The equation x^2 + y^2 = 16 represents a circle centred at the origin with radius r = 4 (since r^2 = 16). The area enclosed by a circle of radius r is pi r^2. So the area = pi (4)^2 = 16 pi square units. (By integration, the area = 4 times the integral from 0 to 4 of sqrt(16 - x^2) dx = 16 pi.)

Area of circle = pi r^2

Separate the variables: dy/(1 + y^2) = dx/(1 + x^2). Now integrate both sides: the integral of dy/(1 + y^2) = the integral of dx/(1 + x^2). This gives tan inverse (y) = tan inverse (x) + C, where C is the constant of integration. This is the general solution of the differential equation.

integral dy/(1 + y^2) = tan inverse (y)

Separate the variables by cross-multiplying: y dy = x dx. Integrate both sides: the integral of y dy = the integral of x dx, giving y^2/2 = x^2/2 + C1. Multiplying by 2: y^2 = x^2 + C (where C = 2 C1). So the general solution is y^2 - x^2 = C, which represents a family of hyperbolas.

A linear differential equation of the first order and first degree has the standard form dy/dx + P y = Q, where P and Q are functions of x only (or constants). It is solved using the integrating factor (IF) = e^(integral of P dx). Multiplying the equation by the integrating factor makes the left side an exact derivative, and the solution is y x (IF) = integral of [Q x (IF)] dx + C.

IF = e^(integral of P dx)

Here P = 1/x and Q = x. The integrating factor IF = e^(integral of (1/x) dx) = e^(ln x) = x. Multiply through by x (or use the solution formula): y x (IF) = integral of [Q x (IF)] dx, that is, y x = integral of (x times x) dx = integral of x^2 dx = x^3/3 + C. Dividing by x gives the general solution y = x^2/3 + C/x.

y (IF) = integral of Q (IF) dx + C

The scalar (dot) product of two vectors is the sum of the products of their corresponding components: a . b = (1)(2) + (2)(-1) + (3)(1) = 2 - 2 + 3 = 3. The dot product is a scalar quantity and also equals |a||b| cos theta, where theta is the angle between the vectors.

a . b = a1 b1 + a2 b2 + a3 b3

The angle theta satisfies cos theta = (a . b)/(|a| |b|). Here a . b = (1)(1) + (1)(-1) + (1)(1) = 1 - 1 + 1 = 1. Also |a| = sqrt(1 + 1 + 1) = sqrt(3) and |b| = sqrt(1 + 1 + 1) = sqrt(3). So cos theta = 1/(sqrt(3) x sqrt(3)) = 1/3. Therefore theta = cos inverse (1/3).

cos theta = (a . b)/(|a| |b|)

The cross product is found by expanding the determinant with rows i, j, k; then the components of a; then the components of b. a x b = i[(1)(1) - (1)(-1)] - j[(1)(1) - (1)(1)] + k[(1)(-1) - (1)(1)] = i(1 + 1) - j(1 - 1) + k(-1 - 1) = 2 i - 0 j - 2 k = 2 i - 2 k. Its magnitude is |a x b| = sqrt(2^2 + 0^2 + (-2)^2) = sqrt(8) = 2 sqrt(2).

a x b = determinant of (i, j, k; a1, a2, a3; b1, b2, b3)

Vector form: the equation of a line passing through a point with position vector a and parallel to a vector b is r = a + lambda b, where lambda is a scalar parameter and r is the position vector of any point on the line. Cartesian form: if the line passes through the point (x1, y1, z1) and has direction ratios a, b, c, then its equation is (x - x1)/a = (y - y1)/b = (z - z1)/c.

r = a + lambda b

If two lines have direction ratios (a1, b1, c1) and (a2, b2, c2), the angle theta between them is given by cos theta = |a1 a2 + b1 b2 + c1 c2| / (sqrt(a1^2 + b1^2 + c1^2) x sqrt(a2^2 + b2^2 + c2^2)). The lines are perpendicular if a1 a2 + b1 b2 + c1 c2 = 0, and they are parallel if a1/a2 = b1/b2 = c1/c2 (their direction ratios are proportional).

cos theta = |a1a2 + b1b2 + c1c2|/(|d1||d2|)

The distance of a point (x1, y1, z1) from the plane Ax + By + Cz + D = 0 is |A x1 + B y1 + C z1 + D|/sqrt(A^2 + B^2 + C^2). Here (x1, y1, z1) = (0, 0, 0), A = 3, B = 4, C = 12, D = -26. Distance = |3(0) + 4(0) + 12(0) - 26|/sqrt(3^2 + 4^2 + 12^2) = |-26|/sqrt(9 + 16 + 144) = 26/sqrt(169) = 26/13 = 2 units.

d = |A x1 + B y1 + C z1 + D|/sqrt(A^2 + B^2 + C^2)

The corner point method is based on the fact that the optimal value of the objective function (if it exists) occurs at a corner point (vertex) of the feasible region. The steps are: (1) find the feasible region determined by the constraints and identify its corner points; (2) evaluate the objective function Z = ax + by at each corner point; (3) the largest of these values is the maximum and the smallest is the minimum. For an unbounded region, the optimal value must be checked for existence.

The feasible region is a triangle bounded by x + y = 4 and the coordinate axes. Its corner points are (0, 0), (4, 0) and (0, 4). Evaluate Z = 3x + 4y at each corner: at (0, 0), Z = 0; at (4, 0), Z = 12; at (0, 4), Z = 16. The maximum value of Z is 16, which occurs at the corner point (0, 4).

Using the conditional probability formula: P(A | B) = P(A intersection B)/P(B) = 0.3/0.6 = 0.5. Similarly, P(B | A) = P(A intersection B)/P(A) = 0.3/0.5 = 0.6. So P(A | B) = 0.5 and P(B | A) = 0.6.

P(A | B) = P(A intersection B)/P(B)

Since A and B are independent, P(A intersection B) = P(A) x P(B) = 0.3 x 0.4 = 0.12. Using the addition theorem, P(A union B) = P(A) + P(B) - P(A intersection B) = 0.3 + 0.4 - 0.12 = 0.58. So P(A intersection B) = 0.12 and P(A union B) = 0.58.

P(A union B) = P(A) + P(B) - P(A intersection B)

Let E1 and E2 be the events of choosing Bag I and Bag II, so P(E1) = P(E2) = 1/2. Let A be the event of drawing a red ball. Then P(A | E1) = 3/7 and P(A | E2) = 5/11. By Bayes' theorem, P(E1 | A) = [P(E1) P(A | E1)] / [P(E1) P(A | E1) + P(E2) P(A | E2)] = [(1/2)(3/7)] / [(1/2)(3/7) + (1/2)(5/11)] = (3/14) / (3/14 + 5/22). Converting to a common denominator (154): 3/14 = 33/154 and 5/22 = 35/154, so the sum = 68/154. Hence P(E1 | A) = (33/154)/(68/154) = 33/68.

P(E1 | A) = P(E1)P(A|E1)/(sum of P(Ei)P(A|Ei))

The mean (expectation) of a random variable is E(X) = sum of x times p(x). Here E(X) = 0 x (1/4) + 1 x (1/2) + 2 x (1/4) = 0 + 1/2 + 1/2 = 1. First we may check that the probabilities add to 1: 1/4 + 1/2 + 1/4 = 1. So the mean of X is 1.

E(X) = sum of x p(x)

This is a binomial experiment with n = 3 trials, probability of success (head) p = 1/2 and failure q = 1/2. The binomial probability of r successes is P(X = r) = nCr p^r q^(n - r). For exactly 2 heads (r = 2): P(X = 2) = 3C2 (1/2)^2 (1/2)^1 = 3 x (1/4)(1/2) = 3 x 1/8 = 3/8. So the probability of getting exactly 2 heads is 3/8.

P(X = r) = nCr p^r q^(n-r)