All Answers 358098 Answers

For this question, you have to factor.

2x^2 - 3x - 2 = 0

2x^2 - 4x + x - 2 = 0

2x(x - 2) + 1(x - 2) = 0

(2x + 1)(x-2) = 0

Now solve both parts.

x = -1/2, 2

@@ End of Day Wrap Wed 15/4/15 Sydney, Australia Time 12:50am   (Really Thurs) ♪ ♫

Good Morning  LOL :)

Our intrepid answerers today were Natthada, Heureka, Arkane, Trincent, Funkygal, MathsGod1, DarkBlaze347, Geno3141, Zacismyname, Queencaitlin_, Alan, Beccaboo1416, Zekkuno, Radio, Jjonesekb and Bertie.  Thanks everyone  

Interest posts:

Thurs 16/4/15

FTJ means "For the Juniors"

500 gallons in 5 hours

equals

100 gallons in 1 hour

300 gallons left to drain = 3 more hours

8 hours in total :)

hi Dakotah :)

A number cube is rolled 20 times and lands on 1 two times and on 5 four times. Find each experimental probability. Then compare the experimental probability to the theoretical probability.

Mmm

The experimental prob that it will land on a 1 is  was  $${\frac{{\mathtt{2}}}{{\mathtt{20}}}}$$ =  $${\frac{{\mathtt{1}}}{{\mathtt{10}}}}$$

The theoretical prob that it will land on a 1 is  $${\frac{{\mathtt{1}}}{{\mathtt{6}}}}$$    So in 20 rolls you would expect to get  $$\frac{20}{6}=\frac{10}{3}=3\frac{1}{3}$$ 

 twos

The experimental prob was lower than the theoretical prob.  :)

The experimental prob that it will land on a 5 is  was  $${\frac{{\mathtt{4}}}{{\mathtt{20}}}}$$ =  $${\frac{{\mathtt{1}}}{{\mathtt{5}}}}$$

The theoretical prob that it will land on a 5 is  $${\frac{{\mathtt{1}}}{{\mathtt{6}}}}$$    So in 20 rolls you would expect to get  $$\frac{20}{6}=\frac{10}{3}=3\frac{1}{3}$$  

 The experimental prob was higher than the theoretical prob.  :

If a wood chuck could chuck wood, a wood chuck would chuck as much wood as a woodchuck could!!! ;)

Ok Bertie - Why is that?      You have to justify your answer or it doesn't count.  LOL  :))

ERROR - COULDN'T CALCULATE THE ANSWER.

Don't worry, I've picked a good definition out of the web for you:

Thanks Bertie, that wasn't so hard after all.  Wish I had thought of it.  LOL   

$$\\1\times \frac{39}{51}\times \frac{26}{50}\times \frac{13}{49}=\frac{13182}{124950}=\frac{6591}{62475}$$

It is the independent because the y variable depends on the x variable

Yep.  

People asked maths questions even when the dinosaurs ruled the Earth!

This is PROOF !!!!!

You think I have competion MG ?  I am shaking in my boots.   LOL     

I do not know radio.  In theory at least I think there are infinitely many primes BUT

You have taken this a step further - I am not so sure your 'yes' would stand up to close scrutiny.  

$${\frac{{\sqrt{{\mathtt{100}}}}}{{\mathtt{4}}}} = {\frac{{\mathtt{5}}}{{\mathtt{2}}}} = {\mathtt{2.5}}$$

Thanks Bertie,

Sorry 315,

Bertie or 315, can you try to explain to me the method that you have used? 

(Just the first one for starters)

Both are correct.

Simply make use of the identity 1+cot^2 = cosec^2 to go from the first form to the second. The -1/4 gets lumped in with the constant of integration.

I have not obtained the identity yet .

I did check your formula.  It seems to me that the 2i needs to be in brackets.  

2x+(x-10)+(2x+15)=180

5x+5=180

5x=175

x=35

The biggest angle is 2x+15=2*35+15=70+15=85 degrees

With your fingers

JJ, firstly, it can not be a question because there is not question mark!

so be careful about who you call names.

Secondly, this is a international forum so this 'question' has no meaning for most people who will read it.

Hi  315  

I do not know WHAT you are doing BUT the second one is NOT correct.

This is how I would do it.

$$\\\int\;CotxCosec^4x\;dx\\\\
=\int\;\frac{Cosx}{Sinx}\frac{1}{sin^4x}\;dx\\\\
=\int\;\frac{Cosx}{Sin^5x}\;dx\\\\
=\int\;CosxSin^{-5}x\;dx\\\\
=\frac{Sin^{-4}x}{-4}+c\\\\
=\frac{-Cosec^{4}x}{4}+c\\\\$$

Picture 52 locations divided into 4 groups of 13.

The first Ace can go anywhere, the second Ace has 39 of 51 possible locations in which to go, the third has 26 of a possible 50 and the fourth 13 of a possible 49.

That works out to about 10.5498%, you should be able to work out the exact fractional value.

thanks anon,

You are right except that the m^3 cancel out :))