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Let (x - k ) / (x - 1) = y x - k = yx - y x = yx - y + k x - yx = y + k x(1 - y) = y + k x = (-y + k ) / (1 - y) x = (y - k ) / (y - 1) It would seem that the function is generally its own inverse...

Depends which Special K you mean....... if your enquiry is in regards to cereal then the...

k ^.3 is the same as saying k ^(3/10) k ^(3/10) is the same as saying tenth_root( k ^3) k ...

k (x) = 2 sin (x+π) / (sin (x+π) - 3) = 2 (- sin x) / (- sin x - 3) = 2 sin x / (sin x + 3) = [2 (sin x + 3) - 6] / (sin x + 3) = 2 - 6 / (sin x + 3) So for k ^-1(x) we get x = 2 - 6 / (sin k ^-1(x) + 3) => 6 / (sin k ^-1(x) + 3) = 2 - x =>...

k ( k - 5) > 0 in two cases: 1) k > 0 and ( k - 5) > 0 2) k < 0 and ( k - 5) < 0 In case 1, k > 0 and k > 5; both of these are true when k > 5. In case 2, k ...

Decomposing 1/[ k ( k +1)( k +2)], you have: A/ k + B/( k +1) + C/( k +2) = 1/[ k ( k +1.... Therefore: A( k +1)( k +2) + B( k )( k +2) + C( k )( k +1) = 1 which must hold true for all valid values of k . To determine...

Note that Σ( k = 5 to ∞) C( k -1, k -5) k ^3/2^ k ...to ∞) (k-1)(k-2)(k-3)(k-4) * k^3 x^ k {at x = 1/2} = (1/24) x^5 * Σ( k ...%29+C%28k-1%2C+ k -5%29+ k ^3%2F2^ k I hope...

k (x) = x^2(x^2+16) k (4 - √2) = (4 - √2)^2 [(4 - √2)^2 + 16] Remember that (4.... (4 - √2)(4 - √2) = 16 - 8√2 + 2 = 18 - 8√2 So then: k (4 - √2) = (4 - √2)^2 [(4 - √2)^2 + 16] k (4 - √2) = (18 - 8√2)(18...