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I think he/she means solving the equations on the calculator...I just tried it.....and it seemed OK........

I do not understand your question  

@@ End of Day Wrap      Sun 1/2/15        Sydney, Australia         Time 10:50  pm

Hi all,

Today our wonderful answerers were CPhill, Heureka, Rosala and Bertie.  Thank you.    

Well the weekend is always a bit quieter and today was no exception but there was a high proportion of both really good and really hard questions.  So I found choosing the interest posts difficult today. 

I always attempt to include a cross section of questions so that most people can find something of interest.   If anyone ever wants a particular question included or if you have a general comment or suggestions please let me know.  It is probably best to do this by private message or on the Lantern post.

Interest posts:

Mon 2/2/15

CPhill just sent me this on continued fractions for anyone who is interested :)

20% of 830=  20/100 of 830 = (830:100)x20 =8.3 x 20=.... 16.60

OK, thanks....

Yes it is  !!!!

Successive iterates are 1, 2/3, 3/4, 8/11, 11/15, 30/41, 41/56, 112/153, 153/209, 418/571, 571/780 ...

They are converging to sqrt(3) - 1 = 0.7320508... .

The last one that I've listed, 571/780 = 0.732051 6dp.

That is really neat Bertie  

Go into Formulary at the top of this page.

Click on sphere

use the formula for volume  

Yep, Melody is correct....Order of Operations....!!!

cosec = 1/sin = Hyp/opp

draw the triangle

find the third side using pythagorean theorem

Read all  the trig functions straight off the triangle. 

Bertie....let me ask you a question about this one.....when I carry it out for a few iterations, I get that the results seem to "oscillate' between both roots of that quadratic, i.e., sqrt(3)±1

Did I do something wrong???

$$\\6736.76=5000(1+r)^{24}\\
(6736.76/5000)=(1+r)^{24}\\
(6736.76/5000)^{1/24}=1+r\\
(6736.76/5000)^{1/24}-1=r\\
r=(6736.76/5000)^{1/24}-1\\$$

$${\left({\frac{{\mathtt{6\,736.76}}}{{\mathtt{5\,000}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{24}}}}\right)}{\mathtt{\,-\,}}{\mathtt{1}} = {\mathtt{0.012\: \!500\: \!029\: \!732\: \!883}}$$

$$\\I=Prt\\
650=8375*0.0625*t$$

Now solve for t

Put the expression equal to N (an integer) and square it.

That gets you N^2 =  t- N  or, t = N^2 + N.

The largest integer value for N that can be substituted so that t is still a 4 digit numbe is 99 and that gets

you t = 9900.

The trick is that you have to multiply first BEFORE you add or subtract anything!

I do not understand this at all.   

thanks Bertie,

I did not have a clue what to make of this question.    

you got a mention in my NewYear 2015 post  

Very nice, Bertie....!!!

What does 115r30 mean?

Evaluate a^2b^7/ a^3b as a fraction in index form when a = (2/5)^4 and b = (5/8)^3.

$$\\\frac{a^2b^7}{ a^3b}=\frac{b^6}{ a}=b^6\div a\\\\
\left(\frac{5^3}{8^3}\right)^6\div \frac{2^4}{5^4}=
\frac{5^{18}}{8^{18} }\times \frac{5^4}{2^4}=
\frac{5^{18}}{8^{18} }\times \frac{5^4}{2*8}=
\frac{5^{22}}{2*8^{19} }=
\frac{5^{22}}{2*(2^3)^{19} }=
\frac{5^{22}}{2*2^{57} }=
\frac{5^{22}}{2^{58} }$$

Put the continued fraction equal to x, and make the substitution further into the fraction to get

$$x=\frac{1}{1+\frac{1}{2+x}}}$$ 

Tidying that up gets you quadratic $$x^{2}+2x-2=0,$$ from which $$x=\sqrt{3}-1.$$