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PtichkaEL [24]
1 year ago
8

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) = ye−xi + e−xj + 2zk
Mathematics
1 answer:
Akimi4 [234]1 year ago
8 0
A three-dimensional vector field is conservative if it is also irrotational, i.e. its curl is \mathbf 0. We have

\nabla\times\mathbf f(x,y,z)=-2e^{-x}\,\mathbf k

so this vector field is not conservative.

- - -

Another way of determining the same result: We want to find a scalar function f(x,y,z) such that its gradient is equal to the given vector field, \mathbf f(x,y,z):

\nabla f(x,y,z)=\mathbf f(x,y,z)

For this to happen, we need to satisfy

\begin{cases}f_x=ye^{-x}\\f_y=e^{-x}\\f_z=2z\end{cases}

From the first equation, integrating with respect to x yields

f_x=ye^{-x}\implies f(x,y,z)=-ye^{-x}+g(y,z)

Note that g *must* be a function of y,z only.

Now differentiate with respect to y and we have

f_y=-e^{-x}+g_y=e^{-x}\implies g_y=2e^{-x}\implies g(y,z)=2ye^{-x}+\cdots

but this contradicts the assumption that g(y,z) is independent of x. So, the scalar potential function does not exist, and therefore the vector field is not conservative.
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The newest invention of the 6.431x staff is a three-sided die. On any roll of this die, the result is 1 with probability 1/2, 2
Gwar [14]

Answer:

<h2>The answer is 0.23(approx).</h2>

Step-by-step explanation:

The given die is a three sided die, hence, there are only three possibilities of getting the outcomes.

We need to find the probability of getting exactly 3s as the result.

From the sequence of 6 independent rolls, 2 rolls can be chosen in ^6C_2 = \frac{6!}{2!\times4!} = \frac{30}{2} = 15 ways.

The probability of getting two 3 as outcome is \frac{1}{4} \times\frac{1}{4} = \frac{1}{16}.

In the rest of the 4 sequences, will not be any 3 as outcome.

Probability of not getting a outcome rather than 3 is 1 - \frac{1}{4} = \frac{3}{4}.

Hence, the required probability is 15\times\frac{1}{16}(\frac{3}{4})^4 = \frac{1215}{4096}≅0.2966 or, 0.23.

4 0
2 years ago
HELP ASAP LIKE WHEN YOU SEE IT ANSWER ALL 7 QUESTIONS *RIGHT*AND I'LL MARK YOU THE BRAINLIST ANSWER !!!!!! WITH 13 POINTS
Vera_Pavlovna [14]

Answer:

1. Her weekly gross is $234

2. $421.20 for federal income tax withholding.

3. $174.096 for social security tax

4. $40.716 for medical tax

5. Her net paycheck is $636.012

6. $2808 for her total gross pay

7. $2171.988 is her total net pay

7 0
2 years ago
On circle OOO below, the measure of \stackrel{\LARGE{\frown}}{FJ} FJ ⌢ F, J, start superscript, \frown, end superscript is 84^\c
lianna [129]

Answer:

100°

Step-by-step explanation:

The angle between chords is the average of the intersected arc angles.

∠FKJ = ½ (84° + 76°)

∠FKJ = 80°

∠HKJ is supplementary to ∠FKJ.

∠HKJ = 180° − 80°

∠HKJ = 100°

7 0
2 years ago
n 2018, homes in East Baton Rouge (EBR) Parish sold for an average of $239,000. You take a random sample of homes in Ascension p
Olenka [21]

Answer:

Conclusion

   There is no sufficient evidence to conclude that the mean of the home prices from Ascension parish is higher than the EBR mean

Step-by-step explanation:

From the question we are told that

   The population mean for EBR is  \mu_ 1  = \$239,000

    The sample mean for Ascension parish  is \= x_2  = \$246,000

   The  p-value  is  p-value  =  0.045

     The level of significance is  \alpha = 0.01

The null hypothesis is  H_o : \mu_2  = \mu_1

The  alternative hypothesis is  H_a  :  \mu_2 > \mu_1

Here \mu_2 is the population mean for Ascension parish

   From the data given values we see that  

          p-value  >  \alpha

So we fail to reject the null hypothesis

So we conclude that there is no sufficient evidence to conclude that the mean of the home prices from Ascension parish is higher than the EBR mean

3 0
1 year ago
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