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If a^n+1 + b^n+1/a^n + b^n is the arithmetic mean of a and b then find n.
Answer:Arithmatic mean of a,b is =(a+b)/2 from the problem(a+b)/2=(a^n+1 +b ^n+1)/(a^n+b^n) then (a+b)(a^n+b^n)=(a^n+1 +b ^n+1)*2a^n(a-b)=b^n(a-b)then (a/b)^n=1=(a/b)^0from this equate the powersthen n=0
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Let u = sin(x). Then du = cos(x) dx. So you can now antidifferentiate e^u du. This is e^u + C = e^sin(x) + C. Then substitute your range 0 to pi. e^sin (pi)-e^sin(0) =0-0 =0
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We require to check the derivative thus let's use v = 60. Plugging it in (2) provides the slope of the tangent line as -1.96, or negative. Thus, for all values of v > 50 we will ha
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