Latus rectum = 2b^2/a and major axis = 2a. Given 2b^2/a = (1/2)(2a) = a, so 2b^2 = a^2. Since b^2 = a^2(1 - e^2), we get 2a^2(1 - e^2) = a^2, so 2(1 - e^2) = 1, giving 1 - e^2 = 1/2, e^2 = 1/2 and e = 1/sqrt2.

b^2 = a^2(1 - e^2)

By the section formula for internal division in ratio m : n, the coordinates are ((m x2 + n x1)/(m + n), (m y2 + n y1)/(m + n), (m z2 + n z1)/(m + n)) with m = 2, n = 3. x = (2(3) + 3(1))/5 = 9/5; y = (2(4) + 3(-2))/5 = 2/5; z = (2(-5) + 3(3))/5 = -1/5. The point is (9/5, 2/5, -1/5).

((m x2 + n x1)/(m+n), ...)

The centroid of a triangle with vertices (x1,y1,z1), (x2,y2,z2), (x3,y3,z3) is ((x1+x2+x3)/3, (y1+y2+y3)/3, (z1+z2+z3)/3). x = (1+3+4)/3 = 8/3; y = (2-1+0)/3 = 1/3; z = (3+5-2)/3 = 6/3 = 2. Centroid = (8/3, 1/3, 2).

G = ((x1+x2+x3)/3, (y1+y2+y3)/3, (z1+z2+z3)/3)

Write (sin 3x)/(sin 5x) = [(sin 3x)/(3x) x 3x] / [(sin 5x)/(5x) x 5x]. As x approaches 0, (sin 3x)/(3x) -> 1 and (sin 5x)/(5x) -> 1, so the limit = (1 x 3x)/(1 x 5x) = 3/5.

lim x->0 sin(kx)/(kx) = 1

Direct substitution gives 0/0. Factorise: x^2 - 9 = (x - 3)(x + 3) and x^2 - x - 6 = (x - 3)(x + 2). Cancelling (x - 3): the expression becomes (x + 3)/(x + 2). Taking x approaching 3 gives (3 + 3)/(3 + 2) = 6/5.

Direct substitution gives 0/0. Rationalise by multiplying numerator and denominator by (sqrt(1 + x) + 1): [(sqrt(1+x) - 1)(sqrt(1+x) + 1)]/[x(sqrt(1+x) + 1)] = (1 + x - 1)/[x(sqrt(1+x) + 1)] = x/[x(sqrt(1+x) + 1)] = 1/(sqrt(1+x) + 1). As x approaches 0 this equals 1/(1 + 1) = 1/2.

(sqrt a - sqrt b)(sqrt a + sqrt b) = a - b

Let u = x^2 + 1 and v = x - 3, so u' = 2x and v' = 1. By the product rule dy/dx = u'v + uv' = (2x)(x - 3) + (x^2 + 1)(1) = 2x^2 - 6x + x^2 + 1 = 3x^2 - 6x + 1.

(uv)' = u'v + uv'

Let u = x + 1 and v = x - 1, so u' = 1 and v' = 1. By the quotient rule dy/dx = (u'v - uv')/v^2 = [1(x - 1) - (x + 1)(1)]/(x - 1)^2 = (x - 1 - x - 1)/(x - 1)^2 = -2/(x - 1)^2.

(u/v)' = (u'v - uv')/v^2

f'(x) = lim h->0 [sin(x + h) - sin x]/h. Expand sin(x + h) = sin x cos h + cos x sin h, so the numerator = sin x cos h + cos x sin h - sin x = sin x(cos h - 1) + cos x sin h. Thus f'(x) = sin x . lim h->0 (cos h - 1)/h + cos x . lim h->0 (sin h)/h. Since lim (cos h - 1)/h = 0 and lim (sin h)/h = 1, we get f'(x) = sin x . 0 + cos x . 1 = cos x.

d/dx(sin x) = cos x

There are 8 observations. Mean = (6+7+10+12+13+4+8+12)/8 = 72/8 = 9. Absolute deviations from the mean: |6-9|=3, |7-9|=2, |10-9|=1, |12-9|=3, |13-9|=4, |4-9|=5, |8-9|=1, |12-9|=3. Sum of absolute deviations = 3+2+1+3+4+5+1+3 = 22. Mean deviation = 22/8 = 2.75.

MD = (sum of |xi - mean|)/n

Mean = (2+4+6+8+10)/5 = 30/5 = 6. Squared deviations: (2-6)^2=16, (4-6)^2=4, (6-6)^2=0, (8-6)^2=4, (10-6)^2=16; their sum = 40. Variance = 40/5 = 8. Standard deviation = sqrt(variance) = sqrt8 = 2 sqrt2 (approximately 2.83).

variance = (sum of (xi - mean)^2)/n; SD = sqrt(variance)

Coefficient of variation (CV) = (standard deviation/mean) x 100. Here SD = 2 sqrt2 which is approximately 2.83, and mean = 6. CV = (2.83/6) x 100, which is approximately 47.1 percent. A higher CV indicates greater relative variability.

CV = (SD/mean) x 100

When two dice are thrown, the total number of outcomes = 6 x 6 = 36. The outcomes giving a sum of 8 are (2,6), (3,5), (4,4), (5,3) and (6,2), which is 5 favourable outcomes. Probability = 5/36.

P(E) = favourable outcomes / total outcomes

Let A = the card is a king and B = the card is a heart. P(A) = 4/52, P(B) = 13/52, and P(A and B) = 1/52 (the king of hearts). By the addition theorem P(A or B) = P(A) + P(B) - P(A and B) = 4/52 + 13/52 - 1/52 = 16/52 = 4/13.

P(A or B) = P(A) + P(B) - P(A and B)

Reflexive: for any integer a, a - a = 0, which is divisible by 3, so (a, a) in R. Symmetric: if (a - b) is divisible by 3, then (b - a) = -(a - b) is also divisible by 3, so aRb implies bRa. Transitive: if (a - b) and (b - c) are both divisible by 3, then their sum (a - b) + (b - c) = (a - c) is also divisible by 3, so aRb and bRc imply aRc. Since R is reflexive, symmetric and transitive, it is an equivalence relation.

One-one: suppose f(x1) = f(x2). Then 2x1 + 3 = 2x2 + 3, so 2x1 = 2x2 and x1 = x2; hence f is one-one. Onto: let y be any real number in the codomain. Solving y = 2x + 3 gives x = (y - 3)/2, which is a real number, and f((y - 3)/2) = y. So every y has a pre-image, and f is onto. Being both one-one and onto, f is a bijection.

Since f is a bijection, its inverse exists. Let y = f(x) = 2x + 3. Solve for x in terms of y: y - 3 = 2x, so x = (y - 3)/2. Therefore the inverse function is f inverse (y) = (y - 3)/2, or writing in terms of x, f inverse (x) = (x - 3)/2. We can check that f(f inverse (x)) = x.

f inverse (x) = (x - 3)/2

First combine tan inverse (2) + tan inverse (3). Using tan inverse (x) + tan inverse (y) = pi + tan inverse ((x + y)/(1 - xy)) when xy > 1: here x = 2, y = 3, xy = 6 > 1, so tan inverse (2) + tan inverse (3) = pi + tan inverse ((5)/(1 - 6)) = pi + tan inverse (-1) = pi - pi/4 = 3pi/4. Adding tan inverse (1) = pi/4: total = pi/4 + 3pi/4 = pi. Hence proved.

tan inverse x + tan inverse y = tan inverse((x + y)/(1 - xy))

To multiply, each element of a row of A is multiplied by the corresponding element of a column of B and summed. AB = [[(1)(2) + (2)(1), (1)(0) + (2)(3)], [(3)(2) + (4)(1), (3)(0) + (4)(3)]] = [[2 + 2, 0 + 6], [6 + 4, 0 + 12]] = [[4, 6], [10, 12]].

(AB)ij = sum of (row i of A)(column j of B)

Write A = (1/2)(A + A') + (1/2)(A - A'). Let P = (1/2)(A + A') and Q = (1/2)(A - A'). Then P' = (1/2)(A + A')' = (1/2)(A' + A) = P, so P is symmetric. And Q' = (1/2)(A - A')' = (1/2)(A' - A) = -(1/2)(A - A') = -Q, so Q is skew-symmetric. Since A = P + Q, every square matrix is the sum of a symmetric matrix P and a skew-symmetric matrix Q.

A = (1/2)(A + A') + (1/2)(A - A')