http://img139.imageshack.us/img139/483/pkfpems0.png
Question:
By using gradients find the value of K.
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http://img139.imageshack.us/img139/483/pkfpems0.png
Question:
By using gradients find the value of K.
Wow i just done maths higher and never had a question like this or even been tought this.
You know, I have no idea why they teach us useless crap like this.
HOW will this graph help us in later life?
Idk but i need the answer :S
To find gradients which are sometimes missing?Quote:
Originally Posted by -:Undertaker:-
It was like 2 years ago since i have done this, so i cant help you as i have forgotten. Sorry!
I managed to get an answer but it is completely wrong... lol.
I got a Higher B in maths and honestly you will not get a question as difficult as this - I think it's hard due to the lack of information given.
Also - how accurate is your diagram? Is it supposed to be a right angle?
My current answer is:-
k = -9x + 29 over 4
Lol it is most likely wrong but it makes sense in the terms I have done it...
its a right angle yeh thats why i put the little angel thing in :P
If it was the correct shape to the question i may be able to help, but from that i cannot tell if the triangles are scaline or equalateral ect.
I'VE GOT IT :)
Lol right...
First of all you find the gadient of QP, which = 1/2 and as PQR is 90degrees you can state that the gradient of QR is therefore -2 (perpendicular).
You then have the gradient of the line QR and you have the point Q so you just find the equation of that line - which is y= -2x + 10.
You then substitute the point R into the equation of the line so you have:-
k = -2(5) + 10
k = -10 + 10
k = 0 :)
That kids got it :) +REPQuote:
Originally Posted by Ostinato
Kid? :oQuote:
Originally Posted by Cheekykarl
Kid/ Adult i dno what you are :PQuote:
Originally Posted by Ostinato
Ontopic - Well done at working it out :)
Lol tar :)Quote:
Originally Posted by Cheekykarl
um, i dont understand how k = 0 when from the picture, it clearly doesnt lol. unless that picture isnt right,
Pmsl hmmm...Quote:
Originally Posted by PTER
what i can find is
QP m = 2/3
y = 2/3x + b
7 = 2/3(-3) + b
7 = -2 + b
b = 7 + 2
b = 9
y = 2/3x + 9
QR since it is perpendicular with QP, its slope would be m = -1.5
y = -1.5x + b
7 = -1.5(-3) + b
7 = 4.5 + b
b = 2.5
y = -1.5x + 2.5
substituting the given value, 5 into the second equation, you'll have
-1.5(5) + 2.5 = k
-7.5 + 2.5 = k
k = -5
and i dont know gradients, sorry
Lol that can't be right - the point R isn't on 0 :)Quote:
Originally Posted by Ostinato
I think that PTER's right.
k is -5 :PQuote:
Originally Posted by PTER
woteva gurlfriend
Don't give in!!Quote:
Originally Posted by Ostinato