if you've read those links already, you'd know what we're doing here.
we'll move the repeating part to the left-side of the dot, by multiplying by "1" and as many zeros as needed, or 10 at some power pretty much.
on 0.13 we need 100 to get 13.13.... and on 0.1234, we need 10000 to get 1234.1234....
![\bf 0.\overline{13}~\hspace{10em}x=0.\overline{13} \\\\[-0.35em] ~\dotfill\\\\ \begin{array}{|lll|ll} \cline{1-3} &&\\ 100\cdot 0.\overline{13}& = & 13.\overline{13}\\ 100\cdot x&& 13 + 0.\overline{13}\\ 100x&&13+x \\&&\\ \cline{1-3} \end{array}\implies \begin{array}{llll} 100x=13+x\implies 99x=13 \\\\ x=\cfrac{13}{99} \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ 0.\overline{1234}~\hspace{10em}x=0.\overline{1234} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%20%5Cbf%200.%5Coverline%7B13%7D~%5Chspace%7B10em%7Dx%3D0.%5Coverline%7B13%7D%0A%5C%5C%5C%5C%5B-0.35em%5D%0A~%5Cdotfill%5C%5C%5C%5C%0A%5Cbegin%7Barray%7D%7B%7Clll%7Cll%7D%0A%5Ccline%7B1-3%7D%0A%26%26%5C%5C%0A100%5Ccdot%200.%5Coverline%7B13%7D%26%20%3D%20%26%2013.%5Coverline%7B13%7D%5C%5C%0A100%5Ccdot%20x%26%26%2013%20%2B%200.%5Coverline%7B13%7D%5C%5C%0A100x%26%2613%2Bx%0A%5C%5C%26%26%5C%5C%0A%5Ccline%7B1-3%7D%0A%5Cend%7Barray%7D%5Cimplies%20%5Cbegin%7Barray%7D%7Bllll%7D%0A100x%3D13%2Bx%5Cimplies%2099x%3D13%0A%5C%5C%5C%5C%0Ax%3D%5Ccfrac%7B13%7D%7B99%7D%0A%5Cend%7Barray%7D%0A%5C%5C%5C%5C%5B-0.35em%5D%0A%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%0A0.%5Coverline%7B1234%7D~%5Chspace%7B10em%7Dx%3D0.%5Coverline%7B1234%7D%0A%5C%5C%5C%5C%5B-0.35em%5D%0A~%5Cdotfill%20)

Answer:
-13 + j*2
Step-by-step explanation:
The additive inverse of a complex number x = a +j*b
is a number y, such that
x + y = 0
This means that
y = -x = - a - j*b
Therefore
The additive inverse of 13 - j*2 is equal to
-(13 - j*2) = -13 +j*2
Answer:
List method: {7,9,11,13,15,17,19}
Set method: {X:N where N is odd, N>5 and N<21}
Sqrt 363 - 3 sqrt27
= sqrt(121* 3) - 3 sqrt (9*3)
= sqrt 121* sqrt3 - 3 * sqrt 9 * sqrt3
= 11 sqrt3 - 3* 3 sqrt3
= 11 sqrt3 - 9 sqrt3
= 2 sqrt3 Answer