it is 99999998900000001
(tan x + cot x)/sec x . csc x The key to solve this question is to turn tan x, cot x, sec x, csc x into the simpler form. Remember that tan x = sin x / cos x, cot x = 1/tan x, sec x = 1/cos x, csc x = 1/sin x The solution is: [(sin x / cos x)+(cos x / sin x)] / (1/cos x . 1/sin x) [(sin x . sin x + cos x . cos x) / (sin x . cos x)] (1/sin x cos x) [(sin x . sin x + cos x . cos x) / (sin x . cos x)] (sin x . cos x) then sin x. sin x + cos x . cos x sin2x+cos2x =1 The answer is 1.
No. Tan(x)=Sin(x)/Cos(x) Sin(x)Tan(x)=Sin2(x)/Cos(x) Cos(x)Tan(x)=Sin(x)
f(x)=x+1 g(f(x))=x f(x)-1=x g(x)=x-1
I get x*x^x-1 + lnx*x^x = x^x + x^xlnx = x^x * (1+lnx) Here, ^ is power; * = times; ln = natural logratithm ( base e)
1 (sec x)(sin x /tan x = (1/cos x)(sin x)/tan x = (sin x/cos x)/tan x) = tan x/tan x = 1
245122011972
999999999
9.99999998 x 10^17
8,699,999,913
99999998900000001
999998999000001
8.88888879E16
1x10153 times 999999999 is 999999999x10153. Normalizing that, you get 9.999999999x10161.
-1 -2 -3 -4 -5 -6 -7 -8 -9 -99 -999 -9999 -99999 -999999 -9999999 -99999999 -999999999
Prime factorization of 999999999 = 34 × 37 × 333667
666,666,666 X 999,999,999 = 6.66666665 × 1017
99999999999999999999999999 9999999999999999999999999999999999999999 99999999999999999999999 99999999999999 99999999 99999 999999999999999999999999999999999999999999999999999999999999999999999999999 99999999999 99999 999999999 999999 9