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the obvious is z =/= 2 because of the 0 denominator, the other is z =/= i as w is purely imaginary.

i love asians. congrats on being asian.

let solutions be: x, -x, y, z x + -x + y + z = -b/a y + z = -5/6 -----------(1) x.-x + x.y + x.z + -x.y + -x.z + y.z = c/a -x^2 + yz = -24/6 -----------(2) x.-x.y.z = e/a -x^2.y.z = 18/6 -----------(3) solve these 3 for the 3 unknowns: firstly, looking at (2) if we...

consider P(x) = (x - a)(x - b), our solutions are a and b, if we expand this we get: P(x) = x^2 - xb - ax + ab P(x) = x^2 - x(b + a) + ab you can see the independant term has factors a and b. lets say we introduce a "c" (a non zero integer) to the coefficient of x^2 to make the...

you can use the formulas i gave before and solve for 3 unknowns: Sum alpha = -b/a Sum alpha.beta = c/a Sum alpha.beta.gamma = -d/a alpha.beta.gamma.delta = e/a might be a little bit more messy, but would still would work.

i) Tk+1 = (-1)^k . a^(n-k) . b^k . nCk Tk+1 = (-1)^k . x^(12 - k) . (2y)^k . 12Ck Tk+1 = (-1)^k . x^(12 - k) . (2^k). y^k . 12Ck general coefficient = (2^k).12Ck (2^k).12Ck / [2^(k-1)].12Ck-1 > 1 (2^k).12Ck > [2^(k-1)].12Ck-1 2.(12Ck) > 12Ck-1 2. 12!/[k!. (12 - k)!] > 12!/[(k-1)!. (12 - k...

i like asian girls

i like asians

alternatively as vds700 suggested: x^2 + 6y^2 = 15 ------(1) y = x/2 + k ------(2) sub (2) into (1): x^2 + 6(x/2 + k)^2 = 15 x^2 + 6(x^2/4 + xk + k^2) = 15 x^2(1 + 6/4) + x(6k) + (6k^2 - 15) = 0 for tangent, delta = 0 (6k)^2 - 4(1 + 6/4)(6k^2 - 15) = 0 36k^2 - 10(6k^2 - 15)...

google "bridge rectifiers"

he meant dy/dx = 1/2, -x/6y = 1/2 ----------(1) x^2 + 6y^2 = 15 ----------(2) solve (1) and (2) simultaneously. this will get you the 2 points on the ellipse where the gradients of the tangents are 1/2.

let, y = 10^(log 3) log y = log [10^(log 3)] (take the log of both sides) log y = log 3 . log 10 (log (a^b) = b log a) log y = log 3 y = 3 but y = 10^(log 3) .'. 10^(log 3) = 3

this is wrong answer: log 25 / log 5 = log (5^2) / log 5 = 2 log 5 / log 5 = 2

1st one.

hmm, not sure your question is correct, im going to assume pi/4 ignoring limits for now: I (tan x)^2n dx = I (tan x)^(2n - 2) (tan x)^2 dx = I (tan x)^(2n - 2) [(sec x)^2 - 1] dx = I (tan x)^(2n - 2)(sec x)^2 dx - I (tan x)^(2n - 2) dx In + I(n-2) = I (tan x)^(2n - 2)(sec x)^2 dx...

(2/6)^x = ln(.75)/ln(.25) ln (2/6)^x = ln [ln(.75)/ln(.25)] x ln (2/6) = ln [ln(.75)/ln(.25)] x = ln [ln(.75)/ln(.25)] / ln (2/6)

the friction force acts to oppose the slipping, hence opposite direction. friction forces are always opposite to motion.

0 < x^2 < 1 -1 < x < 1 test out some values on your calculator and you'll see why

y = 2 sin^-1(1 - x^2) -1 < 1 - x^2 < 1 -2 < - x^2 < 0 2 > x^2 > 0 take 2 cases for +/- square root. y = 2 sin^-1(1 - x^2) sub y = 0 to find x intercepts sub x = 0 to find y intercept

sin^-1 (x) has domain -1 < x < 1 sin^-1 (0) = 0 sin^-1 (1) = pi/2 thats all you need to know