Answer:
[tex]w=100-\frac{(100-k)^2}{100}\\\\z={\frac{100}{\left(1-\frac{k}{100}\right)^2} -100[/tex]
Step-by-step explanation:
[tex]\underline{w \ in \ terms \ of \ z:}[/tex]
[tex]d=\left(1-\frac{k}{100}\right)c\\a=\left(1-\frac{k}{100}\right)d\\ \Rightarrow a=\left(1-\frac{k}{100}\right)^2c...........(1)\\[/tex]
[tex]Again \ a=\left(1-\frac{w}{100}\right)c...........(2)\\[/tex]
from equation [tex](1)[/tex] and [tex](2)[/tex]
[tex]\left(1-\frac{k}{100}\right)^2=\left(1-\frac{w}{100}\right)\\\frac{w}{100}=1-\left(1-\frac{k}{100}\right)^2\\w=100-\frac{(100-k)^2}{100}[/tex]
[tex]\underline{z \ in \ terms \of \ w:}[/tex]
[tex]c=\left(1+\frac{z}{100}\right)a.......(3)\\[/tex]
from equation [tex](1)[/tex] and [tex](3)[/tex]
[tex]a=\left(1-\frac{k}{100}\right)^2\left(1+\frac{z}{100}\right)a\\\left(1+\frac{z}{100}\right)=\frac{1}{\left(1-\frac{k}{100}\right)^2}\\\frac{z}{100}=\frac{1}{\left(1-\frac{k}{100}\right)^2}-1\\z=100\left[\frac{1}{\left(1-\frac{k}{100}\right)^2}-1\right][/tex]
