can anyone answer these
@Michele_Laino
do you know the definition of a rectangle?
yes im sorry thats the wrong one
i meant these questions
we can write this: \[\Large \begin{gathered} {i^2} = - 1 \hfill \\ \hfill \\ {i^4} = {i^2}{i^2} = \left( { - 1} \right) \times \left( { - 1} \right) = 1 \hfill \\ \hfill \\ {i^5} = {i^4}i = ... \hfill \\ \hfill \\ {i^{17}} = {\left( {{i^4}} \right)^4}i = ... \hfill \\ \hfill \\ {i^{265}} = {\left\{ {{{\left( {{i^4}} \right)}^4}} \right\}^4}{i^9} = {\left[ {{{\left\{ {{{\left( {{i^4}} \right)}^4}} \right\}}^4}} \right]^4}{\left( {{i^2}} \right)^4}i = ... \hfill \\ \end{gathered} \] please complete
1. 1^5 , i^17 , I^265 = D 2.7i E 3.-10 G 4. 5/2 I
oops... \[\Large {i^{265}} = {\left[ {{{\left\{ {{{\left( {{i^4}} \right)}^4}} \right\}}^4}} \right]^4}{i^9} = {\left[ {{{\left\{ {{{\left( {{i^4}} \right)}^4}} \right\}}^4}} \right]^4}{\left( {{i^2}} \right)^4}i = ...\]
\[\Large 2i + 5i = i\left( {2 + 5} \right) = ...\]
the second one equal 7i
lol i did all the work i jst want to see if im right
\[\Large \left( {2i} \right) \times \left( {5i} \right) = 2 \times 5 \times i \times i = 10{i^2} = ...\]
please compare my answers with yours
\[\Large \frac{{5i}}{{2i}} = \frac{5}{2}\]
ok thank you very much
\[\Large \left( {3 + 2i} \right) + \left( {15 - 8i} \right) = \left( {3 + 15} \right) + i\left( {2 - 8} \right) = 18 - 6i\]
\[\Large \left( {3 + 2i} \right) - \left( {15 - 8i} \right) = \left( {3 - 15} \right) + i\left( {2 + 8} \right) = - 12 + 10i\]
ok thank i was correct
\[\large \left( {3 + 2i} \right) - \left( {15 - 8i} \right) = \left( {3 - 15} \right) + i\left( {2 + 8} \right) = - 12 + 10i\]
i have to log off thank you so much for the help
:)
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