(1 + i)(2 - i) = 1 x 2 + 1 x (-i) + i x 2 + i x (-i) = 2 - i + 2i - i^2. Since i^2 = -1, this is 2 + i + 1 = 3 + i. So a = 3 and b = 1.

i^2 = -1

The multiplicative inverse of z is 1/z = conjugate(z)/|z|^2. Here conjugate of (4 - 3i) is (4 + 3i) and |z|^2 = 4^2 + (-3)^2 = 16 + 9 = 25. So the inverse = (4 + 3i)/25 = 4/25 + (3/25) i.

z^-1 = conjugate(z)/|z|^2

Using the quadratic formula x = [-b +/- sqrt(b^2 - 4ac)]/(2a) with a = 1, b = 1, c = 1: discriminant = 1 - 4 = -3. So x = [-1 +/- sqrt(-3)]/2 = [-1 +/- i sqrt3]/2. The roots are (-1 + i sqrt3)/2 and (-1 - i sqrt3)/2.

x = [-b +/- sqrt(b^2 - 4ac)]/(2a)

Powers of i repeat with period 4 (i, -1, -i, 1). i^9 = i^(8+1) = (i^4)^2 x i = 1 x i = i. i^19 = i^(16+3) = (i^4)^4 x i^3 = 1 x (-i) = -i. So i^9 + i^19 = i + (-i) = 0.

i^4 = 1

3x - 5 < 7 gives 3x < 12, so x < 4. The solution set is {x : x < 4, x is real} = (-infinity, 4). On the number line, this is shown by an open circle at 4 with shading to the left (all values less than 4).

Write 8! = 8 x 7 x 6!. So 8!/(6! x 2!) = (8 x 7 x 6!)/(6! x 2!) = (8 x 7)/2! = 56/2 = 28.

n! = n x (n-1)!

The hundreds place can be filled in 9 ways (any of 1 to 9), the tens place in 8 ways (remaining digits) and the units place in 7 ways. By the multiplication principle, total numbers = 9 x 8 x 7 = 504.

total = product of choices at each place

Since order does not matter in a committee, we use combinations. Number of ways = 10C3 = 10!/(3! x 7!) = (10 x 9 x 8)/(3 x 2 x 1) = 720/6 = 120.

nCr = n!/(r!(n-r)!)

We use the property that nCa = nCb implies either a = b or a + b = n. Since r is not equal to r + 2, we must have r + (r + 2) = 10, so 2r + 2 = 10, giving 2r = 8 and r = 4.

nCa = nCb => a = b or a + b = n

The binomial theorem states that (a + b)^n = nC0 a^n + nC1 a^(n-1) b + nC2 a^(n-2) b^2 + ... + nCn b^n = sum over r from 0 to n of nCr a^(n-r) b^r. The general term (the (r+1)th term) is T(r+1) = nCr a^(n-r) b^r.

T(r+1) = nCr a^(n-r) b^r

Here n = 6 is even, so there is one middle term, the (n/2 + 1) = 4th term. T4 = 6C3 x^3 (2y)^3 = 20 x^3 x 8 y^3 = 160 x^3 y^3.

middle term = (n/2 + 1)th term when n is even

Here the first term a = 3 and common difference d = 7 - 3 = 4. The nth term is a_n = a + (n - 1)d. So a_20 = 3 + (20 - 1) x 4 = 3 + 76 = 79.

a_n = a + (n - 1)d

Here the first term a = 2 and common ratio r = 6/2 = 3. The nth term of a GP is a_n = a r^(n-1). So a_8 = 2 x 3^7 = 2 x 2187 = 4374.

a_n = a r^(n-1)

AM = (4 + 16)/2 = 20/2 = 10. GM = sqrt(4 x 16) = sqrt64 = 8. Since 10 > 8, we have AM > GM, which agrees with the general result AM >= GM for positive numbers.

AM = (a + b)/2; GM = sqrt(ab)

The slope of a line through (x1, y1) and (x2, y2) is m = (y2 - y1)/(x2 - x1). Here m = (9 - 3)/(5 - 2) = 6/3 = 2.

m = (y2 - y1)/(x2 - x1)

Using the point-slope form y - y1 = m(x - x1) with m = 2 and (x1, y1) = (1, -3): y - (-3) = 2(x - 1), so y + 3 = 2x - 2, giving y = 2x - 5 or 2x - y - 5 = 0.

y - y1 = m(x - x1)

For parallel lines Ax + By + C1 = 0 and Ax + By + C2 = 0, the distance is d = |C1 - C2|/sqrt(A^2 + B^2). Here A = 3, B = -4, C1 = 7, C2 = 5. d = |7 - 5|/sqrt(9 + 16) = 2/5.

d = |C1 - C2|/sqrt(A^2 + B^2)

Compare with x^2 + y^2 + 2gx + 2fy + c = 0: 2g = -4 so g = -2, 2f = 6 so f = 3, c = -12. Centre = (-g, -f) = (2, -3). Radius = sqrt(g^2 + f^2 - c) = sqrt(4 + 9 + 12) = sqrt25 = 5.

centre (-g, -f), radius sqrt(g^2 + f^2 - c)

The equation of a circle with centre (h, k) and radius r is (x - h)^2 + (y - k)^2 = r^2. Here (h, k) = (2, -3) and r = 5, so (x - 2)^2 + (y + 3)^2 = 25. Expanding: x^2 + y^2 - 4x + 6y - 12 = 0.

(x - h)^2 + (y - k)^2 = r^2

The focus is on the positive y-axis and the directrix is y = -3, so the parabola opens upwards with vertex at the origin and form x^2 = 4ay. Here a = 3 (distance from vertex to focus), so x^2 = 4(3)y = 12y.

x^2 = 4ay