All Answers 358087 Answers

No problem.....xvxvxv......English is my "mother tongue" and I'm still strugglig with it !!!!

Ok Chris - I will let you have the last word.  The term is a little ambiguous.

I'm sorry my friend ... 

I forgot it 

and i'm never get angry ... and I am grateful for you and CPhill  

Ican't express my felling because English is not my mother tongue 

Fri 2/8/14  Some of these would make really nice additions to your 'Great questions to learn from' thread Ninja. 

I'll try and categorise them for you.   

 Working with logatithms. Hard one.

Thanks for the reference, Alan....

So it appears that "Dr. Peterson" - if I read Alan's referenced article correctly - tends to side with Melody's "real world" interpretation of "difference" as being poisitve. But, he also admits that some ambiguity exists - that a "textbook" convention of a negative answer is acceptable, as well. Which is exactly what I was trying to imply......that the way we might interpret the left side of the equation may influence our "answer." Thus, as I stated, NDs solution is "an" answer - but not neccessarily the answer anymore than -12/11 is the answer.....

@@ End of Day Wrap :   Thurs 31/7/14     Sydney, Australia     Time 1:00am   (Really Friday morning)   ♬

Hi all,        ♫♪

Our great answers were today provided  by CPhill, Alan, NinjaDevo, Heureka, DragonSlayer554 and AzizHusain.

My thanks to each of you.  

NinjaDevo has been working on this great thread for everyone and now it is finished.  It is a thread of great answers to learn from - just like the title suggests.  This should be great reference material for askers and answerers but it is intended more for the askers.  NinjaDevo has presented it beautifully, it will be constantly added to.  Maybe now much fewer of our wonderful questions and answers will be lost.  If you would like a post added please private message the address of the post to Ninja.  This is Ninja's thread and he will have the final say as  to which threads are included and which are not but he can not find all the good ones himself.  He will need help for this.  Thank you Ninja.  I am very grateful to you for taking care of all our reference material.

Oh, "Great answers to learn from" and "Reference Material" are both Sticky Topics  

Melody is correct...the "antilog" might be thought of in terms of being the "inverse" of a "log."

For instance

log 2.2387211385683396 = .35

So, in essence the "antilog" is an exponential function that raises 10 to the power of .35

Think of it this way.....the original log function takes "A" and outputs "B"

The inverse exponential function takes "B" and outputs "A"

Hey Stu, if you are felling generous you could give Alan some thumbs up. 

Thanks Alan.

Hey, 15x3 if you are feeling really generous you could give CPhill and me some thumbs up.  

If you don't like me calling you 15x3 you'd better say so because otherwise you are stuck with it. 

okay Dragon, you've made up your own question but you answered it corectly - I'll let you have it - thumbs up from me.

You are correct Melody.

Applying the cross multiplication solves for.. meaning what's at the root of it, what does it do exactly?

Imagine you are in a spinning extra-vehicular harness in space (like George Clooney in the film Gravity).  You have to fire your pneumatic boosters to counteract the spin.  The reaction force from the gas jet is a vector, as is the radius arm from your centre of gravity to the gas jet.  The cross-product of these two vectors tells you the magnitude and direction of the torque being supplied to counter your spin (or to increase your spin, or set you tumbling in a different direction, if you get it wrong)!   

There are many more down-to-earth applications of the vector cross-product, often involving rotating objects.

 … if you do not have all the variables for the matrix will it generally come down to a quadratic or similtaneous equation which is solved by a matrix?

I don’t understand what you are asking here.  The matrix determinant form of the cross-product is just a way of remembering which components are multiplied by which in each direction! 

Melody is correct....no solution exists.......here's a graph of both functions....notice that there are no intersection points.....

Because you are xvxvxv

xv is 15 in roman numerals and there are 3 of them so you are 15x3  

yes yes ... now I understand this question ..

very nice idea  

thank you melody 

but why you call me 15×3 !

Because it is equal to an absolute value so it can't be negative!

Do you see what I mean?

sory melody how do you know that  3-x bigger than 0 ! 

Antilog is not a term I have used since I was at school and we wer not allowed to use calculators.

BUT

I think it just means 10^0.35

$${{\mathtt{10}}}^{{\mathtt{0.35}}} = {\mathtt{2.238\: \!721\: \!138\: \!568\: \!339\: \!6}}$$

Please someone correct me if I am wrong.

Here's an interesting article about this:

Hi 15x3 

I didn't even notice that question!

$$|2x-9|&=&3-x\\\\
\begin{array}{rlllrl}
2x-9&=&3-x\qquad &or \qquad 2x-9&=&-(3-x)\\
3x&=&12\qquad &or \qquad 2x-9&=&-3+x\\
x&=&4\qquad &or \qquad x&=&6\\
\end{array}
\\BUT\\
3-x\ge 0\\
-x\ge -3\\
x\le 3\\$$

Therefore there are no solutions.

-----------------------------------------------------------

Heureka please help.

I am just not getting the hang of this aligning properly, I wanted the x up against the equal sign, and its not there!

This is my code.

\begin{array}{rlllrl} 2x-9&=&3-x\qquad &or \qquad 2x-9&=&-(3-x)\\

3x&=&12\qquad &or \qquad 2x-9&=&-3+x\\

x&=&4\qquad &or \qquad x&=&6\\

\end{array} \\

Can you try and explain please?

Here's your answer........just divide it in half, of course!!!

Thanks Alan. Yes, I understand a lot about the vectors but I didn't clearly know the difference between the cross multiplication and dot product/addition of forces. Here was much easier to ask than looking on Kahn. :)

We learnt this week, 3d vectors also solving with a matrix, and the information of the last 1/2 of your post. Your reply is now printed and a good reminder for this week at least while I wait for the text book to arrive. However the screen prints I posted were from last weeks lecture notes before we'd covered this more in depth. I guess that due to missing a lecture I just felt also that I may not have known the information while required knowing it to use in tutorial questions.I think most of the students felt this week a bit like the job was demanding given what we knew.

Additionally, I found that this week when trying to resolve the 3d vector equations, that I was trying to get a single answer in N rather than in the i, j, k form and leaving it in i, j, k. Thankfully it has become more clear now. This was the difference I needed to get my head around and which I was ultimately seeking somewhere within my search for answers, lol. Applying the cross multiplication solves for.. meaning what's at the root of it, what does it do exactly? r is lengths of the values of i j and x right, so then why do we use them to resolve magnitude? Also, if you do not have all the variables for the matrix will it generally come down to a quadratic or similtaneous eqaution which is solved by a matrix?

WE finished the tutorial this week on breaking down a similtaneous equation, into a quadradic and using the quadratic formula. So, we're catching up on the notes from last week as we dive a little deeper. Thanks again. It was not to confusing. It is just that the application is confusingas well as recalling all the information as required both that I have learnt recently or since the beginning of the course. It's becoming simpler though with a special thanks to the forum here and your self.

thank you melody 

what about the second question .؟ 

Aziz, you could probably present a question like this in LaTex next time. 

Only if you are up to it.