You could have picked a BIGGER number!!!???
what is 10% of 4,625,000,000,000 =
[10/100] X 4,625,000,000,000 =462,500,000,000
Just take off one zero.
The expression conservative number can mean. cautious or not out of the norm. not far from the benchmark.
I don't remember exactly but 1984 was about a futuristic time when the government brainwashed and totally controlled its people.
Winston was not as well controlled as most.....
It was written in 1948 when the book was written. The cold war was raging. 'Western' people were terrified of
communism and government control.
I haven't answered your question. I am just rambling on.
Adjacent angles are angles that are next to each other.
They share the same vertex and they have a common arm.
[90.e+6] + [700.e+6] + [91.4e+12] =91,400,790,000,000=9.140079 X 10^13
97877+6677777
9 7 8 7 7
+ 6 6 7 7 7 7 7
1 1 1 1 1 transver
= 6 7 7 5 6 5 4
:- ) !
Becuase everyone deserves to...
Why?
It depends on what you want. Give an example of what you have in mind, and then will show you how to do it.
Use the calculator here and tell us what you got for the answer!.
As follows:
.
that eqauls \({10}^{6656182396877678591} \times 76578465738o74657835678546782686587946\)
ASSUMPTION: The business increases in value 9% every year
FV = 100000(1.09)^3 = 129502.90
3 x 4 x4 x4 x4 x4 = 768
use the web 2.0 calc:
input:
log(10,100)/(1/100)* sqrt3(14661^6)*sqrt3(sqrt(16)*(sqrt(64)+16/sqrt(4)))/( sqrt(log(10,100)^4)/( (1/2)^(-1)) *sqrt( 16^(1/(1/2))) *sin(30)*cos(60) ) + ( sqrt3( ( 0.5/( sqrt((512*0.5)^2) ) )^(-1) ) )^10 - ( 3* (6*sin(30))^3 + sqrt(sqrt((6*0.5)^8) ) /( cos(60) / (21584*cos(60)) )
= 87051515887
23=8 therefore log28=3
i is DEFINED as sqrt(-1) so i^2 = -1 -(-1) = 1
expand –d(d + 5)=- d^2 - d5
P=50,000(1+0.06)-12
P=50,000(1.06)^-12
P=50,000 X 0.49697.......etc.
P=24,848.47
Thank you heureka: The sequence is very simple; just subtract the counting numbers from the sequence and you you get? Twin Primes!!. Or natural counting numbers added to Twin Primes:
1, 2, 3, 4, 5, 6.......+ 3, 5.....5, 7.......11,13..................etc.
-i2= ?
\(\boxed{~ \begin{array}{lcll} z &=& a+b\cdot i \\ \bar{z} &=& a-b\cdot i\\ z\cdot \bar{z} &=& a^2+b^2 \\ && \text{where } \bar{z} \text{ is the complex conjugate of } z \end{array} ~} \)
\(\begin{array}{rcll} -i^2 &=& 0-i^2 \\ &=& (0+i)(0-i) \\ &=& \underbrace{(0+i)}_{=z}\cdot \underbrace{(0-i)}_{=\bar{z}} \qquad a = 0 \qquad b = 1 \\ -i^2 &=& 0^2+1^2 \\ -i^2 &=& 1 \\ \end{array}\)
!
