All Answers 358087 Answers

It is 0.0048

En vinkel som är mindre än 90 grader.

An angle thats less then 90 degree.

x-y=3 => x=3+y.  3+y+y=5 => 2y=2 => y=1

x+1=5 => x=4

xy=4*1=4

x + y + x - y = 2x = 5 + 3 = 8 => x = 4

4 - y = 3 => y = 1

xy = 4*1 = 4

Here's the graphical solution to the last part of Melody's answer using Desmos...

It shows that angles B and C are, indeed, equal.....

Thanks once agaain Heureka 

Not sure what your question is

sqrt(10)/sqrt3(2)=

$$\frac{\sqrt{10}}{\sqrt[3]2}=$$

Is that what you mean?

Square and Multiply Algorithm:

$$64^{235} \mod{391}$$

$$\small{\text{
$
\begin{array}{lr}
Exponent&z=1\\
&64 \mod 391 = 64 = b\\
\hline
235\ odd: &\underbrace{1*64}_{=z*b} \mod 391 = 64 = z\\
&(\underbrace{64}_{=b})^2 \mod 391 = 186 = b \\
:2 & \\
\hline
117\ odd: &\underbrace{64*186}_{=z*b} \mod 391 = 174 = z\\
&(\underbrace{186}_{=b})^2 \mod 391 = 188 = b \\
:2 & \\
\hline
58\ even: &(\underbrace{188}_{=b})^2 \mod 391 = 154 = b \\
:2 & \\
\hline
29\ odd: &\underbrace{174*154}_{=z*b} \mod 391 = 208 = z\\
&(\underbrace{154}_{=b})^2 \mod 391 = 256 = b \\
:2 & \\
\hline
14\ even: &(\underbrace{256}_{=b})^2 \mod 391 = 239= b \\
:2 & \\
\hline
7\ odd: &\underbrace{208*239}_{=z*b} \mod 391= 55= z\\
&(\underbrace{239}_{=b})^2 \mod 391 = 35= b \\
:2 & \\
\hline
3\ odd: &\underbrace{55*35}_{=z*b} \mod 391= 361 = z\\
&(\underbrace{35}_{=b})^2 \mod 391 = 52= b \\
:2 & \\
\hline
1\ odd: &\underbrace{361*52}_{=z*b} \mod 391= \textcolor[rgb]{1,0,0}{4} = z\\
&(\underbrace{52}_{=b})^2 \mod 391= 358= b \\
:2 & \\
0\ end
\end{array}
$
}}$$

@@ End of Day Wrap          Thurs 5/2/15 Sydney,        Australia Time      11:25 pm          ♪ ♫

Hi all,

There were many great questions and answers today. 

Our fabulous answerers were Alan, Mickado, eismax13, CPhill, Hellokittygirl56, Geno3141, Trincent, Happy7, NottheBestMathMaster, soulReaper222, BrittanyJ, Zegroes, Heureka, Gesjls, Awesomenes, TakahiroMaeda, Prodigyinspired and Saseflower.

It is fabulous that so many young people get in and help answer the questions.  This is what I always aspired to.  Thanks guys 

When you are asked "How do I type is a fraction" or how do I do inverse trig, this would be a good post to send them too

long modulo( long b, long x, long n) { // b = basic, x = exponent, n = modulo

   long z=1; b=b % n;

   while (x !=0) {

      if (x%2 !=0) z = (z*b)%n; // Exponent odd

      b =(b*b)% n; x = x/2;

  } // while

  return z;

}

Thanks Heureka 

see: For example:

x*2+8x+16=7   subtract 7 from both sides

x^2 + 8x + 9 = 0   this won't factor  ... using the on-site solver, we have

$${{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{8}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{9}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\sqrt{{\mathtt{7}}}}{\mathtt{\,-\,}}{\mathtt{4}}\\
{\mathtt{x}} = {\sqrt{{\mathtt{7}}}}{\mathtt{\,-\,}}{\mathtt{4}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{6.645\: \!751\: \!311\: \!064\: \!590\: \!6}}\\
{\mathtt{x}} = -{\mathtt{1.354\: \!248\: \!688\: \!935\: \!409\: \!4}}\\
\end{array} \right\}$$

Fri 6/2/15

Is that meant to take me somewhere Heureka??

See:

Square-and-Multiply-Algorithm with modulo for any $$\small{\text{$b^x\mod n$}}$$

Thanks Alan  :)

and

Thanks Chris :)

It's just a different approach....whether it "makes more sense" is debatable....

I think that mine is perfectly valid....

So  they were Chris, sorry Happy I did not notice. 

I have tucked this one away in "reference material"  

Thanks Heureka, I'll try and remember that.  :) 

Thanks Chris but I think Heureka's makes mores sense. :))

Very impressive, Melody and Alan...!!!

Melody chose to answer this using a " fundamental counting principle" approach....I chose a "set counting" approach.......again....

"All roads lead to Rome.'

Melody Sorensen (ca. 2015)

The total ways we can choose any 3 letters out of 8 is given by C(8,3) = 56

But,we only want to choose 3 of the 6 cosonanats  = C(6,3) = 20

So we have 20 / 56  = 5 / 14  = about a 35.7% chance of choosing a consonant..!!!

Page 1316

I have no got this one completely out - Maybe what I have done is wrong