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  1. The image of the new triangle should be the triangle with the coordinates of LHS times 18... Good luck!

    1 Answers · Science & Mathematics · 27/03/2012

  2. I am lucky to understand that Cosign (of an angle) = (adjacent side) / (hypotenuse). Tangent (of an angle)

    1 Answers · Science & Mathematics · 18/04/2008

  3. Easiest way is to multiply the right hand side out. x^4 - x^3 + x^2 + x^3 -x^2 + x + x^2 - x + 1 x^4 + x^2 + 1 (Note: the two cubes cancel out, two of the squares cancel out and the x's cancel out) Jim

    3 Answers · Science & Mathematics · 02/12/2011

  4. left/right hand side

    6 Answers · Science & Mathematics · 02/12/2006

  5. ...) = (sec A - tanA) + (secA + tanA) = 2secA = LHS __________________________ Prove the identity: (secA - 1)/(secA + 1...

    2 Answers · Science & Mathematics · 30/03/2007

  6. LHS> = tan (x+y+z) ................................ let y+z =p = tan (x+p) = (tan x + tan p)/(1 - tanx tan p) = L/...z + tan y+ tan z]/(1 - tan y tan z) ------ (3) ------------------------------------------------ put (3) & (2) in (1) ------------------- LHS = L/M = [tan x+ tan y+ tan z - tanx tan y tan z]/[1 - tan y tan z - tan x tan y - tan x tan z] = ...

    2 Answers · Science & Mathematics · 12/04/2009

  7. I'm not sure what is meant by 1 and 2, but to do 3 you can integrate f(x) over the interval 1-9 and then divide by the length of the interval (which is 9 - 1 = 8). To integrate lnx use integration by parts, you should get xlnx - x.

    2 Answers · Science & Mathematics · 18/03/2013

  8. Parameterize C one segment at a time. (i) r = 2 with θ in [0, π] (clockwise). Parameterize this by x = 2 cos θ, y = 2 sin θ with opposite orientation. So, ∫c₁ (arctan(y/x) dx + ln(x^2 + y^2) dy) = -∫(θ = 0 to π) [θ * (-2 sin θ) + ln(2^2) * (2 cos θ)] dθ = -∫(θ = 0 to π) [2θ...

    1 Answers · Science & Mathematics · 30/10/2013

  9. if that means : left hand side and right hand side, you'll have to be more specific as there are many ways to do so, depending on the situation.

    4 Answers · Science & Mathematics · 27/08/2017

  10. sin3y/siny = sin(y+2y) / siny = ( sinycos2y + cosysin2y) / siny Using sin2y = 2sinycosy = [ sinycos2y + 2siny(cosy)^2 ] / siny = siny[ cos2y + 2(cosy)^2 ] / siny = cos2y + 2(cosy)^2 = cos2y + 2(cosy)^2 - 1 + 1 Using cos2y = 2(cosy)^2 - 1 = cos2y + cos2y + 1 = 2cos2y + 1

    3 Answers · Science & Mathematics · 28/12/2008

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