Melody

Those last 2 are brillliant Chris.  :)

IDENTITY THEFT        LOLLOLLOL  hahaha :D

I'm going to copy that one and post it on other sites :))

I like those faces especially  the real face and the imaginary face.  !!!  :D

But I am stuck on the traslation of h and i :(

I'm gong to try and remember that rhombus one!    LOL    hahaha :)

Very nice Max !

Hi EP, your one might have had its roots exposed, but my one grew square roots   

Why do teachers give out lots of homework?

So that someone in the class might have some hope of solving tomorrows (as in 20 years away) problems.

There are 20 responses here and I cannot see a single one that answers the question 7. 

And it seems to me that question 7 is the one that is highlighted :/

\(\quad \displaystyle\lim_{x \rightarrow 4}\: \frac{\sqrt{5x-4}-4}{4-x}\\ \text{Rationalize the numerator}\\ = \displaystyle\lim_{x \rightarrow 4}\: \frac{(\sqrt{5x-4}-4)(\sqrt{5x-4}+4)}{(4-x)(\sqrt{5x-4}+4)}\\ = \displaystyle\lim_{x \rightarrow 4}\: \frac{(5x-4-16)}{(4-x)(\sqrt{5x-4}+4)}\\ = \displaystyle\lim_{x \rightarrow 4}\: \frac{5(x-4)}{-(x-4)(\sqrt{5x-4}+4)}\\ = \displaystyle\lim_{x \rightarrow 4}\: \frac{5}{-(\sqrt{5x-4}+4)}\\~\\ =\dfrac{-5}{8} \)

My answer has been questioned so I am going to do it a little more main stream and show that the 2 answers are the same. :)

\(i^1=i\\ i^2=-1\\ i^3=-i\\ i^4=1\\ i^5=i\\ i^6=-1\\ i^7=-i\\ \text{and so the pattern continues}\\ i^{4n}=1\\ i^{4n+1}=i\\ i^{4n+2}=-1\\ i^{4n+3}=-i \)

Hence       \(i^3=-1 \qquad and \qquad i^9=i\)

\(\begin{align} \displaystyle\lim_{x\rightarrow i }\; \frac{{z}^{3}+2}{{z}^9+8}& =\frac{-i+2}{i+8}\\ & \text {Rationalize the denominator}\\ &=\frac{2-i}{8+i}\times \frac{8-i}{8-i}\\ &=\frac{16-2i-8i-1}{64--1}\\ &=\frac{15-10i}{65}\\ &=\frac{3-2i}{13}\\ &=\frac{3}{13}-\frac{2i}{13}\\ \end{align}\)

See Majid, the answer is the same       

*

What happened to the plant in the maths class?

ok :))

Hi Majid,    

while z is a complex number

\(\begin{align} \displaystyle\lim_{x\rightarrow i }\; \frac{{z}^{3}+2}{{z}^9+8}& =\frac{-i+2}{i+8}\\ &=\frac{-i+2}{i+8}\times \frac{i-8}{i-8}\\ &=\frac{1+8i+2i-16}{-1-64}\\ &=\frac{10i-15}{-65}\\ &=\frac{2i-3}{-13}\\ &=\frac{3}{13}-\frac{2i}{13}\\ \end{align}\)