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intensity is inversly proportional to distance squared: I = k/d^2 intensity halved, distance tripled: I = [k/(3d)^2]/2 I = k/18d^2 i.e. 1/18th of the original

assume true n = k: √1 + √2 + √3 + ........ + √k + √(k+1) ≤ [(4k + 3)√k / 6] + √(k+1) n = k+1: √1 + √2 + √3 + ........ + √k + √(k+1) ≤ [4(k+1) + 3]√(k+1) / 6 need to show: [(4k + 3)√k / 6] + √(k+1) ≤ [4(k+1) + 3]√(k+1) / 6 (4k + 3)√k + 6√(k+1) ≤ 4k√(k+1) + 4√(k+1) + 3√(k+1) (4k + 3)√k...

PS = PD = (x^2/4a + a)

let A = amount owing I = yearly instalment r = (1 + 15/100 . 1/12)^12 = 1.0125^12 after 1 year: A = 6000r - I after 2 years: A = 6000r^2 - Ir - I after 5 years: A = 6000r^5 - Ir^4 - Ir^3 - ... - I 0 = 6000r^5 - Ir^4 - Ir^3 - ... - I (A = 0 after 5 years) 6000r^5 = I(r^4 +...

haha

no one likes to share their moneys

assume true for n = k cos(x + k*pi) = (-1)^k cos(x) n = k+1: LHS = cos[x + (k+1)*pi], RHS = (-1)^(k+1) cos (x) LHS = cos [(x + k*pi) + pi] = cos(x + k*pi) cos(pi) - sin(x + k*pi) sin(pi) = - cos(x + k*pi) = - (-1)^k cos(x) using assumption = (-1)^(k+1) cos (x) = RHS

-70 = t (85sin 15) - t^2 (9.8 / 2) t^2 (9.8 / 2) - t (85sin 15) - 70 = 0 t ~ 6.6 sec

let roots be A, B, A + B A + B + A + B = -6 A + B = - 3 AB(A+B) = 30 AB = -10 A^2 + 3A - 10 = 0 (A + 5)(A - 2) = 0 roots: 2, -5, -3

just sub x = 1 into (1+x)^n = nCo x^0 + nC1 x^1 +...+ nCn x^n

if we know (b-c) is a factor the thing can be put in the form: a^3(b-c) + b^3(c-a) + c^3(a-b) = (b - c)(.........)(..........)etc subing in c=b will give RHS = (c - c)(.........)(..........)etc = 0 LHS = a^3(c - c) + b^3(b - a) + b^3(a - b) = b^3(b - a) + b^3(a-b) = b - a + a - b = 0...

sub in b = - a a^2(c - a) + a^2(c + a) - ka^2c = 0 (factor theorem) c - a + c + a - kc = 0 2c - kc = 0 k = 2 expanding we get ba^2 + ca^2 + cb^2 + ab^2 + ac^2 + bc^2 + 2abc = b(a^2 + c^2 + 2ac) + b^2(c + a) + ac(a + c) = b(a + c)^2 + b^2(c + a) + ac(a + c) = (a + c)(ab + bc + b^2 + ac) = (a...

sinA + sin(A + 2B) = 2 sin[(A + 2B + A)/2]cos[(A + 2B - A)/2] = 2 sin (A + B) cos B .'. sinA + sin(A + B) + sin(A + 2B) = 2 sin (A + B) cos B + sin(A + B) = sin (A + B) . (2 cosB + 1) <----- numerator of LHS cosA + cos(A + 2B) = 2 cos[(A + 2B + A)/2]cos[(A + 2B - A)/2] = 2 cos(A + B) cos B...

find the gradient of line AP: mAP = (y - 3) / (x - 0) find the gradient of line BP: mBP = (y - 1) / (x + 2) for angle APB to be a right angle both lines have to be perpendicular to each other. this implies: mAP x mBP = - 1 (y - 3)(y - 1) / (x - 0)(x + 2) = - 1 y^2 - 4y + 3 = - (x^2 +...

total number of ways you can choose 2 letters = 20 P(both chosen) = (2/5)(2/4) = 4/20 P(none chosen) = (3/5)(2/4) = 6/20 P(at least 1 chosen) = 1 - (6/20) = 14/20 given at least 1 is chosen, there are now 14 possible outcomes, of which 4 lead to both being chosen. P(both being chosen) = 4/14...

LHS = 2cos[(85 + 35)/2]cos[(85 - 35)/2] + 2cos[(75 + 45)/2]cos[(75 - 45)/2] = 2cos60 cos25 + 2cos 60 cos15 = cos 25 + cos 15 = 2 cos[(25 + 15)/2]cos[(25 - 15)/2] = 2 cos20 cos5 = RHS

[ sup] sup notation [ /sup] = sup notation [ sub] sub notation [ /sub] = sub notation eg. nCr

equation of motion: y = - kx(x - R) ------(1) sub (R/2 , h) to find k: h = - kR/2 (R/2 - R) 4h/R^2 = k sub k into (1): y = - (4h/R^2)x(x - R) y = [4hx(R-x)] / R^2

dy/dx = x/2a mnormal = -2a/x y - y0 = -2a/x0 . (x - x0) normal at P (2a, a): y - a = -2a/2a (x - 2a) y - a = -x + 2a y = -x + 3a ----------(1) normal at Q (4a, 4a): y - 4a = -2a/4a (x - 4a) y - 4a = -x/2 + 2a y = - x/2 + 6a ----------(2) equate (1) and (2): -x + 3a = -x/2 + 6a -x/2 = 3a x...

y = x - (gx^2)/v^2 consider ball being thrown up slope: sub in point (30cosa, 30sina) into y = x - (gx^2)/v^2: 30sina = 30cosa - (g/v^2) (30cosa)^2 divide through by 30cosa: tan a = 1 - (30gcosa)/v^2 -----------(1) consider ball being thrown down slope: same thing but...