🚀 go-pugleaf

Hello Stan,

That guy (ROY VELARDI <royvelardi@bigpond.com.au) is just con artist.
He sent his homework assignment over 4 newsgroups, being sure that
idiots like you will solve all that instead of him while
he is dealing with searching RX3 Tailights (savanna type) for his car.
(which cost about $1700), so he is not poor student!
    Ok,ok, be fool - do assignments for that scoundrel...
He sent his assignment on 24th of May, well, look at his posting
to rec.autos.rotary being sent on 26 of May (below):
=====start quote:===
Message 1 in thread 
From: ROY VELARDI (royvelardi@bigpond.com)
Subject: WANTED: RX3 Tailights 
Newsgroups: rec.autos.rotary
View this article only 
Date: 2002-05-26 01:38:43 PST 
 

Hi All

Im looking for a set of Rx3 Tailights the savanna type either the rotor
shaped ones or hexagon shaped ones. (can provide photos of if you are
unsure)
If you have a pair you want to sell or a friend does please either reply to
this post or email me at the following address royvelardi@bigpond.com.au

Cheers From Australia!

Roy
Post a follow-up to this message

Message 2 in thread 
From: Chris Boehm (zlakto@ihug.com.au)
Subject: Re: WANTED: RX3 Tailights 
Newsgroups: rec.autos.rotary
View this article only 
Date: 2002-05-26 09:03:21 PST 
 

Sorry I can't help you, but I have seen those around and f#$k there
expensive, for a set of series 1 savanna tail lights he wanted $1,700 for
them, good luck

Chris B.

====end quote: ====





qx1741@bigfoot.com (Stan Brown) wrote in message news:<MPG.175811b8bee9472d989a97@news.odyssey.net>...
> [cc'd to previous poster; please follow up in newsgroup not mail]
> 
> ROY  VELARDI <royvelardi@bigpond.com> wrote in uk.education.maths:
> >hi can anyone out there please help me solve these? if you can email me at
> >the following address royvelardi@bigpond.com.au thanks roy.
> 
> In general if you post a question here you should plan to read the 
> answer here, but I'll send you a copy of my follow-up. 
> 
> >1. log (y + 2) = log (y - 7) + log 4
> 
> Hint: Can you combine log(y-7) and log(4) using one of the laws of 
> logarithms? Then you have an equation in the form log A = log B, 
> which means that A = B, so you'll have a simple equation in y.
> 
> >2. 5 (e^2x - 2) = 15
> 
> Treat "e^2x" like a variable, and isolate it on one side of the 
> equation. (This means you'll have to do the usual things to 
> eliminate the factor of 5 and the subtraction of 2.) Once you have 
> e^2x = something, take natural log of both sides and finish the 
> solution.
> 
> >3. solve for x     4/8-x = 3/4
> 
> Add x to both sides, subtract 3/4 from both sides, and simplify. Did 
> you mean to post a different equation? This one seems much easier 
> than the others.
> 
> >4. y - 6/y - 6/y = y + 6/y - 6
> 
> Multiply left and right by y and simplify. (You probably made a 
> typing mistake, "- 6/y - 6/y" seems unlikely to have been part of 
> the actual problem of you're at the level of doing logs and 
> exponentials.)
> 
> If you can't complete any solutions, post again SHOWING YOUR WORK 
> and I'll be happy to help further.