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And one more thing I'll add!  If the Earth were flat, eventually you would walk of the edge and float where ever

Ha ha....if she shared two, then one friend wouldn't like Rebecca!

BTW, I didn't watch the video

Earth isn't flat!  And Earth is constantly spinning!  Did you know that??? 

There's something interesting about this problem that I'd like to bring out.....many people think that there is some "righting" force in nature that tends to bring things "back to center"....in this case........many think that, if they haven't been successful after "N" flips, that there is a greater chance that they WILL be successful on the next flip......this isn't true as shown below...

The probability that a person has one success on any of 6 coin flips  = 1 minus the probaility that he isn't successful on any of them.....

Thus......before any flips are attempted.....the probabilty of success = 1 - (1/2)^6 = 0.984375

Let's suppose he isn't successful after the first toss......then he has five remaining in which to achieve a success and this probability = 1 - (1/2)^5 = 0.96875.....notice that this isn't as good as before the first flip

Look at the remaining probabilities, assuming no successes on previous attempts

After the second toss =  1 - (1/2)^4 = 0.9375

After the third toss =  1 - (1/2)^3 = 0.875

After the fourth toss =  1 - (1/2)^2 = .75

Notice, that as more unsuccessful flips are made, the probability of attaining just one success drops....

And, after the fifth toss......the probabilty is just....1 - (1/2)^1 = 1/2 = .50 ......just what we would expect it to be on one remaining toss !!!

This demonsrates something known as the Gambler's Fallacy......(it's why those big buildings exist in Las Vegas....!!!).......why does this happen???....because, as more "failures" pile up in the sequence of some N trials, the fewer chances there are remaining for a person to achieve a success........!!!

We do order of operations (PEMDAS) Parentheses, Exponents, Multipulication, Division, Addition, Subtraction

Parentheses first, and multipulicatipon in parentheses

13*2=26

Now

74+26=100

100/10=10

That is why

Here's a slightly different approach:

.

(√2011 - 1) / (√2012 - 1) ??  (√2011 + 1) / (√2012 + 1)

Let's cross multiply

(√2011 - 1)(√2012 + 1) ?? (√2012 - 1) (√2011 + 1)

√(2011 * 2012)  -  √2012 + √2011  - 1  ?? √(2011 * 2012)  - √2011 + √2012 - 1

  -  √2012 + √2011   ??  - √2011 + √2012

 √2011 - √2012  ??  √2012 - √2011

2√2011 ?? 2√2012

√2011 < √2012

And since the right side is greater at the end (and we haven't multiplied or divided by any negatives).....then the original right side was greater, too

Whoops!  Meant to post this on my normal account..LOL

4 15 mins in one hour

4*3=12

Thank you，alan!Just making sure that are true.

Yes, they are true.  Examples, using the inbuilt calculator here:

$${\sqrt{{{\mathtt{5}}}^{{\mathtt{2}}}}} = {\mathtt{5}}$$

$${\sqrt{{\left(-{\mathtt{5}}\right)}^{{\mathtt{2}}}}} = {\mathtt{5}}$$

$${{\sqrt{{\mathtt{5}}}}}^{\,{\mathtt{2}}} = {\mathtt{5}}$$

$${{\sqrt{-{\mathtt{5}}}}}^{\,{\mathtt{2}}} = -{\mathtt{5}}$$

.

Assuming standard pressure and temperature.

Gas at standard temperature and pressure is 22.414L/Mole (This is independent of the composition of the gas).

N= (14.01g/mole)

O= (16.00g/mole)

14.01+16.00 =30.01g/mole for Nitric oxide (NO). (Gram Molecular mass of NO)

18.82L/(22.41L/mole) = 0.8397mole

0.8397 mole of (NO) (Number of moles of NO)

(30.01g/mole)*(0.8397mole) =25.198g Nitric oxide (NO). (Mass of NO sample)

_7UP_

Thanks for that reminder about the discriminant, Melody......this should always be kept in mind when searching for "real" roots in a quadratic (it can save us some unnecessary work !!!)

Thank you guys!

@@ End of Day Wrap       Tues 9/12/14       Sydney,   Australia         Time 10:40 pm          ♬

Hi everyone,

There were many great questions the last couple of days and fabulous answers were given by TayJay, GoldenLeaf, geno3141, CPhill, Heureka, Baelovesmaths, Landry, happy7, Rosala, Alan, jomarie and saseflower.       A big thank you to each of you.       

Instructions to questions askers: 

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Older questions tend to lose priority because students often would have found their answers elsewhere or simply lost interest.  If your question is still on the first page, please be patient.  If your question 'falls off' the first page you are welcome to make a new 'priority post'

Perhaps you could title it   "Unanswered question - please can someone help me."

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Copy the address of your original question and paste it into the post.

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Here is an example of a 'bumped' question from today.

Wed 10/12/14

1) Happy's tribute to many on the forum.  Thanks Happy   

$${\frac{{\mathtt{100}}}{{\mathtt{12}}}} = {\frac{{\mathtt{25}}}{{\mathtt{3}}}} = {\mathtt{8.333\: \!333\: \!333\: \!333\: \!333\: \!3}}$$

8 hours and 20 minutes.

Wouldn't a sled be better than a Bicycle?  I am sure that it would     LOL     

250 *   3/5 miles

It is not a fraction - It is a FUNCTION.

How do you evaluate the FUNCTION when x=-2,0 and 3? g(x)=-3*x+5

g(-2) = -3*-2+5 = +6+5 = 11

g(0) = -3*0+5    = 0+5   = 5

Perhaps you can do    g(3)    by yourself ?

one final correction 

I should have written that area 18x as 18x²

If a and b are both negative then 

$$\sqrt a\times \sqrt b=\sqrt{-|a|}\times \sqrt{-|b|}=i^2\sqrt{|a|\times |b|}=-\sqrt{|a| \times |b|}$$

But

 $$\sqrt{a\times b}=\sqrt{-|a|\times -|b|}=\sqrt{|a|\times|b|}$$

i.e. the two are different, so not true when both are negative.

.

For

x+(1/x) = -1

Alan and Heureka are both telling you that there are no real solutions.

Alan has said x cannot equal 0 because you cannot divide by 0

Alan has rearranaged the equation to give a quadratic.

$$x^2+x+1=0$$

For any quadratic you can determine the nature of the roots by examining the discriminate

$$\\\triangle =b^2-4ac\\
\triangle =1-4=-3\\$$

Since the discriminant (which is under a square root) is negative, there are no real roots.