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Melody i love all your wraps!

Yes it is. It's an equations system...

Just to be sure,

are these 

$$\\x^6=y^4+18\\
and\\
y^6=x^4+18$$       ???

@@ End of Day Wrap       Sun 16/11/14        Sydney,   Australia         Time 8:30 pm             ♬

Hi Everybody,

It was a little quiet today - some people don't do maths on the weekend - go figure ?

Anyway we still had great answers from CPhill, BrittanyJ, jillybean, geno3141 and Will85237.  Plus LancelotLink left 315 some cute and furry offerings on loan. 

If you want to find posts by a particular member this is how it is done. Up the top right of your screen is a search box.  Jusy type in  by:LancelotLink   (or the member that you are after) then his/her posts will be retreived:)

Now for a few interest posts:

1) Partial fractions - I haven't seen these on the forum before.  CPhill and Melody  

Chris and I learned/revised how to do them.  They are needed for some integrations.  

This gave us a small sense of achievement.  :)

Mon 17/11/14

1) Let's end the day with a laugh  Thanks Roasla.

Well, 4/3 is equivilent to 4÷3, or 4 x (1/3), and all of these will work on the calculator.

To do it in your head, try dividing an easier number by 3, such as 1, and then multiply by 4.

That just leaves one more

c)

$$\\f(x)=\frac{x^2-10}{(x-2)(x+1)}\\\\
f(x)=\frac{x^2-10}{x^2-x-1}\\\\$$

We need to do an algebraic (or synthetic) division

$$\begin{tabular}{ccccccc} &&&\;1&&& &&&&||&||&||& \\
x^2&-x&-1&\|\;x^2&0&-10\\
&&&\;x^2&-x&-1\\ &&&||&||&||&& &&&&+x&-9&\\

\end{tabular}$$

so

$$\\\frac{x^2-10}{(x-2)(x+1)}=1\;+\;\frac{x-9}{x^2-x-1}\\\\
\frac{x^2-10}{(x-2)(x+1)}=1\;+\;\frac{x-9}{(x-2)(x+1)}\\$$

Now let

$$\\\frac{x-9}{(x-2)(x+1)}=\frac{A}{x-2}+\frac{B}{x+1}\\\\
\frac{x-9}{(x-2)(x+1)}=\frac{A(x+1)+B(x-2)}{(x-2)(x+1)}\\\\$$

Equating numerators we have

$$\\x-9=A(x+1)+B(x-2)\\\\
x-9=(A+B)x+(A-2B)\\\\
so\\
A+B=1\quad\rightarrow\quad A=1-B\\
-9=A-2B\\
-9=1-B-2B\\
-10=-3B\\
B=10/3\\\\
A=1-10/3\\
A=1-3/3-7/3\\
A=-7/3$$

therefore

$$\\f(x)=\frac{x^2-10}{(x-2)(x+1)}\\\\
f(x)=1\;+\;\frac{x-9}{(x-2)(x+1)}\\\\
f(x)=1\;-\;\frac{7/3}{x-2}+\frac{10/3}{x+1}\\\\$$

Ah Chris you beat me to it I was just about to present my answer.

Luckily it is identical to your answer .

so Part B is now officially solved.    

Irrelevant to what?

It is obvioulsy not irrelevenant to Brittany or CPhill. 

Lighten up - they are just chatting.  

Most of us want this forum to be a place of enjoyment as well as a place of learning.

We just have to be careful to get the balance right - that is all.    

that is irrelevant...

b)f(x)=x^2 +5x +7/(x+2)^3 = A/(x+2) + B/(x + 2)^2 + C/(x + 2)^3 ....multiply both sides by (x+2)^3....so we have

x^2 + 5x + 7 =   A(x+2)^2 + B(x+2) + C

x^2 + 5x + 7 = A(x^2 + 4x + 4) +B(x + 2) + C

x^2 + 5x + 7 = A(x^2 + 4x + 4) +B(x + 2) + C     

x^2 + 5x + 7 = Ax^2 +(4A+ B)x + (4A + 2B + C)

So, equating coefficients..

A = 1

4A + B = 5  →  4 + B = 5 →  B = 1

4A + 2B + C = 7  → 4 + 2 + C = 7 →  6 + C = 7   → C  = 1

So we have...

x^2 +5x +7/(x+2)^3  =  1/(x+2) + 1/(x + 2)^2 + 1/(x + 2)^3

I don't think so but you can do them on this one

you forgot to add the 24 Brittany  

I don't think that the web2 calc has a random number generator - it may have but I can't find it.

I think many calcs generate a random number between 0 and 1

I would multiply this generated number by 100 and round it off to get random integers between 0 and 100.

I suppose if you get zero you could ignore that one   

 Hi Brittany

This thread is kept in with the sticky notes - If people aske about fractions on the calc, sending them to this thread is a very good option.  

With each of these terms the denominator is 3 more than the numerator.

You could choose any letter (pronumeral) but they have chose the letter k to be a numerator.

If the numerator is k then the denominator would be k+3

So one way to describe each of these terms is     $$\frac{k}{k+3}$$    so this can be used as the general term.

Now the first term is    $${\frac{{\mathtt{4}}}{{\mathtt{7}}}}$$      so for the first term k=4

The second term is      $${\frac{{\mathtt{5}}}{{\mathtt{8}}}}$$       so for the first term k=5

etc

The last  term is         $${\frac{{\mathtt{10}}}{{\mathtt{13}}}}$$       so for the first term k=10

so it is the sum of all those terms from k=4 to k=10

$$\sum\limits_{\;k=4}^{10}}\;\frac{k}{k+3}$$

Do you understand any better?

This could have been expressed in many ways

For instance

$$\\\sum\limits_{\;k=7}^{13}}\;\frac{k-3}{k}\\\\\\
\sum\limits_{\;a=7}^{13}}\;\frac{a-3}{a}\\\\\\
\sum\limits_{\;g=1}^{7}}\;\frac{g+3}{g+6}\\\\\\$$

these all describe exactly the same sum    

You can not get rid of the x - sorry but that is how it is :)

Thanks Chris - I saw this hours ago when it first popped up.

Gave me a good laugh.  Thank you   

This has no meaning to me.  

This is how it is doen for a 2x2 matrix

Well I suppose

1#2=(1+2)/2=1.5

1.5#3=(1.5+3)/2=2.25

2.25#4=(2.25+4)/2=6.25/2=3.125 = 3 and 1/8

Each day the earth rotates on its own axis once.  That is what a day is!

so   

In 24 hours the Earth rotates 360 degrees (of longitude)

So in 1 hour it must rotate 360/24=15 degrees

that is, in 60 minutes it rotates 15 degrees

so in 60/15minutes it will rotate 15/15 degrees

that is

in    4 minutes it will rotate 1 degrees

SEE

This is why time and degrees are BOTH split into minutes and seconds.  

Hours, minutes and seconds of time  are directly related to the degrees, minutes and seconds that the Earth has rotated.

I think that this is really COOL!!!!!

Let  W = 2L - 3  .....    so we have

(2L - 3)(L) = 317       simplify

2L² - 3L = 317           subtract 317 from both sides

2L² - 3L - 317  = 0      using the onsite calculator and subbing x for L, we have

$${\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{317}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{2\,545}}}}{\mathtt{\,-\,}}{\mathtt{3}}\right)}{{\mathtt{4}}}}\\
{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{2\,545}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}\right)}{{\mathtt{4}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{11.861\: \!998\: \!255\: \!629\: \!438\: \!5}}\\
{\mathtt{x}} = {\mathtt{13.361\: \!998\: \!255\: \!629\: \!438\: \!5}}\\
\end{array} \right\}$$

So...L =  13.4 ft

And W = 2(13.4) - 3 =  23.8 ft

Maybe you can formulate your own answer from this

did you ask this question yesterday?

I think I sent you to another question that I had already answered for you that was almost identical.

Let see if I can find it.

Mmm

Here is where I answered you last time!