Search results

range is between 0 and pi

cos y = sin x from ASTC (taking negative angles in 4th and 3rd quadrants): 1st and 4th quadrants sin (pi/2 - y) = sin x dy/dx = -1 2nd and 3rd quadrants sin (y + pi/2) = sin x dy/dx = 1 edit: should probably note that the derivate at x = pi/2, and x = -pi/2 is undefined as...

consider the quadratic: y = (ax + b)^2 + (cx + d)^2 + (ex + f)^2 (a, b, c, d, e, f are real) since y >= 0, it follows that the quadratic either has 1 or no real roots. this implies the discriminant <= 0. (2ab + 2cd + 2ef)^2 - 4(a^2 + c^2 + e^2)(b^2 + d^2 + f^2) <= 0 (ab + cd...

consider the graph of y = cos x, it cuts the x axis at pi/2, and dips down to -1 at pi. if we have y = cos (2x) now, at x = pi/4, we have y = cos (2 x pi/4) = cos (pi/2) = 0 i.e. it now cuts the x axis at pi/4 instead of pi/2, infact at pi/2 now, y = cos (2 x pi/2) = cos (pi) = -1 is...

cauchy schwarz inequality http://www.artofproblemsolving.com/Wiki/index.php/Cauchy-Schwarz_Inequality

f(x) = 3x + x f(x + h) = 3(x + h) + (x + h) f'(x) = lim h->0 [f(x+h) - f(x)] / h = lim h->0 [3(x + h) + (x + h) - 3x - x]/ h = lim h->0 4h/ h = lim h->0 4 = 4

you can always check the question yourself using series: LHS = (n + 1) + (n + 2) + (n + 3) + ... + (n + n) = (n x n) + (1 + 2 + 3 + ... + n) = n^2 + (n/2)(1 + n) = (2n^2 + n + n^2)/2 = (3n^2 + n)/2 = n(3n + 1)/2 = RHS

assume true for n = k, (k + 1) + (k + 2) + ... + (2k)= k(3k + 1)/2 n = k+1 [(k + 1) + 1] + [(k + 1) + 2] + ... + [(k+1) + (k+1)] = (k+ 1) [3(k + 1) + 1]/2 LHS = [(k + 1) + 1] + [(k + 1) + 2] + ... +[(k+1) + k] + [(k+1) + (k+1)] = k(3k + 1)/2 + k + [(k+1) + (k+1)] (using assumption)*** =...

d/dx (2x^2-3y^2) = d/dx 5 4x - d/dy (3y^2) dy/dx = 0 4x - 6y dy/dx = 0 dy/dx = 4x/6y 3x+4y=10 => m = - 3/4 .'. for normal, gradient of the tangent at the point on the hyperbola has to be equal to 4/3. to find this point on the hyperbola we solve 4/3 = 4x/6y, and 2x^2-3y^2=5...

took log of both sides: 2^(x + y) = 6^y ln [2^(x + y)] = ln (6^y) now used ln (a^b) = b ln a: (x+y) ln2 = y ln6

we have sides: l, l+d, l-d let @ be the angle between l+d and l-d t = 1/2(l-d)(l+d)sin@ => sin@ = 2t / (l-d)(l+d) ---------(1) and from cos rule: cos@ = [(l+d)^2 + (l-d)^2 - l] / 2(l+d)(l-d) ---------(2) (1)^2 + (2)^2 = 1: [2t / (l-d)(l+d)]^2 + {[(l+d)^2 + (l-d)^2 - l^2] /...

1) is it suposed to be 2^(m^n) = (8^n)^m? 2^(m^n) = (2^3n)^m 2^[m^(n-1)] = 2^3n m^(n-1) = 3n m = (3n)^[1/(n-1)] 2) 2^(x + y) = 6^y (x+y)ln2 = yln6 xln2 + yln2 = yln6 x = y(ln6 - ln2)/ln2 x= y ln3/ln2 -----------(1) 3^x = 6(2^y) xln3 = ln6 + yln2 sub in (1): (y ln3/ln2)ln3 =...

let y = e^ln x take log of both sides: ln y = ln (e^ln x) ln y = ln x . ln e (using ln (x^a) = a ln x) ln y = ln x .'. y = x hence x = e^ln x thats derived from sin2x + cos2x = 1 by dividing both sides by cos2x

^_^

P1 = nC1 x (1-x)n-1 P2 = nC2 x2 (1-x)n-2 P3 = nC3 x3 (1-x)n-3 ... notice the similarity to the expansion of: (a + b)^n = a^n + nC1 an-1 b + nC2 an-2b2 etc our a corresponds with (1-x) and our b corresponds with x except that we're missing the first term P0 (which would be...

i like the cartoon. way to attract the fundies :D

haha, nice foot work.

the volume for any curve rotated about the x-axis from a to b is: V = pi I y^2 dx (limits a,b) so our volume for this question: V = pi I (4ax) dx (limits 0,a) ... since y^2 = 4ax, and we're going from 0 to a 4a is just a constant so we can pull him out of the integral: V = pi.4a I x dx...

haha, best unis ever.

ill do you for free :D