Helpp maths (higher) +9 rep

[Galaxay1]

http://img139.imageshack.us/img139/483/pkfpems0.png (http://imageshack.us)

Question:

By using gradients find the value of K.

[VPSwow]

Wow i just done maths higher and never had a question like this or even been tought this.

[-:Undertaker:-]

You know, I have no idea why they teach us useless crap like this.

HOW will this graph help us in later life?

[Galaxay1]

Idk but i need the answer :S

[Wizzdom]

You know, I have no idea why they teach us useless crap like this.

HOW will this graph help us in later life?

To find gradients which are sometimes missing?

It was like 2 years ago since i have done this, so i cant help you as i have forgotten. Sorry!

[Ostinato]

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...

[Galaxay1]

its a right angle yeh thats why i put the little angel thing in :P

[VPSwow]

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.

[Ostinato]

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 :)

[VPSwow]

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 :) +REP

[Ostinato]

That kids got it :) +REP

Kid? :o

[VPSwow]

Kid? :o

Kid/ Adult i dno what you are :P

Ontopic - Well done at working it out :)

[Ostinato]

Kid/ Adult i dno what you are :P

Ontopic - Well done at working it out :)

Lol tar :)

[PTER]

um, i dont understand how k = 0 when from the picture, it clearly doesnt lol. unless that picture isnt right,

[Ostinato]

um, i dont understand how k = 0 when from the picture, it clearly doesnt lol. unless that picture isnt right,

Pmsl hmmm...

[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

[Jõnathan]

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 :)
Lol that can't be right - the point R isn't on 0 :)
I think that PTER's right.

[Galaxay1]

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

k is -5 :P

[Ostinato]

woteva gurlfriend

[Naruto!]

woteva gurlfriend
Don't give in!!