Given that ^nC2 = 66, find the value of n.

You're on a roll — here are the solutions to the remaining problems. 1. Given that ^nC_2 = 66, find the value of n. Step 1: Write down the formula for combinations. ^nC_r = (n!)/(r!(n-r)!) Step 2: Substitute r=2 into the formula. ^nC_2 = (n!)/(2!(n-2)!) Step 3: Expand n! as n(n-1)(n-2)! and simplify. ^nC_2 = (n(n-1)(n-2)!)/(2 × 1 × (n-2)!) = (n(n-1))/(2) Step 4: Set the expression equal to 66 and solve for n. (n(n-1))/(2) = 66 n(n-1) = 132 n^2 - n - 132 = 0 Step 5: Factor the quadratic equation. (n-12)(n+11) = 0 This gives two possible values for n: n-12 = 0 n = 12 n+11 = 0 n = -11 Since n must be a positive integer for combinations, we discard n=-11. The value of n is 12. 2. Using the trapezium rule with five ordinates, evaluate _0^2 (4)/(1+x^2) dx, correct to two decimal places. Step 1: Determine the number of strips (n) and the width of each strip (h). Number of ordinates = 5, so the number of strips n = 5-1 = 4. The interval is from a=0 to b=2. The strip width h = (b-a)/(n) = (2-0)/(4) = (2)/(4) = 0.5. Step 2: Calculate the x-values for each ordinate and their corresponding y-values, y = f(x) = (4)/(1+x^2). • x_0 = 0 y_0 = (4)/(1+0^2) = (4)/(1) = 4 • x_1 = 0.5 y_1 = (4)/(1+(0.5)^2) = (4)/(1+0.25) = (4)/(1.25) = 3.2 • x_2 = 1.0 y_2 = (4)/(1+1.0^2) = (4)/(1+1) = (4)/(2) = 2 • x_3 = 1.5 y_3 = (4)/(1+1.5^2) = (4)/(1+2.25) = (4)/(3.25) ≈ 1.230769 • x_4 = 2.0 y_4 = (4)/(1+2.0^2) = (4)/(1+4) = (4)/(5) = 0.8 Step 3: Apply the trapezium rule formula: _a^b y \, dx ≈ (h)/(2) [ (y_0 + y_n) + 2(y_1 + y_2 + + y_n-1) ] Substitute the values: _0^2 (4)/(1+x^2) dx ≈ (0.5)/(2) [ (y_0 + y_4) + 2(y_1 + y_2 + y_3) ] ≈ 0.25 [ (4 + 0.8) + 2(3.2 + 2 + 1.230769) ] ≈ 0.25 [ 4.8 + 2(6.430769) ] ≈ 0.25 [ 4.8 + 12.861538 ] ≈ 0.25 [ 17.661538 ] ≈ 4.4153845 Step 4: Round the result to two decimal places. ≈ 4.42 The value of the integral is approximately 4.42. That's all for today! What's next?