Question

Algebra 2 Help!

Answer 1

\[\large 6\sqrt[3]{2}\]

Answer 2

\[ \huge 3\sqrt[3]{16}=3\sqrt[3]{2^{4}}=3*2^{\frac{ 4 }{ 3 }}=\]
\[\huge 3*2^{\frac{ 3 }{ 3 }}*2^{\frac{ 1 }{ 3 }} =3*2*\sqrt[3]{2} \]

Answer 3

you're welcome

Answer 4

sure

Answer 5

bottom

Answer 6

yes, the second one

Answer 7

how did you arrive at 4?

Answer 8

-4*

Answer 9

it's not \[\frac{ -4 }{ 4-5i }\]

Answer 10

4i

Answer 11

\[\frac{ 4i }{ 4-5i }\] altough it's possible to simplify it further

Answer 12

multiply everything by the conjugate of the denominator

Answer 13

yes

Answer 14

the final answer still has an i in it

Answer 15

\[\frac{ 4i(4+5i) }{ (4-5i)(4+5i) }\]

Answer 16

close, but one important detail

Answer 17

i*i=i²
and since i=\(\sqrt{-1}\) then i²=-1

Answer 18

41 is correct, double minus is plus

Answer 19

answer is correct too. although you should probably write it \[\frac{ 4 }{ 41 }*(4i-5)\]

Answer 20

alright, no problem

Answer 21

by the way, i may be mistaken on this, but i am pretty sure if you multiply a complex number with it's own conjugate, you can just take the square of both numbers and add them.

for example with (4i-5)(4i+5) = 5²+4² i'm not sure if this trick always works, but i'm 99% sure it does.

Answer 22

i meant (4-5i)(4+5i) obviously

Answer 23

i just used my calculator to check if that rule is always true, and it is.

Answer 24

(a+bi)(a-bi)=a²+b²