Melody

1 digite numbers      1 time

2 digits up to 89          8 times

90 to 98                    9 times

99                             2 times

total so far is 20

100 and something   20times

200 and something   20 times

300 and something   20 times

400 and something   20 times

Total os   5*20 = 100 times

The last one.

\(x^4y^4\sqrt{y}\\ x^4y^4y^{0.5}\\ =x^4y^{(4+0.5)}\\ =x^4y^{4.5}\\ \text{or more normally}\\ =x^4y^{\frac{9}{2}}\)

Well I am glad that you got it. 

Pity there was no 'thank you' thrown in there while you were at it though.

A 'thank you' is always appreciated.

Thanks very much guest, that is indeed the same question only not the desired answer.

There was another one more recent that that did not have a picture with it. It had a few different answers.

I saw this answered not too long ago. Perhaps someone can find the older version?

I seem to recall that it was a long answer, maybe Heureka answered it but I am not sure.   

Yes that sounds good :)

Some of these letters are easy to find so what have you found so far ?

Hi Shreyas1,

You most likely would have had an answer much more quickly if your question was presented in a readable fashion!

But here is your answer  

\(\frac{BD}{DC}=\frac{4}{3}\\ \text{so let BD=4x and DC=3x }\\ Area\;\;\triangle ABD=0.5*4x*h=24\\ \qquad \qquad 2xh=24\\ \qquad \qquad xh=12\\~\\ Area\;\;\triangle ABC=0.5*7x*h\\ Area\;\;\triangle ABC=3.5xh\\ Area\;\;\triangle ABC=3.5*12\\ Area\;\;\triangle ABC=42cm^2\\\)

NOW do the subtraction to get the final answer :)

Thanks gamesmaster

\(\sqrt{x^2}=|x| \qquad not\;\;x\)

If this expression is real then y must be positive.   

Also if an expression is simple then the denominator should be rational. (no Square root on the bottom)

Although your teacher may not care about this.\(\)

\(\sqrt\frac{4x^2}{3y}\\=\frac{|2x|}{\sqrt{3y}}\\ =\frac{|2x|}{\sqrt{3y}}\times \frac{\sqrt{3y}}{\sqrt{3y}}\\ =\frac{|2x|\sqrt{3y}}{3y} \\ =\left| \frac{2x\sqrt{3y}}{3y} \right|\\~\\ = \frac{2x\sqrt{3y}}{3y} \text{ when }x\ge0\\ = \frac{-2x\sqrt{3y}}{3y} \text{ when }x<0\)