9.87654432110999999999999999012345567889 × 10^38
or
987 undecillion, 654 decillion, 432 nonillion, 110 octillion, 999 septillion, 999 sextillion, 999 quintillion, 999 quadrillion, 999 trillion, 12 billion, 345 million, 567 thousand 889.
To figure it, rewrite 999999999999999999999999999 as 1000000000000000000000000000 - 1. Then figure (1000000000000000000000000000 - 1) x 987654432111 =
987654432111000000000000000000000000000 - 987654432111, which after subtracting, you get: 987,654,432,110,999,999,999,999,999,012,345,567,889 (the answer above).
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(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