Melody

I don't know what you mean but there are 8 lots of 15 seconds in 2 minutes.

I am not very familiar with this type of question but I think it is just

$$(x,y)\rightarrow (x-7,y-4)$$

I always find it easier if I write the second one like this   $$y+4=(x+7)^2$$

Look at the graphs.  That will help too.

Do you mean this bit?

$$\begin{array}{rll}
\frac{3500}{T}&=&\left(\frac{3000}{T}+50\right)\\\\
T\times\frac{3500}{T}&=&T\times\left(\frac{3000}{T}+50\right)\\\\

\end{array}$$

I multipled both sides by T so that I could get the T off the bottom.

See, the next line just becomes this;

$$\begin{array}{rll}

3500&=&3000+50T\\\\

\end{array}$$

Because the Ts cancelled out

I'll put in a bit more working to show you.

$$\begin{array}{rll}
\frac{3500}{T}&=&\left(\frac{3000}{T}+50\right)\\\\
T\times\frac{3500}{T}&=&T\times\left(\frac{3000}{T}+50\right)\\\\
\frac{3500\times T}{T}&=&\frac{3000\times T}{T}+50\times T\\\\

3500&=&3000+50T\\\\
500&=&50T\\\\
T&=&10
\end{array}$$

If you still do not understand let me know.   Also tell me if it is still the same spot that is causing you problems because I already have ideas on how I can help you understand.  Just think about it for a little while first though.   

That is great Rosala,

I'd like to chat with you. I'll send you a message.  

@@ End of Day Wrap - Saturday 24/5/14   Sydney, Australia   Time 21:30

Hi everyone,  

I trust that you are all enjoying your weekend!  It has been a fabulously sunny day in Sydney!

Great answers were given today by Takahiro Maeda, CPhill, Rosala, Alan, Rom, reinout-g, rag3phone, zegroes, nadiyah and Bertie.  Thank you, you are all wonderful.

I haven't heard from Andre Massow but I saw no more evidence of the hacker today so presumably Andre has taken care of the problem.  Only Chris and I were complaining yesterday.  It was driving us nuts.  I am wondering if we were the only ones affected. Doesn't matter, hopefully the problem is finished with.

Here are some posts that may be of interest.

Speed questions.  Good (High level) junior high problems.  Students often find questions relating to speed very difficult.  I think that they are often quite interesting.  Rosala presented a few today.

Can anyone answer anonymous's question?

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"But I see that it can also give wrong estimates sometimes

Can someone inform me about these cases? please :D"

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Thanks you Bertie.    

Hi Stacy

You need to use the quadratic formula.

x=c    {c: r1 < c < r2 }

maybe?

where c is a given constant.

I'll answerthe question asked but you do need to use brackets AllSmilesAlways to make your meaning clear.

$$\begin{array}{rll}
12^x-2&=&20\\\\
12^x&=&22\\\\
log(12^x)&=&log22\\\\
xlog12&=&log22\\\\
x&=\frac{log22}{log12}\\\\
x&\approx &1.244 \mbox{ correct to 3 decimal places}
\end{array}$$

Can someone please tell me why this did not align properly?

\begin{array}{rll}

12^x-2&=&20\\\\

12^x&=&22\\\\

log(12^x)&=&log22\\\\

xlog12&=&log22\\\\

x&=&\frac{log22}{log12}\\\\

x&\approx &1.244 \mbox{ correct to 3 decimal places}

\end{array}