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Alexeev081 [22]
1 year ago
5

Determine whether each of the following functions is a solution of Laplace's equation uxx + uyy = 0. (Select all that apply.) u

= e^(−x) cos(y) − e^(−y) cos(x) u = sin(x) cosh(y) + cos(x) sinh(y)
Mathematics
1 answer:
Naddika [18.5K]1 year ago
4 0

Answer with Step-by-step explanation:

We are given that Laplace's equation

u_{xx}+u_{yy}=0

We have to determine given function is  solution of given laplace's equation.

If a  function is solution of given Laplace's  equation then  it satisfy the solution.

1.u=e^{-x}cosy-e^{-y}cosx

Differentiate w.r.t x

Then, we get

u_x=-e^{-x}cosy+e^{-y}sinx

Again differentiate w.r.t x

u_{xx}=e^{-x}cosy+e^{-y}cosx

Now differentiate u w.r.t y

u_y=-e^{-x}siny+e^{-y}cosx

Again differentiate w.r.t y

u_{yy}=-e^{-x}cosy-e^{-y}cosx

Substitute the values in given Laplace's equation

e^{-x}cosy+e^{-y}cosx-e^{-x}cosy-e^{-y}cosx=0

Hence, given function is a solution of given Laplace's equation.

2.u=sinx coshy+cosx sinhy

Differentiate w.r.t x

u_x=cosx coshy-sinx sinhy

Again differentiate w.r.t x

u_{xx}=-sin x coshy-cosxsinhy

Now, differentiate u w.r.t y

u_y=sinx sinhy+cosx coshy

Again differentiate w.r.t y

u_{yy}=sinx coshy+cosx sinhy

Substitute the values then we get

-sinx coshy-cosxsinhy+sinxcoshy+cosx sinhy=0

Hence, given function is a solution of given Laplace's equation.

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Suppose the area that can be painted using a single can of spray paint is slightly variable and follows a nearly normal distribu
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Answer:

Step-by-step explanation:

Hello!

You have the variable

X: Area that can be painted with a can of spray paint (feet²)

The variable has a normal distribution with mean μ= 25 feet² and standard deviation δ= 3 feet²

since the variable has a normal distribution, you have to convert it to standard normal distribution to be able to use the tabulated accumulated probabilities.

a.

P(X>27)

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Now that you have the corresponding Z value you can look for it in the table, but since tha table has probabilities of P(Z, you have to do the following conervertion:

P(Z>0.67)= 1 - P(Z≤0.67)= 1 - 0.74857= 0.25143

b.

There was a sample of 20 cans taken and you need to calculate the probability of painting on average an area of 540 feet².

The sample mean has the same distribution as the variable it is ariginated from, but it's variability is affected by the sample size, so it has a normal distribution with parameners:

X[bar]~N(μ;δ²/n)

So the Z you have to use to standardize the value of the sample mean is Z=(X[bar]-μ)/(δ/√n)~N(0;1)

To paint 540 feet² using 20 cans you have to paint around 540/20= 27 feet² per can.

c.

P(X≤27) = P(Z≤(27-25)/(3/√20))= P(Z≤2.98)= 0.999

d.

No. If the distribution is skewed and not normal, you cannot use the normal distribution to calculate the probabilities. You could use the central limit theorem to approximate the sampling distribution to normal if the sample size was 30 or grater but this is not the case.

I hope it helps!

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If a pack of paper costs 3.75 including tax, and there is a $20 budget, then the equation to find out the answer would be 3.75p is greater or equal to $20. So the answer would be B. 3.75 would be multiplied by the tax and then applied up to $20. 
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A bag contains 100 marbles which are red, green, and blue. Suppose a student randomly selects a marble without looking, records
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Answer:

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billy is creating a circular garden divided into 8 equal sections. The diameter of the garden is 12 feet. what is the area, in s
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