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i^i

\(\begin{array}{|rcll|} \hline i^i &=& [(-1)^{\frac12}]^i \qquad | \qquad i = \sqrt{-1} = (-1)^{\frac12}\\ \mathbf{i^i} & \mathbf{=} & \mathbf{(-1)^{\frac i2}} \\ &=& e^{\frac i2 \cdot ln(-1) } \\\\ && \ln(z) = \ln(|z|) +i\cdot arg(z)\\\\ && \ln(-1+0\cdot i) = \ln(|-1+0\cdot i|) +i\cdot arg(-1+0\cdot i)\\ && \ln(-1) = \ln(|\sqrt{(-1)^2+0^2}|) +i\cdot \arctan\left(\frac{0}{-1} \right)\\ && \ln(-1) = \ln(|1|) +i\cdot \pi \qquad | \qquad \ln(|1|)=\ln(1) = 0 \\ && \ln(-1) = 0 +i\cdot \pi \\ && \ln(-1) = i\cdot \pi \\\\ i^i &=& e^{\frac i2 \cdot \ln(-1) } \\ &=& e^{\frac i2 \cdot i\cdot \pi } \\ &=& e^{\frac {i^2}2 \cdot \pi } \qquad | \qquad i^2 = -1\\ &=& e^{\frac {-1}2 \cdot \pi } \\ &=& e^{\frac {- \pi}2 } \\ \mathbf{i^i} &\mathbf{=}& \mathbf{ e^{-(\frac {\pi}2) } } \\ \mathbf{i^i} &\mathbf{=}& \mathbf{0.207879576350761908546955619834978770033877841631769608075\dots }\\ \hline \end{array} \)

Math have sense if you understand it! Just go check my newest math Aztekium App and have fun with counting ! :D

Negative integer? Negative improper fraction? rational number? Real number? Whole number? 

\(15z+45=25z-5\)

\(15z+45-25z=-5\)

\(-10z=-5-45\)

\(-10z=-50\)

\(z={-50\over{-10}}=5\)

A rhombus has a vertex a (alpha). It is known that SQRT3*sina+cosa=2. It is also known that the longer diagonal is 24 (units). Find the area of the rhombus (the answer is 96*SQRT3) Please somebody smart enough to solve this give me a step-by-step guide on how it works.

6(x+4)=3(x-2)

6(x + 4) = 3(x-2)         [multiply 6 and 3

6x + 24 = 3x - 6          [subtract 3x, subtract 24

3x = - 30                     [divide by 3

x = - 10

896598756522*7?

\(896598756521\times 7= 6.27619129565\times 10^{12} \)

9.158x10 9/ 2.71x105

\(9.158\times \frac{10.9}{2.71} \times 105= 3867.6498155\)

i^i   ???

\(e^{i\pi}=cos\pi+isin\pi\\ e^{i\pi}=-1\\ \sqrt{e^{i\pi}}=\sqrt{-1}\\ e^{\frac{i\pi}{2}}=i\\ (e^{\frac{i\pi}{2}})^i=i^i\\ (e^{\frac{i^2\pi}{2}})=i^i\\ (e^{\frac{-1\pi}{2}})=i^i\\ i^i=e^{\frac{-\pi}{2}}\\ \)

e^(-pi/2) = 0.2078795763507619    approximately

so,

\(i^i=0.2078795763507619   \;\; approximately\)

ok So you want to make b the subject of the equation  :))

\(A=\frac{1}{2}bh\\ \)

To get b on its own you have to get rid of the 1/2 and the h on the right hand side.

To do you do that??

Well if you multiply the right by 2 you will get rif of the 1/2 BUT if you do it to the right you must do it to the left side as well.

\(A=\frac{1}{2}bh\\ A*2=\frac{1}{\not{2}}bh*\not{2}\\ 2A=bh\\ \)

Good - Now you want to get rid of the h on the right if you divide by h that will work but divide the other side by h as well!

Also the fraction line IS a divide sign   :))

\(2A=bh\\ \frac{2A}{h}=\frac{b\not{h}}{\not{h}}\\ \frac{2A}{h}=b\\ \mbox{Now swap sides}\\ b=\frac{2A}{h}\)

There is no question here

That is just an equation :)

It means 'power of'

6^2

means

\(6^2\)

and that equals 36

^ is  a caret   (not carrot lol)

sometimes is it called a  hat

DO NOT REPEAT QUESTIONS THAT YOU ONLY JUST ASKED 10 MINUTES AGO    !!!!

IT IS VERY RUDE !

(This is the first one the second has been answered.)

Rick chews on a piece of bubble gum for 120 minutes. About every 1 1/4 minutes he blows a bubble. how many bubbles did Rick make?

120/(1+1/4) = 96 bubbles

Start by dividing both sides by -6  then it should look like a normal equation to you that you should be able to solve. :)

Hi Htown,

Welcome to the forum :)

what plus what = 7 and what times what = -8 only using two numbers

Look at the times first.

The two numbers mult multiply to a negative number - so one will be positive and the other negative.

Mmm

They also have to add to a positive so the bigger one of the two numbers will be the positive one

They times to 8 so

-1 * +8   = - 8

-2 *  +4  =  -8

They are the only two choices so far (sincethe big one has to be the positive one)

-1+8=7

-2+4=2

So the numbers have to be     -1 and +8

Let me know if that helps :)

a+b=7  

a*b =-8

a=8, b=-1

b=8, a=-1

Thats what I need help on

hello friend, for a quadratic equation is solved by three methods ; complementation of the square, by the factorization and the general formula, in particular need a method ?

If you know how use the X-Box method, its specifficaly made for problems like that. Hope it works!!!

Hola amigo, primero debes resolver lo que se encuentra dentro del parentesis, luego proceder a la división ( puedes trabajar tanto con las divisones como tambien con decimales )  para el ultimo paso resolver el ejercicio es decir +

1/2(6-14)-1/3(6-27)-3

0.5-8-0.33-21-3=

-29,13

1,079 mi