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# Alice and the wonderland

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This link shows the Highlighted question.

Question 2 & 5

MathsGod  Mar 6, 2015

#52
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Wow never knew it could be solved so easily when you just find the equal opposite of it and/or making sure it adds up to 180.

Thanks CPhill & Melody.

I will try this and hopefully get most of it right.

MathsGod1  Mar 22, 2015
#1
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Q2

$${\frac{{\mathtt{9}}{!}}{\left({\mathtt{2}}{!}{\mathtt{\,\times\,}}{\mathtt{3}}{!}\right)}} = {\mathtt{30\,240}}$$      permutations.

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

Q5

KING      Big square photo of himself and Smaller square on of wife      Totally = 1865cm^2

QUEEN     Big square phot of herself and tiny square on of husband       Totally = 1865cm^2

The lengths of all the photos is an exact number of cm.

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

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.

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

okay so the two close together will be near 30

Try 30

$${\sqrt{{\mathtt{1\,865}}{\mathtt{\,-\,}}{{\mathtt{30}}}^{{\mathtt{2}}}}} = {\mathtt{31.064\: \!449\: \!134\: \!018\: \!134\: \!1}}$$

Try 29

$${\sqrt{{\mathtt{1\,865}}{\mathtt{\,-\,}}{{\mathtt{29}}}^{{\mathtt{2}}}}} = {\mathtt{32}}$$

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

Melody  Mar 7, 2015
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Wow!

I'm still puzzled at Q.2 though...

>.<

MathsGod  Mar 7, 2015
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Ik Q2

I'll do a simple example to show you

A1,A2,A3,B

how many ways can these be placed in a line   4*3*2*1=4! ways.

Now lets forget about the B for a moment, how many ways can the A's be ordered. that would be 3!=6 ways

So if you treat the As like they are all the same there will be   4!/3! ways of arranging them.  That's 4 ways.

BAAA, ABAA, AABA, AAAB     THAT IS IT.

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

now marmalade is made of      m m a a a r l d and  e

9 letters so if they were all different there would be 9! ways.

But

there are 2 Ms and 3 A's

so that is      9!/(2!*3!)

Melody  Mar 8, 2015
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Cool, thanks.

I understand now.

But " 9!/(2!*3!)" means= 9*2*3

???

Thanks

MathsGod  Mar 8, 2015
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Hi MathsGod,

No,

9! means 9*8*7*6*5*4*3*2*1

Say you had the letters  A,B,C,D,E,F,G,H,I

There are 9 letters.

So there are 9 letters to chose from for the first letter.

then there are 8 letters to choose from next

then 7 and so on untill there is just 1 then 0 letters left

so the number of ways that they can be ordered is

9*8*7.....*1     as I wrote above.

9!/(2!3!) =   $$\frac{9!}{2!3!}=\frac{1*2*3*4*5*6*7*8*9}{1*2\;\;*\;\;1*2*3}$$                      $${\frac{{\mathtt{9}}{!}}{\left({\mathtt{2}}{!}{\mathtt{\,\times\,}}{\mathtt{3}}{!}\right)}} = {\mathtt{30\,240}}$$

that page of questions that you found are far to hard for you, that is why you cannot understand my explanations.

If you would like to extend yourself that is excellent but you need to find something just a little different from what you are  doing in class.  Not something that is years above you.  :)

How about if I find something for you.

Maybe you will enjoy this video clip.

Melody  Mar 9, 2015
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So altogether there would be 15?

2: 1*2 (2 options)

3: 1*2*3 (3 options)

9: 9*8*7*6*5*4*3*2*1 (9 options)

9+3+2=15? Sorry I'm not familiar with this working but I hope I'm getting it.

Also, thanks for the support it's much appreciated.

:)

MathsGod  Mar 10, 2015
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But then 15 seems so small.

Is it 15 per letter. So 15*9=135

Sorry for all the Q.'s :(

MathsGod  Mar 10, 2015
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Hi MathsGod

I do not mind you asking questions but i do not understand what your question is.

A * sign means multiply.    :)

Melody  Mar 11, 2015
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since it's 9!/(2!*3!) if i was to simplify it into one number. I thought it'll be 15 ? But it seems to small so for each letter? 15*9=135

MathsGod  Mar 11, 2015
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no

$${\frac{{\mathtt{9}}{!}}{\left({\mathtt{2}}{!}{\mathtt{\,\times\,}}{\mathtt{3}}{!}\right)}} = {\mathtt{30\,240}}$$

Melody  Mar 11, 2015
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That many possibilities?! O.O

MathsGod  Mar 11, 2015
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It does seem like an awful lot but I think that it is correct. :)

Melody  Mar 11, 2015
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Why don't you try a little one.

Say

AABBB

how many permutations can you make with those letters?  Just copy my technique.

Can you name them all?

Melody  Mar 11, 2015
#14
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A1 A2 B1 B2 B3

5 letters

2A

3B

so

5!/(2!*3!) ?

MathsGod  Mar 11, 2015
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how would "!" be written on paper?

MathsGod  Mar 11, 2015
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Yes great.

5!      Is 5 factorial and it means

1x2x3x4x5

so what would  5!/(2!×3!) equal?

Melody  Mar 11, 2015
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5!/(2!×3!)

1*2*3*4*5=120

1*2=2

1*2*3=6

is it 120 divided by 6 * 2 ? $${\frac{{\mathtt{120}}}{\left({\mathtt{6}}{\mathtt{\,\times\,}}{\mathtt{2}}\right)}} = {\mathtt{10}}$$

MathsGod  Mar 11, 2015
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Excellent

now see if you can write out all those permutations.

there are exactly 10 of them 😉

Melody  Mar 11, 2015
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Since theirs different types of A does that mean AABBB and AABBB could be different.

AABBB

ABBBA

BBBAA

BABAB

BAABB

BABBA

ABABB

ABBAB

BBABA

um... >.<

MathsGod  Mar 11, 2015
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You are missing one MathsGod  -  you have to be very methodical

I am going to move the A's about in a very consistent way and hopefully I will get all 10 of them.

AABBB        A is first

ABABB

ABBAB

ABBBA

BAABB        B is first and A is second

BABAB

BABBA

BBAAB         B is 1st and second and A is 3rd

BBABA

BBBAA         B is 1st 2nd and 3'rd     A is 4th.

And that is 10

Melody  Mar 12, 2015
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Here is a new one if you would like to try it.

The word is    "Look"

How many permuations of the letters in this word are there?

Can you list them?

Melody  Mar 12, 2015
#22
+8

I am MathsGod P.S

Look

ookl

4!/(2!)

1*2*3*4=24

1*2=2

24/2=12

Look

Lkoo

Loko

Okol

Olok

Oolk

Oklo

Ookl

Olko

Kool

Kloo

Kolo

Thanks.

Is this right?

Guest Mar 12, 2015
#23
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Yes, that 'looks' really good MathsGod

Here is a different question.

Dora has 2 brothers and 2 sisters.  How many Permutation of the family can there be (from Dora 's point of view)

For instance Dora could have 2 older brothers and 2 younger sisters. That is one possibility.  :)

When you get sick of answering these questions you can either tell me or you can just stop answering.

I won't mind :)

Melody  Mar 13, 2015
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I don't mind answering this last one

2 brothers , 2 sisters ... is it BBSS?

Or since they could be either older or younger OB,YB,OS,YS

This is harder

BBSS=

$${\frac{{\mathtt{4}}{!}}{\left({\mathtt{2}}{!}{\mathtt{\,\times\,}}{\mathtt{2}}{!}\right)}} = {\mathtt{6}}$$

?

MathsGod1  Mar 14, 2015
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Older=O , Younger=Y

OB,OB,OS,OS

OB,OB,YS,YS

OB,OB,OS,YS

OB,YB,OS,YS

OB,YB,OS,OS

OB,YB,YS,YS

?

MathsGod1  Mar 14, 2015
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Ha,just realised i had only done the OB section

OB,OB,OS,OS

OB,OB,YS,YS

OB,OB,OS,YS

OB,YB,OS,YS

OB,YB,OS,OS

OB,YB,YS,YS

YB,YB,YS,YS

YB,OB,OS,OS

YB,OB,yS,YS

YB,OB,OS,YS

YB,YB,oS,YS

YB,YB,OS,OS

I think that's all.

MathsGod1  Mar 14, 2015
#27
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OB,YB,OS,YS

$${\frac{{\mathtt{4}}{!}}{\left({\mathtt{1}}{!}{\mathtt{\,\times\,}}{\mathtt{1}}{!}{\mathtt{\,\times\,}}{\mathtt{1}}{!}{\mathtt{\,\times\,}}{\mathtt{1}}{!}\right)}} = {\mathtt{24}}$$

Or

BBSS

$${\frac{{\mathtt{4}}{!}}{\left({\mathtt{2}}{!}{\mathtt{\,\times\,}}{\mathtt{2}}{!}\right)}} = {\mathtt{6}}$$

But i think i got all the combinations right.?

MathsGod1  Mar 14, 2015
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Hi MathsGod,

You have stumbled on a little problem.  I put it there to confuse you  :))

I think you have missed out on some ot the answers.

We have B B S S and Dora

so I think that  the number of permutations is         $$\frac{5!}{2!2!}=30$$

Can you find them all?

Notice that I keep using the word PERMUTATIONS.  This is a very specific maths word.

When doing probablilty it is very important to use the correct language, otherwise people will often misinterprete your meaning.  :)

Melody  Mar 14, 2015
#29
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And Dora ...

How did I not notice?

Ha!

So,BBSSD

$${\frac{{\mathtt{5}}{!}}{\left({\mathtt{2}}{!}{\mathtt{\,\times\,}}{\mathtt{2}}{!}\right)}} = {\mathtt{30}}$$

Hmm...Is there a quick way to find the 'Permutations'

MathsGod1  Mar 14, 2015
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Have you had enough yet or so you want another?  (I will let you finish this one first anyway)

Melody  Mar 14, 2015
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I will do another one.

I want to try it without help :D

Also, is their best way to find the permutations just by doing it each letter at a time, carefully?

MathsGod1  Mar 14, 2015
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There is not really a quick way exactly - you need to be methodical.

I would make a list of the 4 without Dora first (olders to youngest)    That is only  4!/(2!2!) = 6

then I woud repeatedly cut and paste that list putting Dora in each posible position.

so for the fist cut an paste Dora can be eldest.

for the next cut and past Dora can be second and so one.

So there will be 5 cut and  pastes

5*6=30

so that would be pretty quick.   Do you understand what I mean?

Melody  Mar 14, 2015
#33
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Well kind of.

4!/(2!*2!)=6

1.BBSS

2.SSBB

3.SBSB

4.BSBS

5.SBBS

6.BSSB

DBBSS

BDBSS

BBDSS                                       1

BBSDS

BBSSD

DSSBB

SDSBB

SSDBB                                       2

SSBDB

SSBBD

DSBSB

SDBSB

SBDSB                                      3

SBSDB

SBSBD

DBSBS

BDSBS

BSDBS                                       4

BSBDS

BSBSD

DSBBS

SDBBS

SBDBS                                       5

SBBDS

SBBSD

DBSSB

BDSSB

BSDSB                                       6

BSSDB

BSSBD

Is that all ?

I hope so >.<

MathsGod1  Mar 14, 2015
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Well 5*6 is 30 and that is what you were expecting so YES you got them all ant it was not really that hard either was it?

Here is one that is just a little different.

Suzie has 10 different coloured beads.  She uses these beads to make a bracelet.

This bracelet does not have a clasp so there is really no beginning or end (but there is clockwise and anticlockwise)

How many different permutations of beads are possible ?

Melody  Mar 14, 2015
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I'm going to have to go after this question.

:)

digit=

iidgt

$${\frac{{\mathtt{5}}{!}}{{\mathtt{2}}{!}}} = {\mathtt{60}}$$

MathsGod1  Mar 14, 2015
#37
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Wow, i am shocked how much with permutations i have improved on because of you Melody.

### Thanks SO Much!

MathsGod1  Mar 14, 2015
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I wrote another question for you a few posts back.

I whited it our because it is a little different from the others.

Maybe you would like to look at it.  I shall unwhite it for you.

It is the one about the bracelet. :)

Melody  Mar 14, 2015
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Sorry i guess i didn't refresh the page ii had only seen the link.

So 10 different coloured bead . A,B,C,D,E,F,G,H,I,J. (Just to make it easier :) )

Is it just... Since non of the letters are the same.

$${\mathtt{10}}{!} = {\mathtt{3\,628\,800}}$$

and if i was do 1!,10x it would be same

3628800.Too much >.<

MathsGod1  Mar 16, 2015
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This was the question:

Here is one that is just a little different.

Suzie has 10 different coloured beads.  She uses these beads to make a bracelet.

This bracelet does not have a clasp so there is really no beginning or end (but there is clockwise and anticlockwise)

How many different permutations of beads are possible ?

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

If these beadswere placed in a row your answer would have been correct, but when you put the beads in a circle there is no distict start point so you have to put the first bead down to indicate first place.  Now there are 9 beads (not 10 ) that need to be organised around this initial bead.

so the answer is 9!             $${\mathtt{9}}{!} = {\mathtt{362\,880}}$$ so if you go clockwise (or anticlockwise) from any specific point there will be 362 880 possible permutations.

Melody  Mar 17, 2015
#41
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Oops!

$${\mathtt{10}}{!} = {\mathtt{3\,628\,800}}$$

MathsGod1  Mar 20, 2015
#42
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Hi Melody,

I'm having some trouble trouble with angles with my 'basic' level work at school.

I'm wondering how I'd figure out these angles on my online Homework.

I took some screen shots of it.

http://imgur.com/ZbPme7o

http://imgur.com/i4UbIPQ

PS. Don't worry about cheating each time the angle changes

If you could help me it would be much appreciated.

Thanks,

MathsGod1  Mar 20, 2015
#43
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Question 2

Here are the answers for all of these

For the 1st one (with the 126 degree angle), x = 180 - 126 = 54 degrees (consecutive interior angles)

For the one with the 113 degree angle, x = 113 degrees (corresponding angles)

For the one with the 122 degree angle, x = 122 degrees (alternate interior angles)

For the one with the 80 degree angle, x = 180 - 80 = 100 degrees (x is supplemental to the vertical angle formed by the 80 degree angle.....thus this angle is a consecutive interior angle to x)

For the one with the 136 degree angle, x = 180 - 136 = 44 degree (x is supplemental to the vertical angle formed by the 136 degree angle.....like the last one, this angle is a consecutive interior angle to x)

For the one with the 61 degree angle, x = 180 - 61 = 119 degrees (same side exterior angles are supplemental)

CPhill  Mar 20, 2015
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Hi again MG1,

I only just saw you question and CPhill has already answered you.  :)

I have found that CPhill useds slightly different language for these than me.  It is a country based difference.  He is from the US and I am from Australia.  So his terms are probably better for you but I will run through it anyway.

1st on top row)

Cointerior angles   (C shaped)  These are supplementary angles which means that they add to 180 degrees

126+x=180

x=180-126

x= 54 degrees

2nd on top row )

Corresponding angles  (F shaped)  If you think about it you will see that they are corresponding because they are in the same position.

Corresponding angles on parrallel lines are congruent (the same)

x=113 degrees

last on top row.

Alternate  (z shaped)  if you think about it they zig zag.     Alternate angles are congruent.

x=122

Second row.

I don't have specific names for these I would work them out with 2 of the above ones each.

The first one I would use corresponding then adjacent supplementary.

The answer would be 180-80=100 degrees

the second one I'd use the same, the answer will be 180-136= 44degrees

The third one id do the same, the answer will be 180-61=119 degrees

Melody  Mar 21, 2015
#45
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Well I understand better now but the questions that iim really struggling at are the ones involving Algebra.

http://imgur.com/i4UbIPQ

MathsGod1  Mar 21, 2015
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y=64      corresponding angles on parallel lines (F shape)

The one opposite y in the quadrilateral is 75 degrees because it is corresponding on parallel lines to the other 75.

x+75=180  because they are adjacent and supplementary

x=105        Adjacent because they share a common vertex and a common arm.

Adjacent is a normal English word, it means next to, so the property that is adjacent to yours is the one that shares the same fence line.

Supplemenary just means that they add to 180.  (because they form a strght line)

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

the one opposite the x in the quadrilateral is 111 because it is a corresponding angle to 111 on parallel lines.

The one in the triangle next to it is 180-111=69 because they are adjacent supplementary angles.

the other one in the triangle is 180-125 = 55 because it is an adjacent supplementary angle to 125

y is the 3rd angle in the triangle so  69+55+y=180    angle sum of a triangle

$${\mathtt{69}}{\mathtt{\,\small\textbf+\,}}{\mathtt{55}}{\mathtt{\,\small\textbf+\,}}{\mathtt{y}} = {\mathtt{180}} \Rightarrow {\mathtt{y}} = {\mathtt{56}}$$

Now consider the big triangle.  you can get the top angle because it is adjacent supplementary to 111

and now you can get x from angle sum of a triangle.

See if you can do the other 2 on your own but if you can't post back and I will help.

I might just do posts with the pics incase you need help since I have already copied them. :)

Melody  Mar 21, 2015
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Melody  Mar 21, 2015
#48
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Melody  Mar 21, 2015
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Finish this question totally but then I think that you should start a new post in your new name :)

Melody  Mar 21, 2015
#50
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In the little triangle on the right containing x, the angle at the top right = 40 degrees because it is equal to the opposite angle in the parallelogram formed by the intersection of the parallel transversals with the double and triple arrows. And the other angle in this triangle = 67 degrees. It's the corresponding angle to the other 67 degree angle formed by the intersection of  parallel transversals with the single and triple arrows. Thus. x = 180 - 67 - 40 = 73 degrees.

And in the lttle triangle at the top....the angle at the bottom right is the vertical angle to the 67 degree angle in the triangle containing x. And the angle at the bottom left of this small triangle is the corresponding angle to the 40 degree angle formed by the intersection of the parallel transversals with the double and triple arrows. So y = 180 - 67 - 40 = 73 degrees, as well.

CPhill  Mar 21, 2015
#51
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x is supplemental to the 56 degree angle = 124 degrees

And y is the same measure as the angle between the 41 and 56 degree angles at the top left. And these three angles sum to 180 degrees. So, y  = 180 - 56 - 41 = 83 degrees.

CPhill  Mar 21, 2015
#52
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Wow never knew it could be solved so easily when you just find the equal opposite of it and/or making sure it adds up to 180.

Thanks CPhill & Melody.

I will try this and hopefully get most of it right.

MathsGod1  Mar 22, 2015
#53
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Well i got the first bit all right.

But the algebra bit i got 2 wrong.

But atleast i learnt something.

Also, Melody i will start a new post the second i got a new question.

MathsGod1  Mar 22, 2015
#54
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You are very welcome MathsGod.  I am glad that we are helping you to learn. :)

Melody  Mar 23, 2015
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MathsGod1  May 29, 2015
#56
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MathsGod1  May 29, 2015