annie

The redundant node can be defined as a node which links at least one branch which doesn’t contain any elements.

Fermat's Little Theorem is that a^(k-1) = 1 mod k. a and k must be co-prime. K must be prime

Could you please explain in detail. I am lost. :(

Example: 4^-1 mod 5

4^(5-1) congruent to 1 mod 5

4 * 4^3 congruent to 1 mod 5

The inverse is 4^3 mod 5

Is this correct?

Then use the squaring method to find the solution in the range?

I understand that to find the big-oh estimate, one should find the function that grows the fastest.

Example: The big-oh estimate for 5n + 3n^2 + 3 is 0(n^2)

However, I am not sure what to do in cases such as

1) n^3*4^n + 5^n + 16*sqrt(n) I understand that the sqrt(n) will grow the slowest. However, I do not know how to evaluate the expression n^3*4^n. I assumed the answer to be O(n^3*4^n) but it was O(5^n)

2)1+(1/logn)

3)17sqrt(n) + n*3^n*logn + 4^n

Can someone please explain in detail. Thanks

Can some one please explain the answer to me. Thanks in advance.

How do I set up a wireless network with just a Ethernet port? Can I just connect the Ethernet cable from the wireless router's WAN port to the Ethernet port? When I looked up the system requirements for several wireless routers, it said that I need a modem. Do I need one in this situation? Please help.

Marginal Analysis of Sweatshirt Production

Quantity of Shirts Total Benefit Total Cost

0 $0 $0

1 16 9

2 31 20

3 45 33

4 58 48

5 70 65

I need help with answering the following questions:

1. Given the information in the table, the optimal quantity of sweatshirts to produce is ____

2. Given the information in the table, the net benefit at the optimal quantity of sweatshirt is _______.

3. The marginal cost of the fourth sweatshirt is ________.

Please explain your answer.

const void printData (const int array[])const;

The first const indicates that the the value returned by the function cannot be changed.

The second const (inside the argument) means that I cannot change the explicit parameter.

The third const means that I cannot change any of the class's member data.

Is this correct?

Does the third const have something to do with the implicit parameter?

Also, is there any restriction on how many implicit or explicit parameters a member function can have?

struct Point{

double x;

double y;

};

1. How many total double locations are allocated by the above struct definition?

a. 2 b. 1 c. 3 d. 0 e. 4

I think the answers is d (0) a struct definition defines a new type; it is the variable definition of that type that creates memory for each of the struct's members

2. How many total double locations are allocated by the following Point declarations?

a. 2 b. 1 c.3 d.0 e.4

I think the answer is e(4)

Am I right? Could someone explain in detail please.

Isn't global function, a function that can be called anywhere in the program? And a member function, functions that are declared as member of a class? However is there any difference between the number of implicit or explicit parameters that either of two function take? Which is the implicit parameter in a global and member function?

Thanks

Is this the correct way to approach the problem?

Parameterize the polar curve

x = t*cos(t)

y = t*sin(t)

derivative of x = cos(t) - tsin(t)

derivative of y = sin(t) + tcos(t)

length of curve is integral from 0 to 2pi of square root of ( cos(t) - tsin(t))^2 + (sin(t) + tcos(t))^2 dt

x = t^2

y = 1-cos(t)

0 < t < 2pi

I don't care if you solve the problem. But can you please help me setup the equation.

The formula for find the area is integral from a to b of y dx.

y in this case is 1-cost

dx in this case is 1-cost

so area should be integral from 0 to 2pi of (1-cost)(1-cost)dt

Is this correct?

The figure on page 1179 of your text shows a pendulum with length L (meters) and angle theta (radians) from the vertical to the pendulum. It can be shown that theta as a function of time satisfies the differential equation:

d^2theta/(dt^2)+gLsin(theta) =0

where g=9.8 m/sec^2 is the acceleration due to gravity. For small values of we can use the approximation sin(theta) = theta, and with that substitution, the differential equation becomes linear. A. Determine the equation of motion of a pendulum with length 2 meters and initial angle 0.2 radians and initial angular velocity dtheta/dt 0.4 radians/sec.

I figure out the solution to part A by solving the equation m^2 + g/L = 0

The solution is .2cos(sqrt(4.9)t)+.18070sin(sqrt(4.9)t)

I need help with the following two parts

At what time does the pendulum first reach its maximum angle from vertical? (You may want to use an inverse trig function in your answer)

What is the maximum angle (in radians) from vertical?

I would appreciate if you don't give me an answer; instead, explain in detail how to approach the problem.

(y^2+x)*dy/dx - y = 0

dy/dx = 1/(dx/dy)

(y^2+x)(1/(dx/dy)) - y = 0 STEP ONE

y^2+x - y*(dx/dy) = 0 STEP TWO

(dx/dy) - (x/y) - y = 0 STEP THREE

Can someone clearly explain how to get from step one to two to three.

y^-1*dy/dx - (2/x)ln(y) = (1-2ln(x))/x

dy/dx - (2/x)ln(y)*y = y * (1-2ln(x))/x

This would have been a Bernoulli equation. However, what throws me off is the ln(y) in (2/x). Would I include it in my integrating factor e^(integral of (-2/x)ln(x)) ?

(x+4y^2)dy + 2ydx = 0

I understand it is not linear in y because of the x in x+4y^2. Can you please explain how to rewrite the equation so that it is linear in x.

This is what I have so far:

Base case: Let n = 4

4*3*2*1 >? 2^4

24 > 16

Assume that the formula holds true for n = k

k! > 2^k

Want to prove that the formula holds true for k+1

(k+1)! > 2^(k+1)

(k+1)! > 2 * 2^k

I understand that I need to some how use the assumption to prove that the formula holds true for k+1.

Could someone please explain in detail

Someone had post this question on yahoo answers before, but I don't understand this portion of best answer:

" by our induction hypothesis, since k! > 2^k, it means 2^k< k!, and

2 * (2^k) < 2 * k!

But 2 < k + 1, because k is greater than 4, so it follows that

2 * k! < (k + 1)k! "

Where did the "2 < k+1" come from. Is that just an observation. Thank You

A U (A ∩ B) = A

(A U A) ∩ (A U B) = A

A ∩ (A U B ) = A

I keep going in circles.

I also tried to prove it the following way

e: such that

{x | x e A or x e A, x e B}

{x | x e A, x e A or x e A, x e B}

Solve the separable differential equation

dx/dt =x^2+1/49 and find the particular solution satisfying the initial condition x(0)=−8

Here is what I have done so far

dx/dt = (49x^2+1)/49

49dx = (49x^2+1)dt

dx/(49x^2+1) = dt/49 take antiderivative of both sides

arctan(7x) = t/49 +C

arctan(7*-8) = 0 + C

arctan(-56) = C

arctan*7x = t/49 + arctan(-56)

Could someone in detail please explain how to rewrite the equation, y'(y prime) = y^2*x^3 in the differential form x^3* dx -(1/y^2)* dy = 0