Is it possible to place eight Queens on a chessboard, so that none of the Queens occupy the same row, column, or diagonal?

Take a number whose square root is to be calculated, any positive number.

Take a guess at the numbers square root.

Calculate the square root by improving on the current guess as indicated:Next guess = (number/ current guess + current guess)/ 2

Repeat this process until the difference between the next guess and the current is within the accepted level of accuracy. The better your guess, the fewer the number of iterations needed to get the square root. A good first guess is typically half the number whose square root is to be calculated.

The process is ten repeated until the desired accuracy is achieved.

Determine a root of the equation f(x) = x^3-x^2-9x+9 = 0 using the Newton-Raphson method if the initial guess is x1 = 1.5.

Solve the following set of linear simultaneous equations using the Gauss-Seidel method:

3.If illegal, pick the next position.

4.

If no legal position is found, back up to one row.

If legal positions are found for all eight rows, the problem is solved.

Search the current node value to see if it equals the search value.

If the search value is smaller than the current value, make the current node the left child node.

Make the current node the right child node.

Searches as far down the left side of the binary tree.

When it encounters, NULL, the search switches to the bottom-most right child and resumes.

Remove a node from the queue. This becomes the current node.

Place all child nodes of the current node onto the queue.

Get a positive number whose square root is to be calculated from the user.

While more numbers remain, calculate firs guess, x0.

Xn = 0.5 * (X (n-1) + Number/ X (n-1))

Until abs (Xn – X (n 1)) *= Desired precision

Get a positive number whose square root is to be calculated from the user.

1.Set number of iterations num_iter to zero.

2.Set previous estimate of the root (x_prev) to initial guess (x_init).

3.Set current estimate of the root (x_curr) to initial guess (x_init).

4.While num_iter * max_iter do the following

a.Compute the value of the derivative at x_prev from derx = df (x_curr);

b.If derx is less than epsilon (a value close to zero) return (1)

c.

Compute new estimate of the root from x_curr = x_prev f (x_prev)/ df (x_prev)

e. Set new value of root equal to x_curr

f.If estimate of root is within desired tolerance then return (0)

The second approximation to the solution is

X1 (2) = (b1 a12x2 (1) a13x3 (1))/a11 =

X2 (2) = (b2 a21x1 (2) a23x3 (1))/a22 =

X3 (2) = b3 a31x1 (2)-a32x2 (2)/a33 =

Substituting the above values, we obtain the third approximation from

X1 (3) = b1-a12x2 (2) a13x3 (2)/a11 =

X2 (3) = b2-a21x1 (3) a23x3 (2)/a22 =

20 1(1.256) 2(-3.947)/ 5 = 5.328

X3 (3) = b3 a31x1 (3) a32x2 (3)/a33 =

-12 3(1.

256) 2(5.328)/6 = -4.404

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