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x = A sin (nt) v = An cos (nt) when t = 0, v = V .'. V = An A = V/n since the particle momentarily comes to rest at the amplitude of the motion, the particle first comes to rest after traveling the distance of the amplitude i.e. V/n

treat the 3 as 1 person so the total arrangements will be 17! and also include the 2 possibilities: 1) john, sue, ray 2) ray, sue, john so the total number of ways = 17! x 2 probability = 17! x 2 / 19! = 1/171

right angled triangle 40t in x direction, (60t - 60) in y direction (note: wont work for t < 1) D = [(60t - 60)^2 + (40t)^2]^(1/2) from pythagoras dD/dt = (1/2)[(60t - 60)^2 + (40t)^2]^(-1/2) . {[2(60t - 60)].60 + 2(40t).40} sub t = 2, R = (1/2)(3600 + 6400)^(-1/2) . (7200 + 6400) R =...

(x + y)^2 - 2xy = x^2 + y^2 3^2 - 2xy = 5 2xy = 4 (x - y)^2 = x^2 - 2xy + y^2 (x - y)^2 = 1 x - y = + 1 1/a : 1/(3a/2) : 1/(3a/5) 1/a : 2/3a : 5/3a 3 : 2 : 5

here we just need to get h in terms of r. at x = r, h = -10r/3 + 10 .'. V = pi r^2 (-10r/3 + 10) = (pi)r^2(30 - 10r)/3 = 10(pi)r^2(3 - r)/3 V = 10(pi)r^2 - 10(pi)r^3/3 dV/dr = 20(pi)r - 10(pi)r^2 0 = 20(pi)r - 10(pi)r^2 0 = r(2 - r) r = 2 (check for max...

w/(3+w) . (w-1)/(2+w) + 3/(3+w) . 2/(2+w) = 1/2 2 (w^2 - w + 6) = (3+w)(2+w) 2w^2 - 2w + 12 = 6 + 5w + w^2 w^2 - 7w + 6 = 0 (w - 6) (w - 1) = 0 based on sue's case w=6 or w=1 w/(3+w) . w/(3+w) + 3/(3+w) . 3/(3+w) = 5/8 8 (w^2 + 9) = 5 (3+w)^2 8w^2 + 72 = 45 + 30w + 5w^2 w^2 -...

little different working to vds700: N = N0ekt N0ek3 / N0ek0 x 100 = 120 (population increases by 20%) ek3 = 12/10 k3 = ln(12/10) k = (1/3) ln(12/10) let T denote time it takes for population to double: N0ekT / N0ek0 = 2 (population doubles) ekT = 2 kT = ln 2...

sub it into the equation for V and it should equal 0 (i.e. satistfies the equation). V ' (x) = 144 - (54/5)x^2 0 = 144 - (54/5)x^2 x = 12 /√(54/5) x = 2√30/3 check second derivative for maximum, then sub it into V Vmax = (18/5) (2√30/3) (40 - (2√30/3)^2) = (12√30/5) (40 - 40/3) =...

because you can flip it round, and its still the same bracelet

T1/2 = ln 2 / k 1600 = ln 2 / k k = ln 2 / 1600 N = 100 e-k500 ~ 80.52%

truely sharper than any double-edged sword, it penetrates even to dividing soul and spirit, joints and marrow; it judges the thoughts and attitudes of the heart.

haha, oh leave her alone shes a nice girl

such as?

let (x0, y0) denote a point which lies on the circle. i.e. x0^2 + y0^2 = 16 -----(1) the circle cuts the x axis at A (+4, 0) midpoint has parameters: x = (x0 + 4) /2 => x0 = 2x + 4 -----(2) y = y0 /2 => y0 = 2y -----(3) sub (2) and (3) into (1): (2x + 4)^2 + (2y)^2 = 16

let x0 denote the horizontal distance from the wall to the base of the ladder, and y0 denote the vertical distance from the ground to the top of the ladder. length of the ladder is 6m which implies: x02 + y02 = 62 -----(1) the midpoint has parameters: x = x0/2 => x0 = 2x -----(2) y = y0/2...

Re: 回复: Re: Trigonometric Equations needed LHS = sin(a - @) + sin(b - @) = 2 sin[(a - @ + b - @) / 2] cos[(a - @ - b + @) / 2] (using sums to product formula) = 2 sin[(a + b)/2 - @]cos[(a - b)/2] RHS = sin(a + b - 2@) = sin{2[(a+b)/2 - @]} = 2 sin[(a + b)/2 - @] cos[(a+b)/2 - @] (using...

Re: 回复: Re: Trigonometric Equations needed for RHS use the sums to products formula for sines sin A + sin B = 2 sin[(A + B) / 2] cos[(A - B) / 2] for LHS factorise the 2 out and use double angle formula

JOSH WILSON

(2x + 3)^15 Trb = 15C(r-1) 2^(16-r) 3^(r-1) (from beginning) Tre = 15C(16-r) 2^(r-1) 3^(16-r) (from end) Trb / Tre = 8/27 8/27 = 15C(r-1) 2^(16-r) 3^(r-1) / 15C(16-r) 2^(r-1) 3^(16-r) = [15C(r-1) / 15C(16-r)] . [2^(16-r) / 2^(r-1)] . [3^(r-1) / 3^(16-r)] = {15!/[(r-1)!(16-r)!] /...

LHS = sin2(x/2) - 2 sin(x/2) cos(x/2) + cos2(x/2) - 1 = - 2 sin(x/2) cos(x/2) (red parts add to give 1) we know: sin(2x) = 2 sinx cosx, sub x = a/2 sin a = 2 sin (a/2) cos (a/2) hence - 2 sin(x/2) cos(x/2) = - sinx = RHS