All Answers 358116 Answers

Your welcome

thank you. i apprecite it

$${\frac{{\mathtt{5}}}{\left({\frac{{\mathtt{2}}}{{\mathtt{3}}}}\right)}} = {\frac{{\mathtt{15}}}{{\mathtt{2}}}} = {\mathtt{7.5}}$$

Yes it is....there is the proof

no is not

5/(2/3) Swap 2/3 to 3/2 and instead of dividing, multiply

5/1*(3/2)  New sentence

5*3=15

1*2=2

15/2=7 1/2

That is the answer

Didn't I already answer this in my poetry?

Simplify the radical expression :

1.)  (-1-√2)/(1+√2)

$$\frac{(-1- \sqrt{2} )}{ (1+ \sqrt{2} ) }
=-\frac{(1+ \sqrt{2} )}{ (1+ \sqrt{2} ) } = -1$$

2.)  (√8)/(√2+3)

$$\frac { \sqrt{8} } { \sqrt{2} + 3} = \frac { 2\sqrt{2} } { \sqrt{2} + 3} *\frac{ \sqrt{2}-3 }{ \sqrt{2}-3 } =-\frac{2}{7}(2-3\sqrt{2})$$

3.)  (√3)/(√27-√3)

$$\frac{ \sqrt{3} }{ \sqrt{27}-\sqrt{3} } = \frac{ \sqrt{3} }{ 3\sqrt{3}-\sqrt{3} } = \frac{ \sqrt{3} }{ 2\sqrt{3} } = \frac{1}{2}$$

4.)  (√2-√8)/2(√2-1)

$$\frac{ \sqrt{2}-\sqrt{8} }{ 2(\sqrt{2}-1) } =\frac{ \sqrt{2}-2\sqrt{2} }{ 2(\sqrt{2}-1) } =\frac{ -\sqrt{2} }{ 2( \sqrt{2}-1) } * \frac{ \sqrt{2}+1 }{ \sqrt{2}+1 } = \frac{ -\sqrt{2} } {2} ( \sqrt{2} + 1) = - ( 1 + \frac{ \sqrt{2} } {2} )$$

5.)  (3√3-2√2)/(√3-√2)

$$\frac{ 3\sqrt{3}-2\sqrt{2} } { \sqrt{3}-\sqrt{2} } * \frac{\sqrt{3}+\sqrt{2}}{ \sqrt{3}+\sqrt{2} } = ( 3\sqrt{3}-2\sqrt{2} ) * ( \sqrt{3}+\sqrt{2} ) = 9 + \sqrt{6} - 4 = 5 + \sqrt{6}$$

6.)  (√2-√8)/(2√2-1)

$$\samll{\text{
$
\frac{ \sqrt{2}-\sqrt{8} }{ 2\sqrt{2}-1 }
=\frac{ \sqrt{2}-2\sqrt{2} }{ 2\sqrt{2}-1 }
=\frac{ -\sqrt{2} }{ 2\sqrt{2}-1 } * \frac{ 2\sqrt{2}+1 }{ 2\sqrt{2}+1 }
= \frac{ -\sqrt{2} } {7} ( 2\sqrt{2} + 1)
= -\frac{ 1 } {7} ( 4 + \sqrt{2} )
$
}}$$

7.)  (1-√2)/(1+√2)

$$\small{\text{
$
\frac{ 1-\sqrt{2} } { 1+\sqrt{2} } *\frac{ 1-\sqrt{2} }{ 1-\sqrt{2} }
= -( 1-\sqrt{2} )( 1-\sqrt{2} ) = -(1-2\sqrt{2}+2) = -(3-2\sqrt{2}) = 2\sqrt{2}-3
$
}}$$

again 5.) (3√3-2√2)/(√3-√2)

There are many different types of problems that can be factored.

Could you give us an example of the type of problem that you want factored?

5/12  =  0.416666...

1/3  doubled is  1/3 + 1/3  =  2/3

This can also be calculated as  2 x 1/3  =  2/3

1/3 doubled is also 1/3+1/3

2/3 is that answer

Hey,  wait!

What'd you say?

That math sucks?

Well, you better look up

Math is study

Might get you a buddy

Learn about those shapes

Instead of eating grapes

Multiply, add, subtract, divide

Math can help you spread wings and fly!

Fun to learn

It's actually a privelage to earn

Math is important

Why don't you get in for it?

Just read the books

It's easier that in looks

Do you math

It will make a path

Become smart

In the study of math's art

Have some fun

Come on!

Because she's moving upward ...

Her work must be expressed upward because she is moving upward against the force of gravity which is pulling her downward.

3+3=6

And you know this

Hi Lancelot,

Your introductory post is not on the forum anymore. I was hoping to make a proper copy. I made an image on my phone, it’s only a thumbnail sized one, and not very readable. :(

I was in class still recovering from the Banana Phone post, when the translation of CPhill’s Wile e Coyote’s Latin popped up. You know how hard it is for us chimps to suppress laughter. My professor interrupted his lecture to ask if I was all right: He thought I was stroking out.

Are you looking for new talent? I would be pleased and honored to join your troop!

For me, maths used to cause abhorrent abominations and abstract algebraic afflictions of agony. Now, after studying this forum for months, I found the root of the problem, and even though my math skills are still down at the base, near the roots, I am well on my way to climbing the evolutionary maths tree.

Right now, I am studying Lagrange formulas in an attempt to find a solution to convert three-body libration points into two-body libration points. Making the orbital relationship binary should make things much simpler and more stable. Wouldn’t you agree?

Here are two equations that seem to work well at reducing this complexity.

$$\\ 9x-7i>3(3x-7u)\\
\ 9x-7i>9x-21u \\
\ -9x -9x -7i>-21u \\\
\ -7i \div -7 >-21u \div-7 \\\
= i<3u \\

\ And \\

\noindent
\ x^2 + (y-\sqrt[\leftroot{-2}\uproot{2} 3]{x^2})^2 = 1$$

Though my math skills are still wanting, my artistic and philosophy skills are paramount. I also make a great banana daiquiri, so I think I would fit in somewhere in your troop.

A proposal submitted for your consideration:

Perhaps we could have libations while studying librations and other complex relations?

Ginger

P.S. I posted this shortly after your cuddle-worthy kitten post, but it disappeared. I waited to repost because I thought it might return. I don’t think it will, but I hope you do soon.

****

Edit:

To simplify these, multiply both the numerator and denominator of the fraction by the conjugate of the denominator.

The conjugate of   x + √y  is x - √y.

The conjugate of   x - √y  is x + √y.

The conjugate of  x + a√y  is x - a√y

As an example:

To simplify  (2 + √3) / (4 - √5)  multiply both the numerator and denominator by 4 + √5:

     (2 + √3) / (4 - √5) x (4 + √5) / (4 + √5)

Numerator:  (2 + √3)(4 + √5)  =  8 + 4√3 +2√5 + √15

Denominator:  (4 - √5)(4 + √5)  =  16 - 4√5 + 4√5 - √25  =  16 - 5  = 11

Answer:  ( 8 + 4√3 +2√5 + √15 ) / 11

Any questions about this example?

Percentage is Part ÷ Whole x 100:     £4 / 16£ x 100  =  0.25 x 100  =  25% 

3x² + 7x + 4 + ( ??? )  =  11x² - 15

Take these one term at a time:

The x² term:  3x² + ? = 11x²   --->  must be 8x²

The x term:  7x + ? = 0x (no x-term on the right side)   -->  must be -7x

The number:  4 + ? = -15   --->  must be -19

So the missing part must be:  8x² - 7x - 19

I'm going to assume that the choices are random choices, otherwise, Bill could always choose to win.

I'm also going to assume that the game is played "with replacement", meaning that both players choose from all four posibilities, not that one player gets two of the numbers, while the other player gets the two that are left. 

So, what are the possibilities for Darla?

1 + 2 = 3     1 + 3 = 4     1 + 4 = 5       2 + 3 = 5     2 + 4 = 6     3 + 4 = 7

There are 6 possibilities for Darla (we must count the two ways for Darla to get 5 as separate possibilities).

Now, what are the possibilities for Bill?

1 x 2 = 2     1 x 3 = 3     1 x 4 = 4     2 x 3 = 6     2 x 4 = 8     3 x 4 = 12

Each of the possibilities for Bill has a 1/6th chance of occuring.

In how many different ways can Darla beat Bill when

Bill gets 2?  6 ways  So Carla has a 6/6 chance to win.

Bill gets 3?  5 ways  So Carla has a 5/6 chance to win.

Bill gets 4? 4 ways  So Carla has a 4/6 chance to win.

Bill gets 6?  1 way  So Carla has a 1/6 chance to win.

Bill gets 8?  0 ways  So Carla has a 0/6 chance to win.

Bill ets 12?  0 ways  So Carla has a 0/6 chance to win.

So now we need to multiply the probability that Bill gets a certain result times the probability that Darla beats him and add all those results together:

(1/6)(6/6) + (1/6)(5/6) + (1/6)(4/6) + (1/6)(1/6) + (1/6)(0/6) + (1/6)(0/6)  =  16/36  =  4/9

The probability that Darla beats bill is  4/9.

If the choice is "without replacement" (one of them gets two of the numbers, the other gets the other two numbers) needs a different analysis. Re-post if that is the problem.

HOWL!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Just because it is a math site, doesn't mean we can't have a little bit of poetic fun

I already have bro!

Thank you, Melody.

At certain times I do... sometimes though, it's just about finding the fastest/easiest method for me.

Actually i did not even think of fractions.  

You have a very good mind  for mathematics and logical thinking.