The product of 999,999,999 multiplied by 99,999,999 is 99,999,980,000,000,001. This result can be calculated by multiplying the two numbers together using long multiplication techniques, which involves multiplying each digit in one number by each digit in the other number and then adding up the results. In this case, the final result is a 16-digit number.
(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
999998999000001
99999998900000001
8.88888879E16
1x10153 times 999999999 is 999999999x10153. Normalizing that, you get 9.999999999x10161.
8,699,999,913
-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