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5

5childx7marble=35 + 2 marbles left

I have assgined Sisyphus this task......he says he has completed it.......but......he got careless and left the results behind somewhere in the "Illuminati Triangle"...........

That damned Sisyphus......he's worthless.......!!!!

Make sure your calculator is in "degree" mode......then, just evaluate

13*cos(48)    .....you should get the correct answer......

√x=-2    this isn't possible.....to see why.....square both sides

x = 4      but, √4 = 2    not  -2

Thus, we can't get a "negative" result from a "positive" radical.......

$${\frac{{\mathtt{5}}}{{\mathtt{100}}}} = {\mathtt{0.05}}$$

?*7=42... divide both sides by 7.

Use the formula 180(n-2) to find the sum of the interior angles of an n-gon.

The sum would be 180*81 degrees.

cphill we are still waiting for you

the clock is ticking

(20 + 12) / (4 + 4) = 4

32 / 8   =   4

x^2 -7x + 12 = 0      factor

(x - 4) (x - 3)   = 0      set both factors to 0    and x = 4 or x = 3

how do you convert degrees to radians ?

It does not pay to rush MG

You said you were going to  -4 but then you added 4 instead!

Plus you used the latex for cancelling when there was no cancelling to do!

You need to edit this one  - or redo it and delete the original.   :)

Six points on a circle are given. Four different chords joining pairs of the six points are selected at random. What is the probability that the four chords form the sides of a convex quadrilateral ?

Let the six points 1, 2, 3, 4, 5, 6

All convex quadrilateral  are all selection(4 Numbers)  in ascending order from the set 1, 2, 3, 4, 5, 6

$$\rm{convex~quadrilateral } = \bordermatrix{Points& 1 & 2 & 3 & 4 & 5 & 6 \cr
1. &1 &2 &3 &4 & & \cr
2. &1 &2 &3 & &5 & \cr
3. &1 &2 &3 & & &6 \cr
4. &1 &2 & &4 &5 & \cr
5. &1 &2 & &4 & &6 \cr
6. &1 &2 & & &5 &6 \cr
7. &1 & &3 &4 &5 & \cr
8. &1 & &3 &4 & &6 \cr
9. &1 & &3 & &5 &6 \cr
10.&1 & & &4 &5 &6 \cr
11.& &2 &3 &4 &5 & \cr
12.& &2 &3 &4 & &6 \cr
13.& &2 &3 & &5 &6 \cr
14.& &2 & &4 &5 &6 \cr
15.& & &3 &4 &5 &6 \cr} = \binom64$$

Point-connections:

$$\small{\text{$
\begin{array}{|c|c|c|c|c|c|c|}
\hline
&1&2&3&4&5&6 \\
\hline
1&x&1\Rightarrow2&1\Rightarrow3&1\Rightarrow4&1\Rightarrow5&1\Rightarrow6 \\
2& &x&2\Rightarrow3&2\Rightarrow4&2\Rightarrow5&2\Rightarrow6 \\
3& & &x&3\Rightarrow4&3\Rightarrow5&3\Rightarrow6 \\
4& & & &x&4\Rightarrow5&4\Rightarrow6 \\
5& & & & &x&5\Rightarrow6 \\
6& & & & & &x \\
\hline
\end{array}
$}} = \binom62=15$$

Four  chords from 15 pont-connections

  $$=\binom{\binom62}{4}=\binom{15}{4}$$

the probability that the four chords form the sides of a convex quadrilateral

 $$\small{\text{$
\begin{array}{rcl}
&=&\dfrac{\binom64}{\binom{15}{4}}
=\dfrac{ \dfrac{6!}{4!\cdot(6-4)!} }{ \dfrac{15!}{4!\cdot(15-4)!} }
=\dfrac{6!}{2!} \cdot \dfrac{11!}{15!}
=\dfrac{3\cdot 4\cdot 5\cdot 6}{ 12\cdot 13 \cdot 14 \cdot 15}
=\dfrac{36}{3276}
=0.01098901099
\end{array}
$}}$$

You welcome Mellie.

I have given this question my best shot :)

Thanks Nauseated - I have added this to our LaTex thread  :)

BOXES FROM NAUSEATED      18/5/15

Ok I will try it.

First of all, that is so COOL Nauseated but it must take in alot of Coding.

EDITED

$$\begin{array}{rrlll}\\
-6x+4&=&-2\qquad&(-4\;to\;get\;X\;alone)\\\\
-6x-4&=&-2-4\\\\
-6x&=&-6\qqquad&Divide\;6\;on\;both\;sides\\\\
\frac{-6x}{6}&=&\frac{6}{6}\qquad&simplify\\\\
\frac{-6x}{\not{6}}&=&\frac{6}{\not{6}}\\\\
-x&=&\frac{1}{1}
\end{array}$$

I think...Sorry i had to rush this because i was half way and my sis needs the laptop i'll think about it after.

\begin{array}{rrlll}\\
-6x+4&=&-2\qquad&(-4\;to\;get\;X\;alone)\\\\
-6x-4&=&-2-4\\\\
-6x&=&-6\qqquad&Divide\;6\;on\;both\;sides\\\\
\frac{-6x}{6}&=&\frac{6}{6}\qquad&simplify\\\\
\frac{-6x}{\not{6}}&=&\frac{6}{\not{6}}\\\\
-x&=&\frac{1}{1}
\end{array}

yea

but remember, you gave him the command since a long time ago

commands can make or break questions and answers

Poor CPhill    

Let the number of reds be a where a is at least 1

Let the number of whites be b

Let the number of blues be c

Let the number of greens be d

The probabilities below are all "Order does not count."

P(WRBG)=P(WBRR)=P(BRRR)=P(RRRR)

The denominators of these are all going to be the same so I am going to look at the numerators only

4!*abcd = (4!/2!)bca(a-1)=(4!/3!)ca(a-1)(a-2)=a(a-1)(a-2)(a-3)

24abcd = 12bca(a-1)=4ca(a-1)(a-2)=a(a-1)(a-2)(a-3)

24bcd = 12bc(a-1)=4c(a-1)(a-2)=(a-1)(a-2)(a-3)

let

24bcd                (1)

12bc(a-1)          (2)

4c(a-1)(a-2)      (3)

(a-1)(a-2)(a-3)      (4)

24bcd=12bc(a-1)      (1)=(2)

2d=a-1

$${\mathtt{d}} = {\frac{\left({\mathtt{a}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{{\mathtt{2}}}}$$

4c(a-1)(a-2)=(a-1)(a-2)(a-3)         (3)=(4)

4c=a-3

$${\mathtt{c}} = {\frac{\left({\mathtt{a}}{\mathtt{\,-\,}}{\mathtt{3}}\right)}{{\mathtt{4}}}}$$

12bc(a-1)  = 4c(a-1)(a-2)             (2)=(3)

3b = (a-2)

$${\mathtt{b}} = {\frac{\left({\mathtt{a}}{\mathtt{\,-\,}}{\mathtt{2}}\right)}{{\mathtt{3}}}}$$

SO WE HAVE

$${\mathtt{a}}$$                $${\mathtt{b}} = {\frac{\left({\mathtt{a}}{\mathtt{\,-\,}}{\mathtt{2}}\right)}{{\mathtt{3}}}}$$                  $${\mathtt{c}} = {\frac{\left({\mathtt{a}}{\mathtt{\,-\,}}{\mathtt{3}}\right)}{{\mathtt{4}}}}$$                         $${\mathtt{d}} = {\frac{\left({\mathtt{a}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{{\mathtt{2}}}}$$

NOW all these must be whole numbers.

a has to be odd   (from d)

a maybe 7, 11    ( looking at c and checking with b)    a=11 works

a=11,     b=3,   c=2,    5

Minimum number of marbles is   11+3+2+5 = 21

I have not checked that these numbers work Mellie - you will have to do that!        

Thank you very much. I did get 0,7889... but i forgot to calculate with sinus 

Big thanks from dk, heureka :)

oh wait, i almost forgot,

awesome, you can come too

*makes 2 towers fall by themselves usng the force while making one rise

alright, let's see which one of you dodged a bullet, and which one of you have answered your last question

it's game on or game over

anyways, you got these bad and nuaghty numbers to conquer .123, .123, .112233, .13, -.0123, 0.12233, 1/8=.125

alright, so we got to put them in order (yes, we both are on the same side chris, we both hate the number 0 which is why i just wrote the numbers after 0, so you're welcome)

-.0123, .112233, .12233, .123, .123, .125, .13

overall, the 2nd and 6th number is .112233 and .125

so, add .112233 with .125 and you get: .237233

overall,  in the ultimate battle between Brittany and CPhill vs Titanium and Food, Food won

the corect answer is .237233

yea i'm kidding about food winning

CHRIS, YOU WON TODAY'S BATTLE ROYALE

AND AS A REWARD, YOU ARE FORCED TO WRITE ALL THE DIGITS OF PI IN ROMAN NUMERALS (yes, with the 0 as roman numerals as well)

that was a command given by Queen Guinevere of the Magical Realm Of Camelot AND by the Emperor of GVMS Royality (and yes, Royal society runs by the Emperor, Shogun, Nobles, Lords, Knights, Peasants, and traitors/enemies/slaves[yea, we decided to be nice to peasants and place the enemy at the bottom]) while I don't know what Camelot society ran by. but there's no king and queen in the empire, it's the emperor

well, better get started, Chris everyone will be waiting for you

Brittany and I are gonna go and get ice cream. You can get one too once you're done with your task that you've been assigned since the Ancient World

$$$$$$\setlength{\fboxsep}{20pt}
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I have a triangle with C = 90 degrees and a c = 166 cm and last but not least an a = 130,96 cm 

which formulas and calculations do i have to do to find out the missing lengths (A,b, B) ?

$$\\\small{\text{$
\begin{array}{rcl}
\sin{(A)}&=&\dfrac{a}{c}\\\\
\sin{(A)}&=&\dfrac{ 130.96~\rm{cm} }{ 166.00~\rm{cm} }\\\\
\sin{(A)}&=&0.78891566265\\\\
A&=&52.0842936445\ensurement{^{\circ}}
\end{array}
$}}\\\\\\
\small{\text{$
\begin{array}{rcl}
B&=& 90\ensurement{^{\circ}} - A\\\\
B&=& 90\ensurement{^{\circ}} - 52.0842936445\ensurement{^{\circ}}\\\\
B&=& 37.9157063555\ensurement{^{\circ}}\\\\
\end{array}
$}}\\\\
\\\small{\text{$
\begin{array}{rcl}
\sin{(B)}&=&\dfrac{b}{c}\\\\
b &=& c\cdot \sin{(B)}\\\\
b &=& 166.00~\rm{cm} \cdot \sin{(37.9157063555\ensurement{^{\circ}})}\\\\
b &=& 166.00~\rm{cm} \cdot 0.61450148676 \\\\
b &=& 102.01~\rm{cm} \\\\
\end{array}
$}}$$