All Answers 358122 Answers

I really liked this question Rose.

I'd do it CPhill's way.

the whole circumference it 2pi*r

But you only have 80degrees out of a total of 360 degrees in the whole cirlce.

so the answer must be given by

$$arc\;length= \frac{80}{360}\times 2\pi r => \frac{80}{360}\times 2\pi\times 18$$    etc

Rosela, yeh calculators can be used, but it is maybe very difficult to do without one. For example I have no experience before this year and enter into calculus based engineering degree (first year). The calculator does not provide an answer it manifests one from your manual calculations. ALthough, I do think that it is more benificial to use a CAS which I don't know how which can help in many respects with calculations. Additionally the exam is only short 2 hours, but also the calculator is excluded from exams in later years of the degree depending on the subject. 

I am glad that you worked it out DragonSlayer.  

If you had given data someone would have answered but your question was a bit vague.

It is always good when you work out your own problems anyway.  It gives a you an added sense of achievement. 

how did you cross out those numbers Ninja?

$$(x-1)(2x^3+2x^2+2)\\\\
=x(2x^3+2x^2+2)-1(2x^3+2x^2+2)\\\\
=2x^4+2x^3+2x-2x^3-2x^2-2\\\\
=2x^4-2x^2+2x-2$$

Good work DragonSlayer  

Thanks Alan.  There is a lot of room for confusion there.  It seems another example of 'silly' notation. 

Thanks DragonSlayer

I haven't had such a good laugh in ages!  That's a good thing - I mean it in the nicest possible way. 

Thumbs up from me.  

I've looked more carefully in the book at the equations and the diagram from which they are derived, and I can see that the second equation has a misprint (in the book). The term (100 - V2)/30 should be (100 - V3)/30.  

With this change the solutions given for V2 and V3 in the book are correct.

Your question is a bit confusing - is this what your mean.

This is wrong (unless your question is wrong) I just realised that I used log base e where yours is log base 10 sorry

You can still do it straight on the web 2 calc.

$$\left[1+\frac{1}{-e^{0.05*log(0.05)}}\right]^{-1}\\\\

\left[1+\frac{1}{-e^{log(0.05^{0.05})}}\right]^{-1}\\\\

\left[1+\frac{1}{-(0.05)^{0.05}}}\right]^{-1}\\\\

\left[\frac{-(0.05)^{0.05}+1}{-(0.05)^{0.05}}\right]^{-1}\\\\

\left[\frac{-(0.05)^{0.05}}{-(0.05)^{0.05}+1}\right]^{+1}\\\\
\frac{-(0.05)^{0.05}}{-(0.05)^{0.05}+1}\\\\$$

Now you can just do it on the calculator - or you could have just done it on the calculator to start with.   

$${\mathtt{\,-\,}}{\frac{{\left({\mathtt{0.05}}\right)}^{{\mathtt{0.05}}}}{\left({\mathtt{1}}{\mathtt{\,-\,}}{\left({\mathtt{0.05}}\right)}^{{\mathtt{0.05}}}\right)}} = -{\mathtt{6.188\: \!641\: \!566\: \!681\: \!630\: \!8}}$$

$$\left({\left({\mathtt{1}}{\mathtt{\,-\,}}{{\left({{\mathtt{e}}}^{\left({\mathtt{0.05}}{\mathtt{\,\times\,}}{ln}{\left({\mathtt{0.05}}\right)}\right)}\right)}}^{-{\mathtt{1}}}\right)}^{-{\mathtt{1}}}\right) = -{\mathtt{6.188\: \!641\: \!566\: \!681\: \!631\: \!1}}$$

NOW FOR A VERSION THAT MAY BE THE REAL VERSION.

$$\left({\left({\mathtt{1}}{\mathtt{\,-\,}}{{\left({{\mathtt{e}}}^{\left({\mathtt{0.05}}{\mathtt{\,\times\,}}{log}_{10}\left({\mathtt{0.05}}\right)\right)}\right)}}^{-{\mathtt{1}}}\right)}^{-{\mathtt{1}}}\right) = -{\mathtt{14.877\: \!856\: \!312\: \!827\: \!660\: \!5}}$$

The exact value is 

$$5000000+\sqrt{10}$$

An approximate value is 

$${\mathtt{5\,000\,000}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\mathtt{10}}}} = {\mathtt{5\,000\,003.162\: \!277\: \!660\: \!168\: \!379\: \!3}}$$

year 2000 => $100million

year 2001 = > $100*1.07¹million

year 2002 = > $100*1.07^{2 millon}

year 2015 = > $100*1.07¹⁵ million = 1.07*10¹⁷  million dollars

thanks Ninja,

I have not heard of these terms before - not in this context anyway.  

thumbs up from me!

I think that they mean what line has a gradient of -5/6 and goes through (3,4)

$$\frac{y-4}{x-3}=\frac{-5}{6}\\\\
y-4=\frac{-5}{6}\times (x-3)$$

you can finish it!

Melody , I just found out by myself how to copy on a computer and its pretty easy!

But thanks for all the help!

Nice explanation Chris.  I will try and remember that one.   

According to Wikipedia:

I like your pictures too Ninja, that is why i wanted you to put the progam that you use it the "information pages" thread.  It is unfortuate that putting it there will not automatically mean that we can all immediately use it as you do.  But, it would be a good start. Thumbs up from me too. 

Oops!again a mistake !Thanks for the correction Alan!I'll just correct it!

Actually rosala, question 1 is false, because ">" doesn't mean "not equal to", it means "greater than".  Since 18 isn't greater than 19, then b+10>19 is false when b = 8.

You are right that the answer to question 2 is true; however, note that it is an equality, not an inequality.

DragonSlayer how did you get your answer!Have you learnt all this in this age!

You are right in thinking this is hard DragonSlayer554!  Far too hard to do by hand, so I've used the numerical solution capabilities of some technical software to do it:

So , this time im sure both my answers are absolutely correct!

1.out here im solving this by placing 8 in b's place! so here we go

8 + 10 > 19

18>19  ANSWER

So as 18 isnt greater than 19 the inequality is false!The correct thing should be 19>18

2.Out here we are placing 1 in place of y. So here we go!

2x1x1x1 + 3 = 5 

2+3=5 ANSWER

So this inequality is true!

I hope this helps and i have answered you test!

Yes I appologise, I can see that now that you have pointed it out.  

So, thank you Alan and my appologies to you Chris.  

$$\\i=0.075/4=0.01875\\
n \;\;quarters\\

\begin{array}{rlll}
FV&=&P(1+i)^n&\\\\
7000&=&5000(1.01875)^n&\\\\
7000/5000&=&(1.01875)^n&\\\\
7/5&=&(1.01875)^n&\\\\
log(7/5)&=&log(1.01875)^n&\\\\
log(7/5)&=&nlog(1.01875)&\\\\
log(7/5)log(1.01875)&=&n&\\\\
n&\approx &18.11\;\;quarters&\\\\
n&\approx &18.11/4\;\;years\\\
n&\approx &4.528\;\;years\\\\

\end{array}$$

Since the payments are made in quarters, the best answer is probably 19 quaters which is 4 and 3/4 years.