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  1. By definition: tanh(x) = [e^(2x) - 1 ]/[e^(2x) + 1 ]. So, we want to find the inverse of y = [e^(2x) - 1 ]/[e^(2x) + 1 ... for y: x = [e^(2y) - 1 ]/[e^(2y) + 1 ] ==> x[e^(2y) + 1 ] = e^(2y) - 1 , by multiplying both sides by e^(2y) + 1 ==> x*e^(2y) + x = e^(2y...

    2 Answers · Science & Mathematics · 31/05/2011

  2. 1 / 1 /x = 1 * (x/ 1 ) <- multiply the reciprocal of the denominator = x

    2 Answers · Science & Mathematics · 25/09/2010

  3. 1 + 1 =3 May be 1 + 1 =33 And also 1 + 1 =333333333333333333 or 1 + 1 =anything and everything that one ( 1 ) is Imagination And another one is idiotic if we add these two we can get everything in the world.

    20 Answers · Arts & Humanities · 26/06/2008

  4. 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 - 1 + 1 + 1 + 1 +( 1 x0) 13 Remember PEMDAS? Don't forget that multiply/dividing comes before adding/subtracting;)

    6 Answers · Science & Mathematics · 06/10/2012

  5. (- 1 )^0 = 1 Each term is - 1 times the previous term. (- 1 )^0+(- 1 )^ 1 +(- 1 )^2+(- 1 )^3 = 1 + (- 1 ) + 1 + (- 1 ...

    2 Answers · Science & Mathematics · 18/07/2011

  6. (- 1 ^ 1 )= - 1 , (- 1 ^2)= 1 , (- 1 ^3)= - 1 , ( 1 ^4)= 1 , (- 1 ^5)=- 1 THIS IS BASIC...no matter what the exponent. In this case, you won't change 1 because, no matter how many times you multiply it by itself, it...

    2 Answers · Science & Mathematics · 28/04/2010

  7. 1 +cos4x+cos2x+cos4x * cos2x 1 + 2*(cos2x)^2 - 1 + cos2x + [2*(cos2x)^2- 1 ] * cos2x 2 * (cos2x)^3 + 2 * (cos2x)^2 2 * (cos2x)^2 * [ 1 + cos2x]

    2 Answers · Science & Mathematics · 28/01/2008

  8. 1 /7 - 1 /14 = 1 /y 1 /7 - 1 /14 is equal to 1 /14 (as 2/14 is the same as 1 /7) so 1 /14 = 1 /y so y = 14

    3 Answers · Science & Mathematics · 07/11/2011

  9. a²/(a+ 1 ) + b²/(b+ 1 ) = (a²− 1 )/(a+ 1 ) + (b²− 1 )/(b+ 1 ) + 1 /(a+ 1 ) + 1 /(b+ 1 ) = a− 1 + b− 1 + 1 /(a+ 1 ) + 1 /(b+ 1 ) = − 1 + 1 /(a+ 1 ) + 1 /(b+ 1 ) … (i) By Cauchy...

    1 Answers · Science & Mathematics · 15/01/2016

  10. ( 1 -cosx)/sinx - sinx/( 1 -cosx) = 2cscx ( 1 -cosx)/sinx - sinx/( 1 -cosx) = [sinx*( 1 -cosx)]/[(sinx)^2] - [( 1 +cosx)*sinx]/[( 1 -cosx)*( 1 +cosx)] = [sinx - sinx*cosx]/[(sinx)^2] - [sinx + sinx*cosx]/[(sinx)^2] = [sinx - sinx*cosx - sinx - sinx*cosx]/[(sinx)^2] = [2*sinx*cosx]/[(sinx)^2] = 2*cosx/sinx = 2*cscx

    2 Answers · Science & Mathematics · 23/01/2010

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