n(T) = 150, n(C) = 225, n(T intersection C) = 100. Students drinking at least one = n(T union C) = n(T) + n(C) - n(T intersection C) = 150 + 225 - 100 = 275. Students drinking neither = total - n(T union C) = 600 - 275 = 325.

neither = n(U) - n(A union B)

A union B = {2,3,4,5,6,7,8}, so (A union B)' = U - (A union B) = {1,9,10}. Now A' = {1,6,7,8,9,10} and B' = {1,2,3,9,10}, so A' intersection B' = {1,9,10}. Since (A union B)' = {1,9,10} = A' intersection B', De Morgan's law is verified.

(A union B)' = A' intersection B'

For f(x) to be real, the expression under the square root must be non-negative: 9 - x^2 >= 0, i.e. x^2 <= 9, so -3 <= x <= 3. Hence the domain is [-3, 3]. As x varies over this interval, 9 - x^2 ranges from 0 (at x = +/-3) to 9 (at x = 0), so sqrt(9 - x^2) ranges from 0 to 3. Hence the range is [0, 3].

sqrt() defined only for non-negative values

f(2) = 2^2 + 1 = 4 + 1 = 5. f(-1) = (-1)^2 + 1 = 1 + 1 = 2. For f(x) = 10: x^2 + 1 = 10, so x^2 = 9, giving x = +3 or x = -3.

f(x) = x^2 + 1

360 revolutions per minute = 360/60 = 6 revolutions per second. One revolution corresponds to an angle of 2pi radians. Therefore the angle turned in one second = 6 x 2pi = 12pi radians.

angle = number of revolutions x 2pi

Expand each factor: cos(A + B) = cos A cos B - sin A sin B and cos(A - B) = cos A cos B + sin A sin B. Their product is of the form (p - q)(p + q) = p^2 - q^2 with p = cos A cos B and q = sin A sin B. So it equals cos^2 A cos^2 B - sin^2 A sin^2 B. Writing cos^2 B = 1 - sin^2 B and sin^2 A = 1 - cos^2 A: = cos^2 A (1 - sin^2 B) - (1 - cos^2 A) sin^2 B = cos^2 A - cos^2 A sin^2 B - sin^2 B + cos^2 A sin^2 B = cos^2 A - sin^2 B. Hence proved.

cos(A+B)cos(A-B) = cos^2 A - sin^2 B

Using sin^2 x + cos^2 x = 1: sin^2 x = 1 - (9/25) = 16/25, so sin x = +/- 4/5. In the second quadrant sine is positive, so sin x = 4/5. Then tan x = sin x / cos x = (4/5) / (-3/5) = -4/3.

sin^2 x + cos^2 x = 1

The principal value for which sin x = 1/2 is x = pi/6. The general solution of sin x = sin a is x = n pi + (-1)^n a, where n is any integer. Here a = pi/6, so the general solution is x = n pi + (-1)^n (pi/6), n belongs to the set of integers.

sin x = sin a => x = n pi + (-1)^n a

The principal value for which tan x = sqrt(3) is x = pi/3. The general solution of tan x = tan a is x = n pi + a, where n is any integer. Here a = pi/3, so the general solution is x = n pi + pi/3, n belongs to the set of integers.

tan x = tan a => x = n pi + a

Write sin 3x = sin(2x + x) = sin 2x cos x + cos 2x sin x. Substitute sin 2x = 2 sin x cos x and cos 2x = 1 - 2 sin^2 x: = (2 sin x cos x) cos x + (1 - 2 sin^2 x) sin x = 2 sin x cos^2 x + sin x - 2 sin^3 x. Replace cos^2 x = 1 - sin^2 x: = 2 sin x (1 - sin^2 x) + sin x - 2 sin^3 x = 2 sin x - 2 sin^3 x + sin x - 2 sin^3 x = 3 sin x - 4 sin^3 x. Hence proved.

sin 3x = 3 sin x - 4 sin^3 x

Modulus |z| = sqrt(1^2 + (sqrt3)^2) = sqrt(1 + 3) = sqrt4 = 2. For the argument, tan(theta) = sqrt3/1 = sqrt3, and since both real and imaginary parts are positive (first quadrant), theta = pi/3. Polar form: z = 2(cos(pi/3) + i sin(pi/3)).

z = r(cos theta + i sin theta)

Multiply numerator and denominator by the conjugate of the denominator (1 + 2i): [(3 + 4i)(1 + 2i)]/[(1 - 2i)(1 + 2i)]. Numerator = 3 + 6i + 4i + 8i^2 = 3 + 10i - 8 = -5 + 10i. Denominator = 1^2 + 2^2 = 5. So the result = (-5 + 10i)/5 = -1 + 2i.

multiply by conjugate of denominator

Here a = 1, b = -2, c = 5. Discriminant = b^2 - 4ac = 4 - 20 = -16. So x = [2 +/- sqrt(-16)]/2 = [2 +/- 4i]/2 = 1 +/- 2i. The roots are 1 + 2i and 1 - 2i.

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

Let sqrt(-5 + 12i) = a + i b. Squaring: a^2 - b^2 + 2ab i = -5 + 12i. So a^2 - b^2 = -5 and 2ab = 12 (ab = 6). Also |z| = sqrt((-5)^2 + 12^2) = sqrt(25 + 144) = 13, so a^2 + b^2 = 13. Adding a^2 - b^2 = -5 and a^2 + b^2 = 13 gives 2a^2 = 8, a^2 = 4, a = +/-2; then b^2 = 9, b = +/-3. Since ab = 6 is positive, a and b have the same sign. Hence sqrt(-5 + 12i) = +/- (2 + 3i).

(a + ib)^2 = a^2 - b^2 + 2ab i

5x - 3 >= 3x - 5. Subtract 3x from both sides: 2x - 3 >= -5. Add 3: 2x >= -2. Divide by 2: x >= -1. The solution set is {x : x >= -1} = [-1, infinity). On the number line, a closed (filled) circle at -1 with shading to the right.

First inequality: 2x - 1 < 5 gives 2x < 6, so x < 3. Second inequality: 3x + 2 > -4 gives 3x > -6, so x > -2. The common solution is the intersection: -2 < x < 3, i.e. the interval (-2, 3).

Let the shortest side = x cm. Then the longest side = 3x and the third side = 3x - 2. Perimeter = x + 3x + (3x - 2) = 7x - 2. Given perimeter is at least 61: 7x - 2 >= 61, so 7x >= 63, giving x >= 9. Hence the minimum length of the shortest side is 9 cm.

perimeter = sum of sides

MATHEMATICS has 11 letters in which M appears 2 times, A appears 2 times and T appears 2 times; the rest (H, E, I, C, S) are distinct. Number of distinct arrangements = 11! / (2! x 2! x 2!) = 39916800 / 8 = 4989600.

arrangements = n!/(p! q! r!)

nC2 = n(n-1)/2 = 15, so n(n-1) = 30. We need two consecutive integers whose product is 30: 6 x 5 = 30, so n = 6. (n = -5 is rejected since n must be a positive integer.)

nC2 = n(n-1)/2

The thousands place cannot be 0, so it can be filled in 9 ways (digits 1 to 9). After using one digit, the hundreds place can be filled in 9 ways (including 0 now), the tens place in 8 ways and the units place in 7 ways. Total = 9 x 9 x 8 x 7 = 4536.