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 #1
avatar+3454 
+5

Looks like this question has been answered here

but, you probably already knew that. :)

Jul 19, 2014
 #225
avatar+118723 
+3

@@ End of Day Wrap :Sat19/7/14   Sydney, Australia   Time 10:30 pm    ♬                                       

It was a quite Fri/Sat today but as always there  were some good answers.  These were provided by CPhill, Alan, Rosala, NinjaDevo, Zegroes, Kitty3 and GoldenLeaf.  Thanks all.        

My new avitar has created some interest.  She is Lady (Queen) Guinevere of Camelot from the Brittish TV series "Merlin".  Arthur was the Legendary King of Camelot.  I was given the title of Queen Guinevere by another member, Morgan Tud.  Morgan Tud was the chief Physician to the royal court of Camelot.  

http://www.quest4camelot.com/legend/history.html     

https://www.google.com.au/webhp?sourceid=chrome-instant&ion=1&espv=2&ie=UTF-8#q=TV+series+Merlin

I have not settled on any permanent avitar, I will swap to another at some point in time. 

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

I only really noted one interesting thread tonight:

1)  This higher level one looks interesting. (Thanks Alan)      

http://web2.0calc.com/questions/urgent-solve-in-key-steps-please-v2sin-30-v1cos-30-cos-45-cos45-17#r6

That is it for tonight.  Thanks everyone.  Enjoy your weekend.

                                                                    ♬ ♬ MELODY ♬ ♬

                                     ♬ ♬      LADY GUINEVERE,  QUEEN OF CAMELOT   ♬ ♬                

Jul 19, 2014
 #1
avatar+33661 
+8
Jul 19, 2014
 #2
avatar+118723 
+3

Umm yes,

every x value only has one y value.

The graph has to pass the straight line test.  This means that any vertical line will never cut the graph more than once.

This might help you.

http://www.mathsisfun.com/sets/function.html

Sometimes restrictions a placed on relations so that they become functions.

eg 

$${{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{y}}}^{{\mathtt{2}}} = {\mathtt{16}}$$    Is a circle, radius 4 and centre(0,0)

This IS a relation, because there is a relationship between x and y.

When x=0 y can be 4 or -4,  Since there are 2 possible values of y, this is NOT A FUNCTION.

However, if you restrict the range and say  $$0\le y \le 1$$     it then becomes a FUNCTION

NOTE: 

All functions are relations but not all relations are functions.

To make this easier to understand I will show a comparison.

All squares are rectangles but not all rectangles are squares.

All roses are flowers but not all flowers are roses.

All terriers are dogs but not all dogs are terriers.

Jul 19, 2014
 #1
avatar+118723 
+5
Jul 19, 2014
 #6
avatar+118723 
+5

This is an old post that was our "reference material" sticky note topic.

It is worth looking at!

http://web2.0calc.com/questions/graphing-trig-functions_1#rr2

 

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

Ninja, could you please reference this whole thread underneath it - thank you.  Melody.

Jul 19, 2014

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