Melody

Yea well I can ....now    LOL

Although I still get muddled as to whether it is dio or dia  

BUT then I think - no it is definitely not a diaphram  :))

The hypotenuse would have to be 15

3*3=9

4*3=12

5*3=15

5.2 × 10^7 + 7.25 × 10^6 + 1.8 × 10^7

=5.2 × 10^7 + 0.725 × 10^7 + 1.8 × 10^7

=(5.2  + 0.725  + 1.8 )× 10^7

=7.725*10^7

You showed me Heron's formula.       I still had to look it up though.   

I was going to do it the usual way which would not have been very difficult but then I thought it was the ideal situation for ME to use Heron's formula :))      

Thanks Chris, we both need more practice at these.

Has Mellie posted other diophantine equation questions ?  I think she has ://

Could be 8 kittens or 12 or 16 kittens.

Depends on if 2 adults have the same kittens ;/

It takes 2 to reproduce even if you are a cat :)

If the perimeter is 48 inches then each side is 8 inches.

The angle at the centre that subtends to adjacent vertices is 360/6 = 60degrees.

Therefore the hexagon consists of 6 equilateral triangles of sidelength 8 units.

Using Heron's formula

$$\\Area = 6 \;\; *\;\; \sqrt{s(s-a)(s-b)(s-c)}\qquad where\quad s=(a+b+c)/2\\\\
s=24/2=12\\\\
Area = 6 \;\; * \;\; \sqrt{12(12-8)(12-8)(12-8)}\\\\
Area = 6 \;\; * \;\; \sqrt{12*64}\\\\
Area = 6 \;\; * \;\; 8\sqrt{4*3}\\\\
Area = 6 \;\; * \;\; 16\sqrt{3}\\\\
Area = 96\sqrt{3}\;\;inches^2\\\\$$

Thanks Alan,

Or if you are using a different calculator you can use the change of base rule

$$log_{1.1} (1000) =\frac{log1000}{log1.1}$$

you can use any base you like :)

$${\frac{{log}_{10}\left({\mathtt{1\,000}}\right)}{{log}_{10}\left({\mathtt{1.1}}\right)}} = {\mathtt{72.476\: \!573\: \!784\: \!290\: \!347\: \!8}}$$

Thanks Chris,  I also would like to try this one.

We have this situation 

Let X be the number of scoops in and Y be the number of scoops out,  so X and Y are integers

$$46899X-13567Y=1$$      This is a diophantine equation because the solutions must be integers.

I'll rearrange this

$$\\46899X=13576Y+1\\\\
\frac{46899}{13576}*X=Y+\frac{1}{13576}\\\\
$I need the fraction part to be a proper fraction$
\frac{(3*13576+6171)}{13576}*X=Y+\frac{1}{13576}\\\\
3X+\frac{6171}{13576}*X=Y+\frac{1}{13576}\\\\
\frac{6171}{13576}*X=Y-3X+\frac{1}{13576}\\\\$$

Y-3X is an integer so I am going to replace it with Z  (I only care that it is an integer)

$$\\\frac{6171}{13576}*X=Z+\frac{1}{13576}\\\\
So\\
6171*X = 1 Mod\; 13576\\\\
X = \frac{1}{6171}\; Mod\; 13576\\\\
X = 6171^{-1}\;\; Mod\; 13576\\\\\\
$Now I am going to use the $ \underline{Euclidean\; Algorithm} \\\\
\begin{array}{rllll}
13576&=&\underline{6171}*2+\underline{1234}\qquad&(1a)\quad &1234=13576-6171*2\\\\
6171&=&\underline{1234}*5+\underline{1}\qquad&(2a)\quad &1=6171-1234*5\\\\
\end{array}
$I can stop the Euclidean algorithm now because I have the 1 that I wanted$\\\\
$Now I am going to use the \underline{Extended Euclidean Algorithm} and go 'backwards' :)$\\\\$$

$$\\\frac{6171}{13567}*X=Z+\frac{1}{13567}\\\\
So\\
6171*X = 1 Mod\; 13567\\\\
X = \frac{1}{6171}\; Mod\; 13567\\\\
X = 6171^{-1}\;\; Mod\; 13567\\\\\\
$Now I am going to use the Euclidean Algorithm$ \\\\
\begin{array}{rllll}
13576&=&\underline{6171}*2+\underline{1234}\qquad&(1a)\quad &1234=13576-6171*2\\\\
6171&=&\underline{1234}*5+\underline{1}\qquad&(2a)\quad &1=6171-1234*5\\\\
\end{array}
$I can stop the Euclidean algorithm now because I have the 1 that I wanted$\\\\
$Now I am going to use the extended Euclidean Algorithm and go 'backwards' :)$\\\\
\begin{array}{rllll}
1&=&6171-1234*5\qquad &(2b)\\\\
1&=&6171-(13576-6171*2)*5\qquad &(2a)\\\\
1&=&6171-5*13576+5*6171*2\qquad &(2a)\\\\
1&=&6171-5*13576+10*6171\qquad &(2a)\\\\
1&=&11*6171-5*13576\qquad &(2a)\\\\
&&$now 5*13576 mod 13576=0 so$\\\\
1&=&11*6171\\\\
\end{array}$$

So X=11

The number of scoops in is  11+13576N

But each scoop in is 46899 peices.

46899*(11+13576N)<600000

(11+13576N)<12.79

N has to be 0

SO  

The number of scoops in MUST be 11.        

Here are 3 lollipops, one for me, one for CPhill and one for Mellie.  

BECAUSE WE DESERVE THEM  

I just realized that some otf the LaTex was not displaying.

Hopefully there is no more problems.

What are you after here?  

Are k and i just variables?  

Does   $$i=\sqrt{-1}$$    ????

1/10 ki power 1/3

$$(\frac{ki}{10})^{1/3}=\sqrt[3]{\frac{ki}{10}}$$