Question

Express The square root of negative 72. in i notation.

Answer 1

\[\sqrt{-72}\]

Answer 2

i bet you know this

Answer 3

i72, this is too EZ! and you know the answer to this question?
.....
Do you teach math anywhere?

Answer 4

actually if it was me, i would write
\[6\sqrt{2}i\] since \(\sqrt{72}=6\sqrt2\)

Answer 5

OK, I agree!

Answer 6

Im trying still i did this \[8i \pm \sqrt{3}\]

Answer 7

there is no sum in this

Answer 8

Ohh ok

Answer 9

i forgot how to do that sqrt thing

Answer 10

\[6 \sqrt{2}\]

Answer 11

\(72=36\times 2\)

Answer 12

Ohh and i take sqrt of 36 and i get 6 oh ok thanks again

Answer 13

Oh and Mr. satellite when i simplify \[i ^{18}\]

Answer 14

Is it \[i ^{16}\times i ^{2}?\]

Answer 15

i guess so
i would ask the following simpler question
what is the integer remainder when you divide 18 by 4?

Answer 16

\[-i?\]

Answer 17

let me ask again
forget complex numbers for a second
what is the remainder when you divide 18 by 4?

Answer 18

or -1 i think -1

Answer 19

its 4.5

Answer 20

once more
what is the "remainder" when you divide 18 by 4?

Answer 21

ohh man i fell dumb its 2

Answer 22

whew!

Answer 23

then
\[i^{18}=i^2=-1\]

Answer 24

sorry my head is just acting this stuff is hard to me

Answer 25

similarly \(i^{203}=i^3=-i\)

Answer 26

\(i^{12345}=i^1=i\) and so on

Answer 27

i was looking at that and it seems like odd nubers will be i and even numbers will be 1

Answer 28

not numbers the exponents

Answer 29

but then i just need to find weather its positive or negative

Answer 30

take the integer remainder when you divide the exponent by 4, that is all

Answer 31

\[i^0=1,i^1=i, i^2=-1,i^3=-i\] it has to be one of those

Answer 32

if the remainder is 3, it is \(-i\)
if the remainder is 2, it is \(-1\)
if the remainder is 1, it is \(i\) and if the remainder is 0, i.e. if the exponent is divisible by 4, then it is 1

Answer 33

Ok thanks im worn out now i need sleep gn