All Answers 358134 Answers

We agree, how unusual :)

hi there.

Any questions?

Okay thanks Alan, I think I have digested that.  And I can see why Chris's answer makes sense.

BUT

I still would like to understand why my logic  is wrong :)

I have always thought with probability that it is often fairly easy to understand the correct answer but much harder to understand why an incorrect answer is wrong.

Still, with all the prob questions we have been getting I think my ability is improving :)

$$\\\frac{(4xy^{-3})^{4}}{(2x^{2}y^{-7})}\\\\
\frac{(4^4x^4y^{-3*4})}{(2x^{2}y^{-7})}\\\\
\frac{2*4^3x^4y^{+7}}{1x^{2}y^{+12}}\\\\
\frac{128x^{(4-2)}}{y^{(12-7)}}\\\\
\frac{128x^{2}}{y^{5}}\\\\$$

It might help to think of a simpler situation first.  Consider four people sitting two each side of a two-sided table(!).

The distinct possibilities, excluding rotations, are easily visualised:

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3.25-5 = -(5-3.25)=-(5-3-0.25)=-(2-0.25)=  -1.75 

0,4      1,3     2,2     that is it

3 ways

hi Keven, why don't you get a proper account? 

We'd like that :D

AMA ?

American Medical Association?

Australian Medical Associtation?   

That is a great explanation - thanks Alan :)

I have no idea what you are talking about.

It looks more like computing/programing than mathematics. 

Hi Bertie, I was hoping that you would join in. thank you. 

So, here's a question for Melody. Forget about rotations and seat positions for the moment, suppose that I have eight people and that I'm going to split them into four groups of two. In how many ways can this be done ?

Well I thought that it was

 8C2*6C2*4C2*2C2 but this is obviously not correct - I guess i am double counting again :/

Hi eannehail,

Welcome to the forum,   

I gave you an extra 3 points to help get you on your way :)

Yes that is a much better idea - thanks Alan.      

This question might be related to the following common coin weighing problem.

 You have 10 coins that look identical, but one is counterfeit.  Using a balance (i.e. where you put coins in pans at both ends of a "see-saw", but can't tell the actual weight) what is the minimum number of weighings required to identify the counterfeit coin (and can you tell if it is heavier or lighter than a genuine coin?)?

Thanks Alan,

That sounds like a diistict possibility. 

I guess now we just need and answer. 

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Энэ асуулт нь дараахь нийтлэг зоос жинтэй асуудалд холбоотой байж болох юм.

  Та ижил харагдах 10 зоос, харин нэг нь хуурамч юм. Тэнцвэрийг ашиглах нь (хэрэв та "-г үзнэ үү харсан" хоёр төгсгөлд баригч нь зоос тавьж, өөрөөр хэлбэл, гэхдээ бодит жин нь хэлж чадахгүй байгаа) (хуурамч зоос тодорхойлох шаардлагатай жинлэлтийн хамгийн бага тоо гэж юу болох бөгөөд та хэрэв хэлж болно энэ нь жинхэнэ зоос илүү хүнд эсвэл хөнгөн байдаг вэ?)?

Баярлалаа Алан,

Энэ нь diistict боломжийг сонсогдож болох юм.

Би одоо бид зүгээр л хэрэгтэй, хариулт бодож байна.

Hi Chris and Melody, hope you don't mind me joining the conversation.

I'm with Chris on this one, I think that the 10080 is correct, (I've gone through the calculation in a different way and arrived at the same answer).

So, here's a question for Melody. Forget about rotations and seat positions for the moment, suppose that I have eight people and that I'm going to split them into four groups of two. In how many ways can this be done ?

Let's see If we can get agreement on that to begin with.

This question might be related to the following common coin weighing problem.

You have 10 coins that look identical, but one is counterfeit.  Using a balance (i.e. where you put coins in pans at both ends of a "see-saw", but can't tell the actual weight) what is the minimum number of weighings required to identify the counterfeit coin (and can you tell if it is heavier or lighter than a genuine coin?)?

.

In this case Melody, it might be better to let y = √x  (a positive number) when you get to 0.08x - 1 = √x  so that you have  0.08y² - 1 = y.  Then when you solve for y, if you get a negative number you can discard it immediately:

$${\mathtt{0.08}}{\mathtt{\,\times\,}}{{\mathtt{y}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{1}} = {\mathtt{y}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{y}} = {\mathtt{\,-\,}}{\frac{\left({\mathtt{5}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{33}}}}{\mathtt{\,-\,}}{\mathtt{25}}\right)}{{\mathtt{4}}}}\\
{\mathtt{y}} = {\frac{\left({\mathtt{5}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{33}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{25}}\right)}{{\mathtt{4}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{y}} = -{\mathtt{0.930\: \!703\: \!308\: \!172\: \!535\: \!8}}\\
{\mathtt{y}} = {\mathtt{13.430\: \!703\: \!308\: \!172\: \!535\: \!8}}\\
\end{array} \right\}$$

Taking only the positive value we then have (approximately)

$${\mathtt{x}} = {{\mathtt{13.430\: \!7}}}^{{\mathtt{2}}} \Rightarrow {\mathtt{x}} = {\mathtt{180.383\: \!702\: \!49}}$$

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Hi Gunnar,

Welcome to the forum :)   I gave you 5 points :)

$$\\log(8x)-log(1+x^{1/2})=2\\\\
log\frac{8x}{1+x^{0.5}}=2\\\\
10^{log\frac{8x}{1+x^{0.5}}}=10^2\\\\
\frac{8x}{1+x^{0.5}}=100\\\\
8x=100(1+x^{0.5})\\\\
0.08x=1+\sqrt{x}\\\\
0.08x-1=\sqrt{x}\\\\
0.0064x^2-0.16x+1=x\\\\
0.0064x^2-1.16x+1=0\\\\$$

x is approx  0.866  or  180.384  but these must be checked. :)

$${\mathtt{0.006\: \!4}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{1.16}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\mathtt{125}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{33}}}}{\mathtt{\,-\,}}{\mathtt{725}}\right)}{{\mathtt{8}}}}\\
{\mathtt{x}} = {\frac{\left({\mathtt{125}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{33}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{725}}\right)}{{\mathtt{8}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{0.866\: \!208\: \!647\: \!843\: \!302\: \!2}}\\
{\mathtt{x}} = {\mathtt{180.383\: \!791\: \!352\: \!156\: \!697\: \!8}}\\
\end{array} \right\}$$

log(8x)-log(1+x^1/2)=2

check

$${log}_{10}\left({\mathtt{8}}{\mathtt{\,\times\,}}{\mathtt{0.866\: \!208\: \!643}}\right){\mathtt{\,-\,}}{log}_{10}\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{0.866\: \!208\: \!643}}}^{{\mathtt{0.5}}}\right) = {\mathtt{0.554\: \!996\: \!959\: \!640\: \!612\: \!2}}$$

$${log}_{10}\left({\mathtt{8}}{\mathtt{\,\times\,}}{\mathtt{180.383\: \!791\: \!352}}\right){\mathtt{\,-\,}}{log}_{10}\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{180.383\: \!791\: \!352}}}^{{\mathtt{0.5}}}\right) = {\mathtt{1.999\: \!999\: \!999\: \!999\: \!798\: \!4}}$$

$$So\;\; x\approx 180.184$$

If n is an integer then your series is given by ∑_{k=1 to n(}k² - 2)

k      k² - 2    ∑_{k=1 to n(}k² - 2)

1        -1           -1

2          2            1

3          7            8

4          14          22

5          23          45

...

summation notation of -1+2+7+14+23...

What I noticed is that if I add 2 to each of these terms I get

1+4+9+16+     familiar?

sum of  n^2-2

$$\displaystyle\sum_{n=1}^\infty\;(n^2-2)$$

3x - y = 23         (1)

2x + 3y = 8        (2)

Multiply all terms in (1) by 3 

9x - 3y = 69      (3)

Now there are the same number of y's in equation (3) as in equation (2), so add equation (2) to equation (3) term by term:

11 x = 77

so x = 11

Plug this back into equation (1)

3*11 - y = 23  or 33 - y = 23

Subtract 23 from both sides

10 - y = 0

Add y to both sides

10 = y   (or y = 10)

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The gcf, or greatest common factor, is the largest integer that exactly divides into two other integers. 

For example, 6 is the gcf of 12 and 18.

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