y,z) = 5xz + xy + 4yz + C

And since f(0,0,0)=0, then C=0

Thus,

f = f(x,y,z) = 5xz + xy + 4yz

b) By the Fundamental Theorem of Line Integrals, we know that

\int\limits^a_b F. dr = F[r(b)]-F[r(a)]

Hence,

\int\limits^a_b F. dr = F(1,1,1)-F(0,0,0) =[(5+1+4)-(0+0+0)]=10

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Answer:

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Step-by-step explanation:

Given a parallelogram ABCD. Diagonals AC and BD intersect at E. We have to prove that AE is congruent to CE and BE is congruent to DE i.e diagonals of a parallelogram bisect each other.

In ΔACD and ΔBEC

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a₅ = 10

a₁₀ = 20

a₁₅ = x

-------------------------------------------------------------------------------------------------------------------

Note that from a₅ to a₁₀, there is a addition of 10. This means that for every x(₅) jump, you add 10 to the whole number.

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So in this case:

a₁₀ = 20

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x = 30, or (A) is your answer

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