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• 2/3 = 0.66666.... Rounded to 0.667 = 100*0.667 % = 66.7% So 70% is greater.

  in Maths problem!!???? Comment by whitesox09 June 2021
• x = sin(t) so t = arcsin(x) Plug into y: y = cos(2arcsin(x)) Apply the given trig identity: y = 1 - 2sin²(arcsin(x)) Simplify: y = 1 - 2x²

  in convert from parametric to rectangular? Comment by whitesox09 May 2021
• Let b be the base of the system (note that b ≥ 2): 32 + 135 = (3*b^1 + 2*b^0) + (1*b^2 + 3*b^1 + 5*b^0) = (3b + 2) + (b^2 + 3b + 5) = b^2 + 6b + 7 211 = 2*b^2 + 1*b^1 + 1*b^0 = 2b^2 + b + 1 b^2 + 6b + 7 = 2b^2 + b + 1 b^2 - 5b - 6 = 0 …

  in Numerical base system? Comment by whitesox09 May 2021
• f(x)=2ln[cos(x)] By chain rule: f'(x) = 2 * (1 / cos(x)) * -sin(x) = -2sin(x)/cos(x) = -2tan(x)

  in f(x)=2 ln [cosx]? Comment by whitesox09 May 2021
• f(x)=2ln[cos(x)] By chain rule: f'(x) = 2 * (1 / cos(x)) * -sin(x) = -2sin(x)/cos(x) = -2tan(x)

  in f(x)=2 ln [cosx]? Comment by whitesox09 May 2021
• Let b be the base of the system (note that b ≥ 2): 32 + 135 = (3*b^1 + 2*b^0) + (1*b^2 + 3*b^1 + 5*b^0) = (3b + 2) + (b^2 + 3b + 5) = b^2 + 6b + 7 211 = 2*b^2 + 1*b^1 + 1*b^0 = 2b^2 + b + 1 b^2 + 6b + 7 = 2b^2 + b + 1 b^2 - 5b - 6 = 0 …

  in Numerical base system? Comment by whitesox09 May 2021
• x = sin(t) so t = arcsin(x) Plug into y: y = cos(2arcsin(x)) Apply the given trig identity: y = 1 - 2sin²(arcsin(x)) Simplify: y = 1 - 2x²

  in convert from parametric to rectangular? Comment by whitesox09 May 2021
• 2/3 = 0.66666.... Rounded to 0.667 = 100*0.667 % = 66.7% So 70% is greater.

  in Maths problem!!???? Comment by whitesox09 May 2021