All Answers 358516 Answers

The "mighty" Wolftram/Alpha engine gives these 3 "real" values for x:

x ~~ -0.0558159
x ~~ 0.0740962
x ~~ 0.740807

Hi Alan, 

I am not saying your answer is wrong - I am just not necessarily convinced it is right.

Here is another possibility (probably wrong)

Say we stick an ABC together so that instead of choosing 27 characters we choose 24 plus our one lot of joined letters.

Then the answer could  be

\(\frac{28*62^{24}}{62^{27}}=\frac{28}{62^{3}}=\frac{28}{238328}\)

28/238328 = 0.0001174851465208

So that is about 1 in 10,000

Much more often than what Alan got.

I do not think that this is correct because The ABC could occur in our not prejoined together sequences.  So I tend to think it would happen more often than this - not less.  :///

idk probability is so confusing  LOL

P.S. This is taylor-made for "heureka"!!.

2,44=12500x^2/(3+x)^4*(1-x)*(2-2x)^2

\(2.44=\frac{12500x^2}{(3+x)^4}*(1-x)*(2-2x)^2\\ 2.44=\frac{12500x^2}{(3+x)^4}*(1-x)*4(1-x)^2\\ 2.44=\frac{12500x^2}{(3+x)^4}*4(1-x)^3 \qquad x\ne-3\\ 0.61=\frac{12500x^2}{(3+x)^4}(1-x)^3 \\ 0.61*(3+x)^4=12500x^2(1-x)^3 \\ 0.61*(x^4+4*3x^3+6*9x^2+4*27x+81)=12500x^2(-x^3+3x^2-3x+27) \\ \mbox{ok I'm bored now- got better things to do....} \)

2,44=12500x^2/(3+x)^4*(1-x)*(2-2x)^2

2.44=12500x^2/[x^4+12 x^3+54 x^2+108 x+81]*(1 - x)*[4 x^2-8 x+4], solve for x. OK, guys and gals: I have expanded it as much as I could!  See if you can perform some magic on it!. Thanks.

Hello Alan: Can the font of your software package be made a little larger?

Even with eye glasses, my eyes have a rough time reading your answers! Thanks.

1.414213562373095

I found a whol stack of them.  All posted by the same member at around the same time.

I wonder if it is all of an assignment   :))

Alan, I found a physics one.  How about that   :)

What do you think?

450mm is 0.45m  so it will cost   $(0.45/22)*108.90  = $2.228  or  $2.23 to the nearest cent.

It was your reply but I think it is ok again now - I restarted my modem. 

I am glad you did that one, thanks :)

My reasoning is as follows:

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Do you mean the original question or the image in my reply?  

Both load ok for me.

The following should help:

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It is not loading for me Alan :(

The site seems to be playing up a bit - maybe it will load later ?

Does it load ok for you?

Mix a solution with a .65% acidity with 100ML of a solution with a 3% acidity to receive a final product of .7% - Please show work.

Ok you have 100 ml at 3% so in that 100ml there is  0.03*100 = 3ml of acid

Let y be the amount of the 0.65% solution that is needed.   0.65% = 0.0065

0.7% = 0.007

\(\frac{0.03*100+0.0065*y}{100+y}=0.007\\ \frac{3+0.0065*y}{100+y}=0.007\\ 3+0.0065*y=0.007(100+y)\\ 3+0.0065y=0.7+0.007y\\ 2.3=0.0005y\\ 2.3\div 0.0005=y\\ y=4600ml\;\;or\;\;4.6L\)

woahhhhhh... Could you explain that a bit more simply please? You've made me confused, sorry

As below:

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In how many ways can 4 boys and 4 girls sit around a circle table if all the boys sit together? (Rotations of the same arrangement are still considered the same. Each boy and girl is unique, not interchangeable.)

Sit one child anywhere.

Choose any 4 consecutive seats for the boys.  There are 4!  ways to seat them.

and there are also 4! ways to seat the girls.

So that makes  4!*4! ways = 24*24= 576 ways     

12m⁷ x 5n⁸ x 4p⁴               

_____________          

20m⁵ x 6n⁴

How do i simplify it? I don't see how you can.

\(\begin{array}{rcll} \dfrac{ 12m^7 \cdot 5n^8 \cdot 4p^4 }{ 20m^5 \cdot 6n^4 } &=& \dfrac{12\cdot 5 \cdot 4} {20\cdot 6} \cdot \dfrac{ m^7 \cdot n^8 \cdot p^4 }{ m^5 \cdot n^4 } \\\\ &=& \dfrac{12\cdot 5 \cdot 4} {6\cdot 20} \cdot \dfrac{ m^7 \cdot n^8 \cdot p^4 }{ m^5 \cdot n^4 } \\\\ &=& \dfrac{12}{6}\cdot \dfrac{5 \cdot 4}{20} \cdot \dfrac{ m^7 \cdot n^8 \cdot p^4 }{ m^5 \cdot n^4 } \\\\ &=& \dfrac{12}{6}\cdot \dfrac{20}{20} \cdot \dfrac{ m^7 \cdot n^8 \cdot p^4 }{ m^5 \cdot n^4 } \\\\ &=& 2\cdot 1 \cdot \dfrac{ m^7 \cdot n^8 \cdot p^4 }{ m^5 \cdot n^4 } \\\\ &=& 2\cdot \dfrac{ m^7 \cdot n^8 \cdot p^4 }{ m^5 \cdot n^4 } \\\\ &=& 2\cdot \dfrac{m^7 }{m^5} \cdot \dfrac{ n^8 }{n^4 } \cdot p^4 \\\\ &=& 2\cdot (m^7\cdot m^{-5}) \cdot ( n^8 \cdot n^{-4})\cdot p^4 \\\\ &=& 2\cdot m^{7-5} \cdot n^{8-4}\cdot p^4 \\\\ \mathbf{\dfrac{ 12m^7 \cdot 5n^8 \cdot 4p^4 }{ 20m^5 \cdot 6n^4 } }&\mathbf{=}& \mathbf{2\cdot m^{2} \cdot n^{4}\cdot p^4 } \\\\ \mathbf{\dfrac{ 12m^7 \cdot 5n^8 \cdot 4p^4 }{ 20m^5 \cdot 6n^4 } }&\mathbf{=}& \mathbf{2\cdot m^{2} \cdot (n\cdot p)^4 } \\\\ \end{array}\)

if you mean 5^(-2) its 1/25

You work out hard problems that you will probably not even need for the future.. That's how you maths!

I think it's 36 but I'm not sure. I did it visually btw

This may help you find out how to calculate percentages :D http://www.skillsyouneed.com/num/percentages.html