Question

simplify the expression

Answer 1

\[\left[\begin{matrix}-6+ & i \\ -5+ & i\end{matrix}\right]\]
A. \[\left[\begin{matrix}31+&i \\ 26 \end{matrix}\right]\]
B \[\left[\begin{matrix} 31+ & i \\ 24\end{matrix}\right]\]
C \[\left[\begin{matrix}29+ & i \\ 26\end{matrix}\right]\]
D \[\left[\begin{matrix}31+11i \\ 26\end{matrix}\right]\]

Answer 2

are they fractions?

Answer 3

yeah lol i tried to do that i couldnt find it

Answer 4

\[\frac{-x+i}{-5-i}\] multiply top and bottom by the conjugate of the denominator
the conjugate of \(a+bi\) is \(a-bi\) and this works because \((a+bi)(a-bi)=a^2+b^2\) a real number

Answer 5

\[\frac{-6+i}{-5-i}=\frac{-6+i}{-5-i}\times \frac{-5+i}{-5+i}\]
\[=\frac{(-6+i)(-5+i)}{5^2+1^2}\]

Answer 6

the denominator is \(26\)
your last job is to multiply out in the numerator, i.e. multiply
\[(-6+i)(-5+i)\]

Answer 7

now let me rant
this is not called "simplifying"
it is dividing
and in fact none of your answers are complex numbers in standard for as
\[a+bi\]
the correct answer is
\[\frac{29}{26}-\frac{11}{26}i\]

Answer 8

oops i made a mistake, it is
\[\frac{-6+i}{-5+i}=\frac{36}{26}+\frac{1}{26}i\]

Answer 9

go with A

Answer 10

then as your teacher what he or she means by "simplify" because it means nothing

Answer 11

thank you

Answer 12

yw