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seropon [69]
3 years ago
9

Determine whether or not the vector field is conservative. if it is conservative, find a function f such that f = ∇f. (if the ve

ctor field is not conservative, enter dne.) f(x, y, z) = 6xy2z2i + 4x2yz3j + 6x2y2z2k
Mathematics
1 answer:
sweet [91]3 years ago
3 0
A conservative vector field \mathbf f has curl \nabla\times\mathbf f=\mathbf0. In this case,

\nabla\times\mathbf f=12xy^2z(1-z)\,\mathbf j+4yz^2(2z-3x)\,\mathbf k\neq\mathbf 0

so the vector field is not conservative.
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The table shows the relationship between the number of almonds and the number of cashews in various measures of chocolate fudge.
katovenus [111]

Answer:

A = 9

Step-by-step explanation:

3 0
3 years ago
How to differentiate y=x^n using the first principle. In this question, I cannot use the rule of differentiation. I have to do t
Zarrin [17]

By first principles, the derivative is

\displaystyle\lim_{h\to0}\frac{(x+h)^n-x^n}h

Use the binomial theorem to expand the numerator:

(x+h)^n=\displaystyle\sum_{i=0}^n\binom nix^{n-i}h^i=\binom n0x^n+\binom n1x^{n-1}h+\cdots+\binom nnh^n

(x+h)^n=x^n+nx^{n-1}h+\dfrac{n(n-1)}2x^{n-2}h^2+\cdots+nxh^{n-1}+h^n

where

\dbinom nk=\dfrac{n!}{k!(n-k)!}

The first term is eliminated, and the limit is

\displaystyle\lim_{h\to0}\frac{nx^{n-1}h+\dfrac{n(n-1)}2x^{n-2}h^2+\cdots+nxh^{n-1}+h^n}h

A power of h in every term of the numerator cancels with h in the denominator:

\displaystyle\lim_{h\to0}\left(nx^{n-1}+\dfrac{n(n-1)}2x^{n-2}h+\cdots+nxh^{n-2}+h^{n-1}\right)

Finally, each term containing h approaches 0 as h\to0, and the derivative is

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4 0
3 years ago
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Lesechka [4]
15 = 2Y - 5
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5 0
3 years ago
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Nonamiya [84]

Answer: 1091

Step-by-step explanation: add the numbers

8 0
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Ira Lisetskai [31]

Answer:

y = 8x + 4(6)

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8 friends bought tickets to a hockey game at $x

4 of them also bought a guide book at $6

If y = total cost

8 people bought tickets and 4 bought guides.

Then simplify by multiplying 4 and 6.

8 0
3 years ago
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