LaTeX Form (Part 3)

I'll try.

I just did 20mins worth of LaTeX...It's gone...

$$\\2)\\\\
\frac{15}{10}\\\\
\frac{5*3}{5*2}\\\\
\frac{\not{5}\;*3}{\not{5}\;*2}\\\\
\frac{3}{2}\\\\\\$$

$$\\3)\\\\
\frac{49}{56}\\\\
\frac{7*7}{7*8}\\\\
\frac{\not{7}\;*7}{\not{7}\;*8}\\\\
\frac{7}{8}\\\\
4)\\\\
\frac{8}{12}\\\\
\frac{4*2}{4*3}\\\\
\frac{\not{4}\;*2}{\not{4}\;*3}\\\\
\frac{2}{3}$$

It might be wrong/presentation might be bad, because i rushed it -sorry .

The problem is with Latex is, when you type in a lot.

The Coding only sends the last part.

So, this time i had to send the code into parts.

So far I know:

• Fractions
• Squaring
• Square root
• New Lines
• Colours
• Bigger (lines etc...)
• Array

Maybe other small things, but I've covered the basics...I think?

Here was your question from your last thread that you have not answered yet :)

Can you simplify this one for me ?  Do it with cancelling.  

Can you remember how to do that or do you want me to show you?

$$\frac{21x}{y}\times \frac{8y^2x}{14t}$$

I'm not 100% on how to do this. I'd prefer it if you'd show me.  

I like your way of teaching.

Also, I've seen something like this in my book...

Would you have to factorise to cancel out? (Or am I on the wrong track?)

Hi MG,

Yes basically you do factor then cancel that is.

I want to go back a step and see how much you understand of numbers.  If I was tutoring you face to face this would be so much easier and quicker for both of us but I really need to work out your level.  It is easy to rote learn, you obvioulsy have a good memory, but i need to be certain that you really understand each step and you are just not mimicking what i have told you.  (you may not even realise that you really did not understand)

So 

Please be patient with me and show me how to do these.  If you can't do them say so but it would be good if you have a go as well.

Please do all the cancelling that you can and do Q2 and Q3 in a few steps so I can see how you have done them.    

$$\\1)\;\;\frac{2}{3}\times \frac{5}{7}\\\\
2)\;\; \frac{16}{12}\times \frac{15}{10} \\\\
3)\;\;6\frac{2}{5}\times 1\frac{2}{6}\\\\$$

Ok, Melody. 

    $$\\Q1)\\\\
\frac{2}{3}\times\frac{5}{7}\\\\
Times\;the\;numerator\;by\;the\;numerator\\
Times\;the\;denominator\;by\;the\;denominator.\\\\
2\times5=35\\
3\times7=21\\\\
\frac{2}{3}\times\frac{5}{7}\;=\;\dfrac{35}{21}\\\\
You\;can\;simplify\;\quad\frac{35}{21}\;\div7=\quad\frac{5}{3}\\\\
\frac{5}{7}\;is\;your\;answer.$$

$$\\Q2)\\\\
\frac{16}{12}\times\frac{15}{10}\\\\
Times\;the\;numerator\;by\;the\;numerator.\\
Times\;the\;denominator\;by\;the\;denominator.\\\\
16\times15=(numerator*numerator)\\\\
To\;this\;without\;calculator.\\\\
10\times5=50\\
10\times10=100\\
6\times5=30\\
6\times10=60\\\\
50+100=150\quad150+60=210\quad210+30=240\\\\
12\times10=(denomiator*denominator)\\\\$$

$$\\To\;do\;this\;without\;a\;calculator.\\\\
10\times10=100\\
2\times10=20\\\\
100+20=120\\\\
16\times15=240\\\\
12\times10=120\\\\
Similify\quad\frac{240}{120}\quad\div120\quad=\frac{2}{1}\\\\
Answer=\frac{2}{1}$$

$$\\Q3\\\\
6\frac{2}{5}\times1\frac{2}{6}\\\\
6\times1=6\\\\
Times\;the\;numerator\;by\;the\;numerator.\\
Times\;the\;denominator\;by\;the\;denominator.\\\\
2\times2=4\\\
5\times6=30\\\\
\frac{2}{5}\times\frac{2}{6}=\frac{4}{30}\\\\
Simplify\quad\frac{2}{30}\quad\div2\quad=\frac{1}{15}\\\\
Answer=6\frac{1}{15}\\
(I\;think...)$$

Thanks MG and thanks for being patient :)

Q1 and 2 have the correct answers but I will talk about them.

Q3 is not correct but I do not worry about that too much yet.  :)

Q1  Your answer is right but you working was shaky 

$$\frac{35}{21}$$      If you divide this by 7 then the value would change.

BUT if you divide it by1 then the value will be the same.   for example

$$\\8\div 1=8\\\\
\frac{3}{4}\div 1 = \frac{3}{4}\\\\
$Now any number divided by itself = 1$\\\\
\frac{5}{5}=1 \qquad \frac{12}{12}=1 \qquad \frac{-6}{-6}=1\\\\\\
\frac{35}{21}\div 7 \qquad $will change value BUT $\\\\
\frac{35\div 7}{21 \div 7} \qquad $will keep the same value$\\\\
=\frac{5}{3}
=1\frac{2}{3}$$

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

Q2     I have spread this answer out a real lot so that you can see better what is happening

$$\frac{16}{12}\times \frac{15}{10}\\\\
=\frac{16}{4*3}\times \frac{5*3}{10}\\\\
=\frac{16}{4*\not{3}^1}\times \frac{5*\not{3}^1}{10}\\\\
=\frac{16}{4}\times \frac{5}{10}\\\\
=\frac{16}{4}\times \frac{5*1}{5*2}\\\\
=\frac{16}{4}\times \frac{\not{5}*1}{\not{5}*2}\\\\
=\frac{16}{4}\times \frac{1}{2}\\\\
=\frac{4*4}{4*1}\times \frac{1}{2}\\\\
=\frac{\not{4}*4}{\not{4}*1}\times \frac{1}{2}\\\\
=\frac{4}{1}\times \frac{1}{2}\\\\
=\frac{2*2}{1}\times \frac{1}{2*1}\\\\$$

$$\\=\frac{\not{2}*2}{1}\times \frac{1}{\not{2}*1}\\\\
=\frac{2}{1}\times \frac{1}{1}\\\\
=2\times 1\\\\
=2$$

What I have done in question 2 is called cancelling :)

I'll do a few more for you 

Simlify

$$\\1) \;\;\frac{10}{12} \qquad$the common factor here is 2 so I divide the top and bottom by 2$\\
\frac{\not{10^5}}{\not{12}^6}=\frac{5}{6}\\\\
$I have divided top and bottom by 2 but i didn't actually write the $\div 2 $ there, I just did the cancelling.$\\\\
$Your turn$\\\\
$Simplify these with cancelling$\\\\
2)\;\;\frac{15}{10}\\\\
3)\;\;\frac{49}{56}\\\\
4)\;\;\frac{8}{12}\\\\$$

If you use LaTex     \not{8}^4    is what I might use, it would look like this       $$\not{8}^4$$

I know you want to do higher level maths than this but you need a really good grounding in basic maths and then algebra will be easy, so please be patient  :)

No problems here, it is all really good MG   

When you are trying to learn mathematics try not to care excessively about Latex.  Don't let it get in the way of your maths learning.   :)

Oh, if I write too much LaTex it does do that.

I cut some of my lines until it is working properly.  Then I 'ok' it.

Then I open a new Latex box and paste the cut code in there.  

So I may end up with 2 or 3 consecutive lots of LaTex coding.

It ususally only happens when the code is very long though :/

I might try looking for units that you can do elsewhere and then I can tie my teaching in with it.  It is VERY important that you get grounded properly in basic arithmetic.  Then you will be able to fly ahead.  :)

I would like you to work your way through these tutorials

https://www.khanacademy.org/math/arithmetic/fractions/multiplying_fractions/v/multiplying-fractions-and-whole-numbers

The early ones might seem to be too easy but it might be a good idea to watch them anyway.

Plus, some clips start out very easy but add more difficult stuff as it continues.

There are exercises for you to check your progress on too.

Let me know how you get on.    

I eventually want you to learn to cancel effectively without too many steps.  

Don't skip steps till you are really ready to though.

Don't worry, I'll definitely see the video.

Also, the 'book' thing, is kind of a new thing.

In class, we take it really slow. Really. I started fractions, let's say at the beginning of year 5.

I understand the recapping, as things like this, people tend to forget.

But taking a bit further would be usefull.

Maybe: Dividing, taking away, multiplying, adding them.

I had already known the multiplying and adding.

But the other two seemed trickier

So, that's why I have my book.

It's now pretty easier to do them.

Taking away is the same as adding.

(LCD)

When Dividing and multiplying by a whole number, it's crucial to turn the whole number to an improper fraction. Then convert it back to a whole number. (That was my mistake).

As  for division, something to do with "reciprocals", but you flip the second fraction (one you're dividing by) so that the denominator is the numerator and the numerator is the denominator then times them as normal.

I see the video, once I have my breakfast. 

Yes, that all sounds really good MG   :)

I saw you do an equation on the forum today too.  You did a good job of it.

This book that you are talking about.  Is it an exercise book to take notes in?

Well it's those CGP and Essential Books.

They teach you then give questions.

So, yes you can make notes.

Thanks MG     

I always see these words but never know the meaning.

• Cos
• sin
• tangent
• tan

You can look at this video to get an idea     

https://www.khanacademy.org/math/trigonometry/basic-trigonometry/basic_trig_ratios/v/basic-trigonometry

BUT

It is really important that you concentrate on the earlier work that I have given you.

Without a VERY solid grounding in the fundamentals you will NEVER really be a Maths God.     

Ok :D

I won't go off topic.

I just see everyone writing about those words and a have no single idea.

I will see the video and leave it there,.

Also, the video of fractions i have watched it, it's about multiplying fractions. :D

Now what? 

I'll retry Q3) :D

$$\\6\frac{2}{5}\times1\frac{2}{6}\\\\
6\frac{2}{5}\;as\;an\;improper\;fraction\\\\
5\times6\;=\;30\;+2=32/Denominator\times\;number+numerator\\\\
6\frac{2}{5}=\frac{32}{5}\\\\\\
1\frac{2}{6}\;as\;an\;improper\;fraction\\\\
1\times6\;=\;6\;+2=8/Denominator\times\;number+numerator\\\\
1\frac{2}{6}=\frac{8}{6}\\\\$$

$$\\Now\;\frac{32}{5}\times\frac{8}{6}\\\\
\frac{32*8}{5*6}\\\\
\frac{256}{30}\\\\
Simplify\;fraction\;\frac{256\div2}{30\div2}=\frac{128}{15}\\\\
\frac{128}{15}\;to\;mixed\;number=128\div15=8\;with\;a\;remainder\;of\;8\\\\
answer\;8\frac{8}{15}$$

Good but I want you to cancel BEFORE you multiple.  That way you do not need to worry about such big numbers.

Also when you cancel you can also do it sideways so long as it is multiply and you cancel something on the top with something on the bottom.

eg

$$\\\frac{5}{3}\times \frac{9}{10}\\\\
=\frac{\not{5}^1}{3}\times \frac{9}{\not{10}^2}\\\\
=\frac{1}{3}\times \frac{9}{2}\\\\
=\frac{1}{\not{3}^1}\times \frac{\not{9}^3}{2}\\\\
=\frac{1}{1}\times \frac{3}{2}\\\\
=1\times \frac{3}{2}\\\\
=\frac{3}{2}\\\\
=1\frac{1}{2}\\\\$$

When you get good at it you can do all of this in 1 step :)

Try this one

$$1\frac{1}{14}\times 2\frac{2}{20}$$

Try and do it the easiest way.  Make the best use of cancelling  

Ok, but this is the first time I'm seeing it. I've never seen/learnt this way of cancelling.

Q) Do I cancel before converting the fraction to a mixed number.

$$\\1\frac{1}{14}\times2\frac{2}{20}\\\\\\
1\frac{1}{14}=\frac{15}{14}\\\\
2\frac{2}{20}=\frac{43}{20}\\\\
\frac{15}{14}\quad\frac{43}{20}\\\\
\frac{15}{\not{14}7}\quad\frac{43}{\not{20}10}\qquad(\div2)\\\\
\frac{15}{7}\quad\frac{43}{10}\\\\
\frac{15}{\not7}\quad\frac{43}{10}\\\\$$

Converting after.

$$\\1\frac{1}{14}\times2\frac{2}{20}\\\\
1\frac{1}{\not{14}\;7}\times2\frac{\not{2\;}1}{20}\\\\
1\frac{1}{7}\times2\frac{1}{20}\\\\
1\frac{1}{7}=\frac{8}{7}\\\\
2\frac{1}{20}=\frac{41}{20}\\\\
\frac{8}{7}\times\frac{41}{20}\\\\
\frac{8*41}{7*20}=\frac{328}{140}=\frac{82}{35}=2\frac{12}{35}$$

Ps: I watched the video.

I learnt a lot.

Like in a right angle triangle, there is : Opposite side , Hypotenuse side and adjacent side.

And Soh-Cah-Toa

To find Sine (shorter Sin)=SOH.

S(Sin)O(Opposite)H(Hypotenuse)

So in Sin you do Opposite/Hypotenuse.

To find Cosine(shorter Cos)=CAH

C(Cos)A(Adjacent)H(Hypotenuse)

So in Cos you do Adjacent/Hypotenuse

To find Tangent(shorter Tan)=TOA

T(Tangent)O(Opposite)A(Adjacent)

So in Tan you do Opposite/Adjacent

The longest side is hypotenuse.

A^2 + O^2 = H^2

(I knew this from past HW)

(Just recapping what I learnt, I know you know this) :)

I think that's a lot for a 9min video.

Thanks for the video Melody.

There is a lot of briliant videos out there where you can learn a lot.

This video was good but certainly not one of the very best one.

BUT

Be very careful MG.  You must get a  really good grounding or it WILL catch up with you.

Do fractions decimals and percents (with ordinary numbers) until you can do EVERY question in your sleep.

THEN you will have a good grounding for everything else!

You have made a bit of a mess of the multiplication.

$$1\frac{1}{14}\times 2\frac{2}{20}$$

You changed them inot improper fractions well, although it would have been good if you had noticed that 2/20=1/10 right at the ver beginning.

When you are cancelling.

You must cancel something on the top with something on the bottom.  I don't know what you did!

I'll show you the best way to handle this one, I have put in extra steps that will hwelp you understand.

$$\\1\frac{1}{14}\times 2\frac{2}{20}\\\\
1\frac{1}{14}\times 2\frac{2\div 2}{20\div 2}\\\\
=1\frac{1}{14}\times 2\frac{1}{10}\\\\
=\frac{1*14+1}{14}\times \frac{2*10+1}{10}\\\\
=\frac{15}{14}\times \frac{21}{10}\\\\
=\frac{15\div 5}{14}\times \frac{21}{10\div 5}\\\\
=\frac{3}{14}\times \frac{21}{2}\\\\
=\frac{3}{14\div 7}\times \frac{21\div 7}{2}\\\\
=\frac{3}{2}\times \frac{3}{2}\\\\$$

$$\\=\frac{3*3}{2*2}}\\\\
=\frac{9}{4}}\\\\
=2\frac{1}{4}}\\\\$$

When you finish check with a calculator that your answer is corrrect

$$\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{14}}}}\right){\mathtt{\,\times\,}}\left({\mathtt{2}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{2}}}{{\mathtt{20}}}}\right) = {\frac{{\mathtt{9}}}{{\mathtt{4}}}} = {\mathtt{2.25}}$$             that is good

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

Have you gone through those fraction videos that I told you to look at ?

You have a very good memory MG.  

In many ways this is great     BUT     

it can lead you to think that you understand things that you really do not understand.

You may have copied and learned a method and you may not even realize that you don't understand what you are doing.

Eventually this catches up with you and you cannot advance further.

You will always need to be very aware of this!

The multiplying fraction video is the only video you have given so far.And i have watched it.I'll try keep stable. 

I'll try keep stable. 

I know you're doing this for my good.

Thanks Melody for the support.

That fraction thing had lots of parts I think.  Lots of video clips with exercises inbetween each one.

Did you work through all of it  ??

I just watched the link you gave to me. I'll check now if there was anything else that's related to fractions. :)

EDIT: There are 9 Videos.

I'll go get the popcorn and set up video 2. :)

I watched them all.

Most of them were just a repeat.

Your technique of canceling is different from what I learnt.

Based on the video, i do the sum then simplify, you simplify then do the sum.

:)

Yes that is right MG, you cancel as you go.  You do not have to deal with numbers bigger than necessary that way.  

 (Oh you said so the sum, sum is addition, you meant, do the multiplication, or find the product. It is just a technicality but it is important to use the right words, especially in mathematics )

Plus it will tie in with algebra better too.  Please do it closer to my way :)

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

Here is one that I will make up off the top of my head.

Do it the easiest way, with lots of cancelling.   NO Calculator, except at the end to check you answer

(use the forum calc so I can see and try to make the forum calc display it as I have done, or as near as you can).

Do you remember how to divide by a fraction?  Change it into an improper fraction first and then turn the one you are dividing by upside down (take the reciprocal of the divisor) and change the divide sign to a multiply sign.

Happy calculations   

$$\\34\frac{1}{8}\times \frac{14}{51}\div 1\frac{15}{34}$$

OH did you enjoy the popcorn.  I want some too !!!       

Yes, "sum" was a very bad word to use there! 

I will use the cancelling technique, and I will give this question all I've got!

(PS: I enjoyed the popcorn, I will send you some Unpopped kernels from England - it probably won't be as tasty as the popcorn in Austrailian )

$$\\34\frac{1}{8}\times\frac{14}{51}\div1\frac{15}{34}\\\\
34\frac{1}{8}\times\frac{14}{51}\\\\
=\;34\frac{1}{8\div2}\times\frac{14\div2}{51}\\\\
=\;34\frac{1}{4}\times\frac{7}{51}\\\\
=\;\frac{34*4+1}{4}\times\frac{7}{51}\\\\
=\;\frac{137}{4}\times\frac{7}{51}\\\\
=\;\frac{137*7}{4*51}\\\\$$

$$\\=\;\frac{959}{204}\\\\
=\;\frac{959}{204}\div1\frac{15}{34}\\\\
=\;\frac{959}{204}\times1\frac{34}{15}\\\\
=\;\frac{959}{204\div34}\times1\frac{34\div34}{15}\\\\
=\;\frac{959}{6}\times1\frac{1}{15}\\\\
=\;\frac{959}{6\div3}\times1\frac{1}{15\div3}\\\\
=\;\frac{959}{2}\times1\frac{1}{5}\\\\$$

$$\\=\;\frac{959}{2}\times\frac{1*5+1}{5}\\\\
=\;\frac{959}{2}\times\frac{6}{5}\\\\
=\;\frac{959}{2\div2}\times\frac{6\div2}{5}\\\\
=\;\frac{959}{1}\times\frac{3}{5}\\\\
=\;959\times\frac{3}{5}\\\\
=\;\frac{2877}{5}$$

I gave it everything.

I got an answer in the end.

I'm not sure if I made any mistakes.

But, again with the code i messed up, i wrote this before but when I copied and pasted (just to be quicker with fractions) but I forgot that I copied the code before, then the code doubled, and everything went in between and I had to start again.

CALC.

Umm...How do I do it on the calc again?

I've been using Latex so much I almost forgot how to do a fraction and it was just /.

When you're doing mixed numbers, how do you do it wihout mixing the number with the fraction.

\\34\frac{1}{8}\times\frac{14}{51}\div1\frac{15}{34}\\\\
34\frac{1}{8}\times\frac{14}{51}\\\\
=\;34\frac{1}{8\div2}\times\frac{14\div2}{51}\\\\
=\;34\frac{1}{4}\times\frac{7}{51}\\\\
=\;\frac{34*4+1}{4}\times\frac{7}{51}\\\\
=\;\frac{137}{4}\times\frac{7}{51}\\\\
=\;\frac{137*7}{4*51}\\\\
=\;\frac{959}{204}\\\\
=\;\frac{959}{204}\div1\frac{15}{34}\\\\
=\;\frac{959}{204}\times1\frac{34}{15}\\\\
=\;\frac{959}{204\div34}\times1\frac{34\div34}{15}\\\\
=\;\frac{959}{6}\times1\frac{1}{15}\\\\
=\;\frac{959}{6\div3}\times1\frac{1}{15\div3}\\\\
=\;\frac{959}{2}\times1\frac{1}{5}\\\\
=\;\frac{959}{2}\times1\frac{1*5+1}{5}\\\\
=\;\frac{959}{2}\times\frac{6}{5}\\\\
=\;\frac{959}{2\div2}\times\frac{6\div2}{5}\\\\
=\;\frac{959}{1}\times\frac{3}{5}\\\\
=\;959\times\frac{3}{5}\\\\
=\;\frac{2877}{5}

I am sorry MG I know you tried really hard but you have messed up right from the beginning.  

I will start you off.

$$\\34\frac{1}{8}\times \frac{14}{51}\div 1\frac{15}{34}\\\\
$None of the fraction parts can be simplified so you firstly need to change all the mixed numerals to improper fractions.$\\\\
\frac{8*34+1}{8}\times \frac{14}{51}\div \frac{34*1+15}{34}\\\\
$now simplify what you have got$\\\\
*\\\\
$now invert the last fraction and change the divide to a time$\\\\
$Now you are up to cancelling BUT don't cancel yet$$$

Don't cancel yet.  Get it right up to here first!   Then tell me what you want to cancel.

Don't tell me in the LaTex.  Tell me underneath in normal text.

Good luck  

(I was taking a risk, up there in the Latex, i was going to do this but then i followed this i had 2 ideas planed)

I'm here!

$$\\34\frac{1}{8}\times \frac{14}{51}\div 1\frac{15}{34}$$

So convert.

(34*8+1)/(8) * (14)/(15)  /  (1*34+15)/(34)

Conversion

= (273)/(8) * (14)/(15)  /   (49)/(34)

Simplify

= (273)/(8) * (14)/(15)  /   (49)(34) 

Umm...Changing?

=(273)/(8) * (14)/(15)  *  (34)(49)

Done, it all makes sense now.

(The reason I canceled before changing it from an improper fraction to a mixed number, is because you had shown it in one of your examples)

Ok I am glad that it makes sense so far    

$$\\34\frac{1}{8}\times \frac{14}{51}\div 1\frac{15}{34}\\\\
\frac{(34*8+1)}{(8)} * \frac{14}{51} \div \frac{(1*34+15)}{(34)}\\\\
= \frac{273}{8} * \frac{14}{51} \div \frac{49}{34}\\\\
= \frac{273}{8} * \frac{14}{51} * \frac{34}{49}\\\\
$Done, it all makes sense now.$$$

Well I am glad it all makes sense, What are you going to do next?  (no multiplying!)     

Cancel out. :)

But is there a difference when cancelling between three numbers?

That is a good question MG  :)

There is no real difference.  You must always cancel ONE thing on the top with ONE thing on the bottom.

Eg. you might start by divided the 8 by two on the bottom and the 34 by two on the top.  

That would be one step  

273/8 * 14/15 * 34/49.    /2. (8,34)

273/4 * 14/15 * 17/49    /2. (4,14)

273/2 * 7/15   * 17/49.  /7. (7,49)

273/2 * 1/15   * 17/7.   /3. (273,15)

91/2. *  1/5.    * 17/7.   /7. (91,7)

13/2.  * 1/5.    * 17/1.  

Lol they are all prime numbers now...1 mathematically not but .

Sorry to the cheap presentation I'm using my kindle. :(

I hope it's clear in the brackets is what's being divided by the number  /x.

Yes, that is excellent MG, can you finish it off  and then check it onthe forum calc in the same posts?

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

then

You can work out WHY you have not answered the original question that I asked and answer that one :))   [You made a copy error]

13/2*1/5

13*1=13

2*5 = 10

13/10*17/1

13*17=221

10*1 =10

221/10 is the answer.

That is great MG

BUT

it is an improper fraction I want a mixed numeral for the answer.

Since it is really easy you could give the decimal answer as well.:)

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

this is the question you have answered - it is a little different from the question that I asked.

$$\\34\frac{1}{8}\times \frac{14}{15}\div 1\frac{15}{34}$$

and this is how you would put it into the forum calc.

(34+1/8)*(14/15)/(1+15/34)        

You have to put this in the calculation box of course 

I want you to use the calc to check that your answer is correct  :)

221/10

As a mixed number.

10 fits into 221, 22 times with a remainder of 1.

22+1/10

I'll do the algebraic question (1st comment) once I have access to my computer so I can do latex

$${\frac{\left({\mathtt{34}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{8}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{14}}}{{\mathtt{15}}}}\right)}{\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{15}}}{{\mathtt{34}}}}\right)}} = {\frac{{\mathtt{221}}}{{\mathtt{10}}}} = {\mathtt{22.1}}$$

That is excellent MG

Do you want to do my original question now MG.

You could even do it in LaTex :)

$$\\34\frac{1}{8}\times \frac{14}{51}\div 1\frac{15}{34}$$

When you finish, check it on the forum calc :)

. I'll be happy to do it once I get access to the laptop.

. I think that's the wrong question, I just done it.

$$\\\frac{21x}{y}\times \frac{8y^2x}{14t}\\\\
Cancel\:out...Only\;Like\;Terms!!!\\\\
\frac{21x}{y}\times \frac{8y^2x}{14t}\\\\\\
$Umm...\;I can't find anything.\\\\$
$I could divide 21 and 14 by 7, but they're not like terms.\\\\$
$So, it's already canceled?\\\\$
\frac{21x\times8y^2x}\times{y\times14t}\\\\
\frac{21x^2+8y^2}{14ty}\\\\
\frac{168xy^4}{14ty}$$

I think.

No MG 

You only usually talk about like terms when you add or subtract algebra.

Plus you cannot suddenly introduce addition into a multiply and divide only question.

This is the one I wanted you to do.

It is different from the one you did before because you made a copy error.

In the other one you did 51/14    NOT   14/51       

so do this one please.

When you can do everything with numbers I will introduce you to algebra slowly so that you will understand.

BUT I want you to be competant with numbers first!

You can use LaTex too, but not if it makes the question too hard :/

$$\\34\frac{1}{8}\times \frac{14}{51}\div 1\frac{15}{34}$$

For the + sign, it was just to seperate the two otherwise it would have been too squashed.

As for the question, i just done it a few comments ago, in steps. Do you want me to do it all together?

Yes I would prefer you do it altogether, like I do it :)

If you need to spread your LaTex out, use   \;\;

That would be much better 

It is after 4am for me so I finishing here for the evening, I'll see you later :)

Goodnight Melody.

(I don't know how late you can stay up!!!)

I'd fall asleep on the laptop at 1.

Not that I stay up that late!

Yes I used to be like that but I spend so much time socializing with people in different time zones that I  usually finish for the day quite late.  

BUT

1 am is still pretty late.   

Before i start my latex.

I have a question, you said   16 post's ago (all post not including this one)

you said:

None of the fraction part can't be simplified so change the mixed number to an improper fraction.

However i could've divided 8 and 14 by 2...?

I will start you off.

$$\\34\frac{1}{8}\times \frac{14}{51}\div 1\frac{15}{34}\\\\
$None of the fraction parts can be simplified so you firstly need to change all the mixed numerals to improper fractions.$\\\\
\textcolor[rgb]{1,0,0}{\mbox{You cannot cancel the 8 and the 14 by 2 because the 1/8 is a part of (34+1/8).}}\\\\
\textcolor[rgb]{1,0,0}{\mbox{ You cannot cancel mixed numeral, they have to be made into improper fraction first}}\\\\
\frac{8*34+1}{8}\times \frac{14}{51}\div \frac{34*1+15}{34}\\\\
$now simplify what you have got$\\\\
$now invert the last fraction and change the divide to a time$\\\\
$Now you are up to cancelling BUT don't cancel yet$
\textcolor[rgb]{1,0,0}{}$$

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

this is what I was referring to

$$\\2\frac{6}{8}\div 1\frac{8}{10} \\\\
$first I can simplify the fraction part that I have in the mixed numerals$\\\\
=2\frac{\not{6}^3}{\not{8}^4}\div 1\frac{5}{22} \\\\
\textcolor[rgb]{1,0,0}{=2\frac{3}{4}}\div 1\frac{5}{22} \\\\
=\frac{2*4+3}{4}\div \frac{1*22+5}{22} \\\\
=\frac{11}{4}\div \frac{27}{22} \\\\
$What is next here?$$$

DO YOU UNDERSTAND?

Yes i do. 

4 and 22 it can divide by 2.

i got confused because on  May 29, 2015 2:16:15 PM 

You cancelled out mixed numbers but only the one.

Ok then,

please do this one - you have not done it yet.

Do it with the most allowed canceling so that you do not need to multiply big numbers any more than necessary.  Also, keep it all together please like I do.  I would like you to use Latex please.   

$$\\34\frac{1}{8}\times \frac{14}{51}\div 1\frac{30}{68}$$

Hi MG

This is the Alice in Wonderland post that you were talking about

http://web2.0calc.com/questions/alice-and-the-wonderland#r1

I will show the trick of referencing the exact post that you want within a thread AFTER you get that fraction one above correct to my satisfaction.       I think it will be a very handy thing for you to know :)

You can call it bribery if you want.   

If it gets me answers! You can call it bribery all you want.

$$\\34\frac{1}{8}\times\frac{14}{51}\div1\frac{30}{68}\\\\
\frac{34\times8+1}{8}\times\frac{14}{51}\div\frac{1\times30+68}{68}\\\\
\frac{273}{8}\times\frac{14}{51}\div\frac{98}{68}\\\\
\frac{273}{8}\times\frac{14}{51}\times\frac{68}{98}\\\\
\frac{273}{8}\times\frac{14\div14}{51}\times\frac{98\div14}{68}\\\\
\frac{273}{8}\times\frac{1}{51}\times\frac{7}{68}\\\\
\frac{273}{8\div4}\times\frac{1}{51}\times\frac{7}{68\div4}\\\\
\frac{273}{2}\times\frac{1}{51}\times\frac{7}{17}\\\\
\frac{273}{2}\times\frac{1}{51\div17}\times\frac{7}{17\div17}\\\\
\frac{273}{2}\times\frac{1}{3}\times\frac{7}{1}\\\\$$

$$\\\frac{273\div3}{2}\times\frac{1}{3\div3}\times\frac{7}{1}\\\\
\frac{91}{2}\times\frac{1}{1}\times\frac{7}{1}\\\\
\frac{91\div7}{2}\times\frac{1}{1}\times\frac{7\div7}{1}\\\\
\frac{13}{2}\times\frac{1}{1}\times\frac{1}{1}\\\\
\frac{13}{2}\times1\times1\\\\
\frac{13\times1}{2\times1}\\\\
\frac{13}{2}\\\\
\frac{13\times1}{2\times1}\\\\
\frac{13}{2}\\\\
6\frac{1}{2}$$

(The end part was just to show my workings when multiplying by 11 it's obviously the same.)

$${\frac{\left({\mathtt{34}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{8}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{14}}}{{\mathtt{51}}}}\right)}{\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{30}}}{{\mathtt{68}}}}\right)}} = {\frac{{\mathtt{13}}}{{\mathtt{2}}}} = {\mathtt{6.5}}$$

Great latex there MG.

there is an error in the second row (typo perhaps)

Plus you could have simplified a little before that. 

So fix the 2nd row

then just say what you should have done before that.

I have not looked at the rest yet but at a glance it looks good. :)

$$\\\frac{34\times8+1}{8}\times\frac{14}{51}\div\frac{1\times30+68}{68}$$

This row?

I have no idea what the problem is?

Yes I could have simplified the first bit. . But then that would mean I'd have  to start again. 

Hi MG,

Now I have looked properly I see there are many mistakes.  :(

I shall think before  giving you more instruction. :/

I know you are trying hard and I really appreciate that.  :)

Hmm...Wow.

I honestly, don't see any problems (in my eyes, if you say there is then there must be), I got the correct answer in the end?

I even checked with a calculator to be sure that the numbers are divisible.

The problem must be my technique of cancelling there must've been a quicker/better way to find the solution.

(The only problem that I know is that at the beginning I could've cancelled but I didn't.)

Hi MG,

I am so pleased that you can use LaTex, it makes it so much easier to check your work. 

When you cancel fractions you can only do it with multiply (no mixed numerals either)  I think you already know that,

BUT when you cancel it has GOT to be something on the TOP with something on the BOTTOM

It cannot be 2 things on the top  OR two things on the bottom, it doesn't work that way.

That was a repetitive mistake that you made all the way through.  

It worries me a little that you cannot see the error on the second line too.  It does not effect the 3rd line but that is just a lucky conincidence.  Do you want to take a look again and see if you cannot spot the error :)

What do you want to do.   Do you want to leave it and do easier examples.

Maybe you could redo it but just add one or two lines at a time so I can check it as you go :/

Actually if you really can't see what is wrong with the second line then I should concentrat on that first.

It is important that you not only can get the answers and working correct but that you also completely understand why you are doing things and why they work.

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

Now I am going to show you have to reference a specific post withing a thread.  :)

At the top right of each post is a date stamp.  If you click on that the address in the address bar will change from being the address of the question at the top to the address of the specific post you clicked into.

Now you can just copy and paste the specific post address that you want.  Next time you want to reference the 15th post in your thread you will  be able to send me straight to it and I won't have to count anything.

Do you understand?    Use it a few times so you wont forget :)

Argh!

I just looked at that post rather then the entire latex.

I multiplied by 30 instead of 68 and added 68 instead of 30.

As for the latex, i can only divide the numerator and denominators and not numerator by numerator?

(The post count is useful thank you)

"I multiplied by 30 instead of 68 and added 68 instead of 30.        

YES

"As for the latex, i can only divide the numerator and denominators and not numerator by numerator?"

YES

Please think about this example

$$\\\frac{3}{4}\times \frac{7}{8}\\
$You'd want to cancel the 4 with the 8 but you can't!! $\\
$you do know that it equals$\\
\frac{3*7}{4*8}=\frac{21}{32}\\\\
$21 and 32 have no common factors, this cannot be simplified!! $\\
$If this answer cannot be simplified then the working you used to get there cannot be simplified either!$$$

Do you want to try that question over again ?

Also, whenever you see fractions on the forum I would like you to race in and try to answer them.

You probably already do that as much as you can anyway :)

Ok Melody, thanks for the example and racing to answer questions is a great opportunity, missing it would be dreadful. 

$$\\34\frac{1}{8}\times\frac{14}{51}\div1\frac{30}{60}\\\\
34\frac{1}{8}\times\frac{14}{51}\div1\frac{30\div30}{60\div30}\\\\
34\frac{1}{8}\times\frac{14}{51}\div1\frac{1}{2}\\\\
\frac{8\times34+1}{8}\times\frac{14}{51}\div\frac{2\times1+1}{2}\\\\
\frac{273}{8}\times\frac{14}{51}\div\frac{3}{2}\\\\
\frac{273}{8}\times\frac{14}{51}\times\frac{2}{3}\\\\
\frac{273\div3}{8}\times\frac{14}{51\div3}\times\frac{2}{3}\\\\$$

$$\\\frac{91}{8}\times\frac{14}{17}\times\frac{2}{3}\\\\
\frac{91}{8\div2}\times\frac{14\div2}{17}\times\frac{2}{3}\\\\
\frac{91}{4}\times\frac{7}{17}\times\frac{2}{3}\\\\
\frac{91}{4\div2}\times\frac{7}{17}\times\frac{2\div2}{3}\\\\
\frac{91}{2}\times\frac{7}{17}\times\frac{1}{3}\\\\\\
\frac{91\times7}{2\times17}\\\\
=\;\frac{637}{34}\\\\
\frac{637\times1}{34\times3}\\\\
=\;\frac{637}{102}\\\\
=\;6\frac{25}{102}$$

Argh 

Did 60 instead of 68, I hope you don't mind...

(I tried to do it this time right!)

$${\frac{\left({\mathtt{34}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{8}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{14}}}{{\mathtt{15}}}}\right)}{\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{30}}}{{\mathtt{60}}}}\right)}} = {\frac{{\mathtt{637}}}{{\mathtt{30}}}} = {\mathtt{21.233\: \!333\: \!333\: \!333\: \!333\: \!3}}$$

And i got it... Wrong. 

\\34\frac{1}{8}\times\frac{14}{51}\div1\frac{30}{60}\\\\
34\frac{1}{8}\times\frac{14}{51}\div1\frac{30\div30}{60\div30}\\\\
34\frac{1}{8}\times\frac{14}{51}\div1\frac{1}{2}\\\\
\frac{8\times34+1}{8}\times\frac{14}{51}\div\frac{2\times1+1}{2}\\\\
\frac{273}{8}\times\frac{14}{51}\div\frac{3}{2}\\\\
\frac{273}{8}\times\frac{14}{51}\times\frac{2}{3}\\\\
\frac{273\div3}{8}\times\frac{14}{51\div3}\times\frac{2}{3}\\\\
\frac{91}{8}\times\frac{14}{17}\times\frac{2}{3}\\\\
\frac{91}{8\div2}\times\frac{14\div2}{17}\times\frac{2}{3}\\\\
\frac{91}{4}\times\frac{7}{17}\times\frac{2}{3}\\\\
\frac{91}{4\div2}\times\frac{7}{17}\times\frac{2\div2}{3}\\\\
\frac{91}{2}\times\frac{7}{17}\times\frac{1}{3}\\\\\\
\frac{91\times7}{2\times17}\\\\
=\;\frac{637}{34}\\\\
\frac{637\times1}{34\times3}\\\\
=\;\frac{637}{102}\\\\
=\;6\frac{25}{102}

(In Lated i do copy and paste when repeating to save time, apart from that this was what i typed in)

Ok MathsGod, lets look at what you have done :)

Ok I can see the problem.

Perhaps you have a very slight problem with dyslexia MG.

This is probably not the right word and every suffers from what I am talking about a little bit, I know I do.

But I do think you need to be extra careful.  You have swapped the digits over AGAIN.

This time you did it when you put the numbers into the calculator. 

So your calculations are perfect        (except that you copied my orriginal question down wrong)

It is your calculator answer that is incorrect!

ALSO I would have prefered that you finished it like this (plus you did not put your equal signs in )

$$\\=\frac{91}{2}\times\frac{7}{17}\times\frac{1}{3}\\\\
=\frac{91\times7\times 1}{2\times17\times 3}\\\\
=\frac{637}{34\times 3}\\\\
=\frac{637}{102}\\\\
=6\frac{25}{102}$$

You can take as many steps as you want but it is better if it is all kept together :)

Here is another question for you MathsGod :))

$$3+8\div16\times(8+5)$$

(It might help if you remember that a division sign is the same as a fraction line)

Do you know about order of operation?  Maybe you call it PEDMAS or BODMAS or BEDMAS ?

I'll remind you / teach you  anyway :)

When doing a question like this there are strict rules about the order you have to do it in.

Most counties seem to call it     PEDMAS

1) Parentheses (they are brackets )   

2) Exponents  (they are powers)

3) Dividsion and multiplication are EQUALLY important.  Do them from left to right.

4) Addition and subtraction are EQUALLY important.  Do them from left to right.

Now use this order  to answer the question.  Use LaTex please :))

When I use LaTex I almost always copy the line of just done and then past it underneath, then I just change it a little and I keep going down the page like this.  Maybe you already do this :))

Dyslexia. That's the first I've heard that.

I don't know whether it's true or not.

There is grammar check on this website. I use it a lot, sometimes a double a word or miss one out. It's because i plan what i write before writing it, so I miss out small words like "for" and "know" etc. Also letters.

(I searched the symptoms of it that's why I'm knowing this).

Also for the 15 and 51.

I mixed the 51 and 15 in my Latex but luckily i noticed it before i posted it.

The reason why I done this (i think) is because i worked on a question before it with 14/15 and this question was 14/51.

Everyone does it MG.  I think I do it more than most others.

I just think you need to on guard for this type of error the same as I do :)

I do use the copy & paste feature for latex. :)

$$\\3+8\div16\times(8+5)\\\\
=3+8\div16\times\mathbf{(8+5)}\qquad(Parentheses)\\\\
=3+8\div16\times13\\\\
=3+8\div16\times13\qquad(There\;are\;no\;Exponents\;)\\\\
=3+8\div16\times13\\\\
=3+\mathbf{8\div16}\times13\qquad(Division)\\\\
=3+2\times13\\\\
=3+\mathbf{2\times13}\qquad(Multiplication)\\\\
=3+26\\\\
=\mathbf{3+26}\qquad(Addition)\\\\
=29$$

What is 8 divided by 16 MG

Gah, i don't pay enough attention. 

The second i saw  divide with 8 & 16 my first thought was one.

$$\\3+8\div16\times(8+5)\\\\
=3+8\div16\times\mathbf{(8+5)}\qquad(Parentheses)\\\\
=3+8\div16\times13\\\\
=3+8\div16\times13\qquad(There\;are\;no\;Exponents\;)\\\\
=3+8\div16\times13\\\\
=3+\mathbf{8\div16}\times13\qquad(Division)\\\\
=3+0.5\times13\\\\
=3+\mathbf{0.5\times13}\qquad(Multiplication)\\\\
=3+6.5\\\\
=\mathbf{3+6.5}\qquad(Addition)\\\\
=9.5$$

It was quick, because i copied and paste then edit. 

Yes, that looks excellent MG,

I do want you to get used to checking your answers though.  It makes it easier for me to check as well!

For these number questions I want you always to run the question through the forum calc at the end so that we can both see that the answers are the same.   

New one:

$$1\frac{3}{4}\times 7 +8^2-6\div \frac{4}{10}$$

1\frac{3}{4}\times 7 +8^2-6\div \frac{4}{10}

I have not worked this one through so I have no idea how nicely it will turn out,it is probably horrible   

Definetly horrible. 

$$\\1\frac{3}{4}\times 7 +8^2-6\div \frac{4}{10}\\\\
=1\frac{3}{4}\times7+8^2-6\div\frac{4}{10}\qquad(There\;are\;no\;Brackets)\\\\
=1\frac{3}{4}\times 7 +8^2-6\div\frac{4}{10}\\\\
=1\frac{3}{4}\times7+\mathbf{8^2}-6\div\frac{4}{10}\qquad(Exponents)\\\\
=8\times8=64\\\\\$$

$$\\=1\frac{3}{4}\times7+64-6\div\frac{4}{10}\\\\
=1\frac{3}{4}\times7+64-`\mathbf{6\div\frac{4}{10}}\qquad(Division)\\\\
=\frac{6}{1}\times\frac{10}{4}=15\\\\
=1\frac{3}{4}\times7+64-15\\\\
=\mathbf{1\frac{3}{4}\times7}+64-15\qquad(Multiplication)\\\\
=\frac{7}{4}\times\frac{7}{1}=\frac{49}{4}\\\\
=\frac{49}{4}+64-15\\\\
=\mathbf{\frac{49}{4}+64}-15\qquad(Addition)\\\\
=\frac{49}{4}+\frac{256}{4}=\frac{305}{4}\\\\
=\frac{305}{4}-15\\\\$$

$$\\=\mathbf{\frac{305}{4}-15}\qquad(Substraction)\\\\
=\frac{305}{4}-\frac{60}{4}=\frac{245}{4}
=\frac{245}{4}$$

$$\frac{245}{4}$$

I had to make a new comment.

I coundn't ris losing 20minues of latex.

$$\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{3}}}{{\mathtt{4}}}}\right){\mathtt{\,\times\,}}{\mathtt{7}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{8}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\frac{{\mathtt{6}}}{\left({\frac{{\mathtt{4}}}{{\mathtt{10}}}}\right)}} = {\frac{{\mathtt{245}}}{{\mathtt{4}}}} = {\mathtt{61.25}}$$

YES!!!!!!!!!!!!!!!!!!

I did double check, didn't have to edit it though.

As for the bit where there is a - 

I got confused whether is was negative 6 or taking away 6.

At first i worked it out as a negative then it didn't work out...

Underneath each bracket collum i worked it out. 

I skiped all the workings and went to the final stage (because it's really long and I didn't want the latex to look messy).

If you want i can show my workings and how i got there in the next post - i don't mind! 

YES that looks really good MG.

There are a couple of points I'd like to make though.  

First, you did not simplify  $$\frac{4}{10}$$    right at the very begining!

second, you are still splitting it into bits - i don't wnat you to do that!  

I accept that not ALL your working will be shown, some will be scribbled on bits of paper and it does not need to be included.  

Also, when you are adding and subtracting mixed numerals do not change them into improper fractions.

It is NOT necessary and it makes the calculations worse!

$$\\1\frac{3}{4}\times 7 +8^2-6\div \frac{4}{10}\\\\
1\frac{3}{4}\times 7 +8^2-6\div \textcolor[rgb]{1,0,0}{\frac{2}{5}}\\\\
=1\frac{3}{4}\times7+\textcolor[rgb]{1,0,0}{64} -6\div\frac{2}{5}\qquad\\\\
=\textcolor[rgb]{1,0,0}{\frac{4*1+3}{4}}\times7+64 -6\div\frac{2}{5}\qquad\\\\
=\textcolor[rgb]{1,0,0}{\frac{7}{4}}\times\textcolor[rgb]{1,0,0}{\frac{7}{1}}}+64 -6\div\frac{2}{5}\qquad\\\\
=\textcolor[rgb]{1,0,0}{\frac{49}{4}}+64 -6\div\frac{2}{5}\\\\
=\frac{49}{4} +64-6\textcolor[rgb]{1,0,0}{\times\frac{5}{2}}\\\\
=\frac{49}{4} +64-\textcolor[rgb]{1,0,0}{\frac{6}{1}\times\frac{5}{2}}\\\\
=\frac{49}{4} +64-\textcolor[rgb]{1,0,0}{\frac{3}{1}\times\frac{5}{1}}\\\\
=\frac{49}{4} +64-\textcolor[rgb]{1,0,0}{\frac{3*5}{1*1}}\\\\$$

$$\\=\frac{49}{4} +64-\textcolor[rgb]{1,0,0}{15}}\\\\
=\textcolor[rgb]{1,0,0}{12\frac{1}{4}} +64-15\\\\
=\textcolor[rgb]{1,0,0}{12+\frac{1}{4}} +64-15\\\\
=\textcolor[rgb]{1,0,0}{64+12-15+\frac{1}{4}}\\\\
=\textcolor[rgb]{1,0,0}{64-3+\frac{1}{4}}\\\\
=\textcolor[rgb]{1,0,0}{61+\frac{1}{4}}\\\\
=61\frac{1}{4}$$

Now that was an effort!

Look though how I have done it MG and learn!!   I am sure that you will :)

Especially look at the mixed numeral addition!

I'm getting confused near the end.

On row 14 is where it starts.

Weren't you meant to do 64+12 then take away 15? 

I don't get how you got 3 on row 15?

12-15 = -3 ?

But you're meant to do addition first ..

YES you are correct MG

BUT

12+64-15 = 12-15+64 = 64-15+12

there is an invisable + in front of the 12

you can add     +12,     +64     and      -15     in any order and the answer will be the same.

So I arranged them in the order that I thought was easiest but you may have chosen a different order.

I saw that 10-15=-3    and 64-3=61

I just thought that looked easy :)

so I rearranged    12+64-15      to      12-15+64   

Does that make sense?

For your next question you need to start a new thread again :)

Remember to start it with the address of this thread so that you keep your continuity :)

Yes thank you, adding first is more simple as it follows PEDMAS.

Also what did you mean by same address.

LaTeX Form (Part 4)

Where did I say "same address"  ?

I meant start a new thread but in your initial post give the address of this thread.

Kind of like having book1, book2, book3, etc where on the first page of each book you write down where to find the last book.

So if you ever want to refer back it is easy :)

Oh ok.

But don't I already do that?

Or am i mistaken for something else.

So...

Could you tell me the trick for the Alice and the Wonderland post?

http://web2.0calc.com/questions/alice-and-the-wonderland#r1

Hi MG

"Don't I already do that?"

Yes you do - it was just a reminder:)

----

I think the maths in the Alice in wonderland problem is above your head.  Do you understand any of it?

Oh I almost forgot.

That is excellent use of that address referencing!

I understand how you the Kings image.

But I'm still slightly confused on the Queens.

(Also Thank you for Address reference comment)

EDIT: For the post underneath, i 'd just noticed that you had done the same on the AIN post.

KING = Big Portrait of himself and slightly small portrait of the Queen.

QUEEN = Big Portrait of herself and extremely small portrait of the King.

TOTAL = 1865cm

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

Your Answer :

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

$$x^2+y^2=1865$$

We  can solve this with just a little sensible trial and error.

$${\sqrt{{\mathtt{1\,865}}}} = {\mathtt{43.185\: \!645\: \!763\: \!378\: \!368\: \!2}}$$

So for the  2 that are a long way apart one is are going to be just a little smaller than this number and the other will be little.

Try 43

$${\sqrt{{\mathtt{1\,865}}{\mathtt{\,-\,}}{{\mathtt{43}}}^{{\mathtt{2}}}}} = {\mathtt{4}}$$                                          ok so the first 2 queens ones are    4cm and 43cm

Now lets think about the 2 that are close together.  

think about   

$${\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}} = {\mathtt{1\,865}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{{\sqrt{{\mathtt{1\,865}}}}}{{\sqrt{{\mathtt{2}}}}}}\\
{\mathtt{x}} = {\frac{{\sqrt{{\mathtt{1\,865}}}}}{{\sqrt{{\mathtt{2}}}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{30.536\: \!862\: \!969\: \!204\: \!940\: \!9}}\\
{\mathtt{x}} = {\mathtt{30.536\: \!862\: \!969\: \!204\: \!940\: \!9}}\\
\end{array} \right\}$$

Bingo, the Kings ones are   29cm and 32 cm

The King's pictures were    32cm x 32cm       and      29cm x 29cm

and

The queen's were              43cm x 43cm       and      4 cm x 4cm

I'm actually confused on King:

King: 

When you Sqrt 1864/Sqrt 2 .   Why did you have a negative one and a positive one?

Couldn't you just do one, round it up and down to find the two that are close. Or is there a reason for having the negative.

That's it...

But there is one more question that is bugging me a bit I understand it partially.

Why do you square root? 

I understand that it's to find 'like the centre'

(I'm bad with these terms I felt that the word middle would've been wrong, otherwise you'd just half 1865)

EDIT: (Isn't the Kings phot meant to 29cm x 31 cm , to 30 rounds to 31?)

IDK...?

Here are the King's pictures.  (We know the total area is 1865cm^2 but ignore this to start with)

I have used x to be the length of the King's picture

and I have used y to represent the length of the Queen's picture.

Do you understand what I mean by these last 2 sentences?

What is the area of them ?

We are told that they are nearly the same size.

If they were EXACTLY the same size what could you say about the lengths and the areas then?

I understand those two sentence.

So the total is 1865^2

( I probably would've never thought of this, but i have already seen your post)

To find a near middle of 1865 find the sqrt of it and round it then take it away from 1865^2 . The smaller one would be the smaller portrait.

If they were the same wouldn't the answer be ( JUST FOR QUEEN , IF THERE WAS NO KING) 1865

=1865^2

1865*1865

Sorry MathsGod but that is not correct.

The total area is   $$1865 cm^2$$    THAT IS NOT  $$1865^2$$    !!!!

1)  If the sides of a square are 8cm  What is the area of that square ?

2)  If the area of a squara is 25cm^2 .   What is the length of the side?

If both sides of a square is 8cm then the area is 64cm

8*8

If the total is 25cm^2 = 625

Each side is 25 

25*25

Oh dear MG,

We have stubbled on a concept that you absolutely need to understand.

Length measures distance. The units might be cm, m, yards, km etc

The perimeter or the length or bredth or height of am object is a length.

You can measure theses with a ruler or with a measure tape etc

Area measures how many SQUARES fit INSIDE a shape.  

So the units might be   $$cm^2\;\; or\;\;m^2\;\;or \;\;miles^2$$

The units are square units because they are counting squares.

Work through this page and think really hard about what you are being shown.

https://www.mathsisfun.com/geometry/area.html

NOW

You know that   $$7^2=49$$     

What would be the inverse of this square operation.   

I mean if you have 49 what can you do to it to get 7

You know that   $$8^2=64$$

If you have 64 what can you do to it to get an answer of 8?

The two answers above should be the same.     

Square root 49 and 64 to get the number that you square to get it.

Yes that is correct MathsGod   

ok so square root 'undoes' square!

$$\sqrt{9^2}=\sqrt{81}=9$$

If the length of the square is $$9 cm$$ then the area will be $$9^2cm^2=81cm^2$$

If the area of the square is   $$81cm^2$$  then the length will be   $$\sqrt{81}cm = 9cm$$

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

NOW

I went to a lot of trouble to draw that diagram.

So can you answer the questions at the bottom please?   I am trying to make sure that you understand an important concept here.

Did you go through the maths is fun page that I sent you to?

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

When you get sick of my questions then you can just say you have had enough for now

OR you can just stop answering.  

Either way you can start again when and if you want to. 

I like teaching you but I honestly will not mind.       

Q1)

= 16cm^2

It's a square all sides are the same and finding the area times both sides.

Q2)

The area is 9^2 = 81

It's a square meaning all sides are the same.

So two SAME numbers that times together to make 81

9cm

Hi Melody, :)

My End of Year assessment is in 2 days.

These are the topics I've studied since September

1. Place value and powers of 10

• Write digits in numbers and words
• Multiply and divide by 10, 100 ,1000
• Ordering positive, negative and decimal numbers on a number line

2. Rounding and estimation

• Rounding to the nearest whole number, 10, 100 and 1000
• Approximate answers to calculations by providing correct estimates

3. Adding and subtracting decimals

4. Mental calculations

• Know up to 12 x 12 times table
• Use BIDMAS/PEDMAS Correctly

5. Written calculations of positive and whole numbers

• Multiply and divide positive whole numbers

6. Factors, multiples and primes

• List factors of a given number
• Find the HCF and LCM of two numbers
• Understand what we mean by a prime number and identify all prime numbers between 1 and 10

7. Proportional reasoning

• Spot the difference between direct and inverse proportion
• Find missing quantities using direct and inverse proportion

8. More calculations

• And and subtract negative number
• Multiply and divide negative numbers
• Add and subtact decimal
• Multiply and divide decimal

9. Introduction to algebra

• Add, subtract, multiply and divide algebraic terms
• Expand brackets in algebra
• Substitution

10. Solving Equations

• Solve simple one step equation (EG) x+5=7
• Solve equation with two steps (EG) 2x-8=10
• Solve equations with unknowns on both sides
• Solving equation with brackets

11. Angles

• Draw and measure lines and angle
• Understand what we mean by paralles and perpendicualr lines
• Find missing angles in triangles and rectangles
• Use angle facrs to find algebra in a paralles lines 
• Calculate bearings and solve problems involving bearings

12. Proporties of shape

• Understand properties of triangle, quadrilateral and circle
• Use algebra to find missing lenghths
• Undestand properties of 3D shapes and draw plan, front and side elevation froma given 3D shape
• Find volumes and surface area of a shape

13. Fractions 

• Shade fractions and find equivalent fractions
• Ordering fractions 
• Find fractions of amounts
• Find a fraction of another fraction

14. Percentages

• Convert between fraction, decimals and percentages
• Make equivalent fractions with denominator over 100
• Find a percentage of a number
• Convert fractions and decimals greater than 100% to decimals.

Hi Melody.

Sorry for not replying for long and also my work is very messy and hasn't had enough effort put into it.

But I'm really nervous about Monday (Todays Saturday here) all my tests about the entire year of my learning.

My final Year 7 grade.

If I mess up I won't be happy.

Maths...I always make dumb mistakes but It's mainly english and Science I'm good in both but I find it very hard and English is so much work it's hard.

Again. sorry. 

Hi Melody.

Sorry for not replying for long and also my work is very messy and hasn't had enough effort put into it.

But I'm really nervous about Monday (Todays Saturday here) all my tests about the entire year of my learning.

My final Year 7 grade.

If I mess up I won't be happy.

Maths...I always make dumb mistakes but It's mainly english and Science I'm good in both but I find it very hard and English is so much work it's hard.

Again. sorry. 

You do not have to appologize to me MG.

I am very proud of you and I know that you try very hard.

If you are really busy with school or something else then just let me know and I will know not to push so hard.

Sometimes I get too pushy, I know that :))

You should do very well at maths.  

Just be super careful because you do tend to make a lot of careless mistakes.

I expect that your knowledge base is excellent :))

I am sure that you work really hard at your other subject as well so you will do well in those exams too :)

Having said all that is there any maths topics that you would like me to look at with you?

I mean ones that relate to Monday's exam?

All of the topics above are what I've been learning for the past year so, that will be in the exam.

1.PLACE VALUE AND POWERS OF 10: Seems like an easy thing to do.

2. ROUNDING AND ESTIMATION:Rounding I know, but sometimes i need to catch up on it.

3.ADDING AND SUBTRACTING DECIMAL: Is easy

4.MEANTAL CALCULATION: I can do it.

5. WRITTEN CALCULATIONS OF POSITIVE AND WHOLE NUMBER: Yep.

6. FACTORS MULTIPLES ND PRIMES:I can do it.

7. PROPORTIONAL REASONING: (I had an idea of what that was but now i know) I can do that.

http://teachmath.openschoolnetwork.ca/grade-6/proportional-reasoning/

8. MORE CALCULATIONS: Ehh... I always mess up with Negatives! 

9.INTRODUCTION TO ALGEBRA: I can do that.

10. SOLVING EQUATIONS: I can do that.

11. ANGLES: Partially, I know it but it's not polished.

12. PROPERTIES OF SHAPES: I hate shapes, i don't understand how to find things like the volume and surface area of a 3d shape and others.

13. FRACTIONS: I can do it. 

14. PERCENTAGE: Yes

2,8, 11, 12  Is what I'm struggling.

Here is a video on perimeter and area that I think it might be really good for you to watch.Don't give up on it to easily. it only goes for 8.5 minutes and I think you might get a fair bit out of it.

If you are sure you understand ALL of this then we'll look at some harder ones.

https://www.khanacademy.org/math/basic-geo/basic-geo-area-perimeter/basic-geo-area-perimeter-polygon/v/perimeter-and-area-basics

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

Actually there are whole heap of other ones on he right hand side of that page.  

So maybe you might like some of those :/

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

Volume:   Look at the area first but I found this one on volume.

https://www.youtube.com/watch?v=6nJgDehGmI4

Hi MG,

I decided I would persist with this today because you have length and area in your test tomorrow so it might help.

FIRST SQUARE

What I wanted you to say here is that the area of the first square is $$16cm^2$$    

DO NOT FORGET THE SQUARED UNITS.   This is very important because it is the number of SQUARES that fit inside the plane (flat) shape.

I asked for 2 reasons 

Reason 1):  There is 16 squares inside that shape.  Each square has a side length of 1cm.  

So there are 16     1cm squares.      That is   $$16cm^2$$

Reason 2:    A=L*L = 4*4 = $$16cm^2$$

SECOND SQUARE

The area of this square is $$9cm^2$$   I know that because I can COUNT the 9 squares AND because I am told.

The side length is 3cm     I know the side is 3 cm because i can see it on the picture.

I also know the side is 3cm because the area is   $$9cm^2$$    and area is  Length*length = $$L*L=L^2$$

 $$\mathbf{If\;\;\;\;L^2=9\qquad then\; \; L=\sqrt{9} = 3cm}$$

Here is a good post that might help you with negative numbers

http://web2.0calc.com/questions/positive-and-negative-numbers-directed-numbers

The answers have been whited out but I will un-whiten them for you after you answer the questions so that you can check if your answers are correct :)

Ok!

I get i, but whats the point of the square sign or is it just a math rule?

I remember learning this but I needed a recap, I will watch the video.

Also I'm having a huge problem with the ones i said on the post above.

Especially Shapes.

Hi MG

ALL THE SYMBOLS THAT I USE HAVE A PURPOSE.   

Do you mean this line

$$\mathbf{If\quad L^2=9\qquad then\; \; L=\sqrt{9} = 3cm}$$

rectangle

Area = Length*Breadth

A square is just a special rectangle where     length=  breadth    So

Area of a square = L*L

but L*L can be written as     $$L^2$$

So with the little rectangle 

$$\mathbf{If\quad L^2=9\qquad then\; \; L=\sqrt{9} = 3cm}$$

This is because the inverse of square is square root.

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

Isn't this area stuff a part of what you mean by shapes?

$$\\6*6=6^2\\
8*8=8^2\\
L*L=L^2\\$$

That is what squared means 

To answer your question.

I'm mainly talking about 3D Shapes.

To answer your question.

I'm mainly talking about 3D Shapes.

To understand the volume and the surface area of 3D shapes it is essentual that you understand the area of 2D shapes first.

You see, to find the surface area of 3D shapes you just add up the areas of all the sides.

To find the volume of any prism you just find the area of the side that stays the same all the way through and then multiply it by the depth.    (some people call the depth the height, it doesn't matter what you call it, if you UNDERSTAND what you are doing then the different names will not confuse you.)

The bottom clip on this post covered volume of prisms.  

http://web2.0calc.com/questions/latex-form-part-3#r100

I've watched the video read the post on Negative and Positive numbers.

Thanks it's really helped.

So to find volume : Find the area times by the depth.

Hi MG

Good luck in all your tests today    

Yes the volume of a right prism (all the ones you are asked to deal with are right prisms)

Is just the area of the face that stays the same  x  how far back it goes.  

Hi Melody,

My maths tests are later this week!

Todays was only English test.

:D

Hi Melody,

(Test tomorrow!!!)

:)

In my test theres something that i don't get.

I said ( some time ago) that i have 2 maths teacher, ones bad ones good.

The teacher who teaches us bad taught us about Bearings in angles...

I didn't get anything... 

:/

Hi MG,

Here is an introductory video on bearings for you to watch :)

https://www.youtube.com/watch?v=lyYqvI3MUME

Thank you Melody I will watch this.

But before I do. 

I just REALLY want to thank you, really you've being doing a lot for me and it is much appreciated.

Thank you Melody, thank you very much.   :D

You are very welcome MG :))

Well i learnt two different ways to find the bearings!

I hope it'll be usefull for the test.

I want to learnt this because, 

here: http://web2.0calc.com/questions/latex-form-part-3#r95

11, last bulletpoint.

$$\\$So, what now?$\\$$

Here is something you can play with for a bit MG

http://web2.0calc.com/questions/number-plane-co-ordinates-interactive-practice-for-beginners

Here is a whole stack of clips on angles.

https://www.khanacademy.org/math/basic-geo/basic-geo-angles/basic-geo-angle-basics/v/angle-basics

You could skip to the measuring one if you want but I think they are probably all good for you.

https://www.khanacademy.org/math/basic-geo/basic-geo-angles/basic-geo-angle-basics/v/angle-basics

Once you have gone through the measuring angles try to use your protractor and check that you can do it yourself.  When you use your own protractor, there are two sets of numbers, one inner and one outer.  This is so it is easy to measure angles no matter whether they are facing left or right.

When you measure your angle always use the nmbers that start from 0   (not from 180) 

Hopefully you will understand what I mean when you try to use your protractor.

If you are still confused there a lots of other protractor help sites,  I might even take a look now :))

Here is another good site 

https://www.mathsisfun.com/geometry/protractor-using.html

If you press the arrow in the blue circle it will show you how to construct an angle :))

Thaks Melody, if there is a new topic we start I'll create a new post.

Just so that the post actually follows it's name I'll do it in LaTeX.

Please start a new thread MG. This one is driving me n**s:)

You can reference this one from the beginning AND from the end.

Pretty Please. 

And tell me how you get on with angles. :)

Here is part 4  :)))

http://web2.0calc.com/questions/latex-form-part-4