The Student Room Group
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>

Physics/Maths - [Gradient - Help]

Posting an image below:

I done the usual but it saids my answer is wrong.
The usual as in m = change in y/change in x
I think it has something to do with the graph not starting from 0.
I put the y intercept as 10 but that is wrong too. I doen all the other ones when it starts at 0 so im guessing thats the problem.

Thanks
Reply 1
Avatar for AdeptDz
AdeptDz
OP
21
Capture.PNG
change in y=−2,y = -2, , change in x=50x = 50, m=−250=−125m = \frac{-2}{50} = \frac{-1}{25}

The y intercept doesn't matter at all when you are finding the gradient. If you are trying to find the equation of the line theny=mx+c,m=−125,c=10y = mx + c, m = \frac{-1}{25}, c = 10

therefore y=−125x+10 y = \frac{-1}{25}x + 10
(edited 7 years ago)
Original post by AdeptDz
Capture.PNG


maybe the change in x needs to be negative?

no idea what you posted - it doesn't make any sense
Reply 4
Avatar for AdeptDz
AdeptDz
OP
21
Original post by Maths is Life
maybe the change in x needs to be negative?

no idea what you posted - it doesn't make any sense


Basically:
Gradient and y-intercept of tht graph i just posted..
And I done that
Reply 5
Avatar for AdeptDz
AdeptDz
OP
21
Original post by Shipreck
change in y=−2,y = -2, , change in x=50x = 50, m=−250=−125m = \frac{-2}{50} = \frac{-1}{25}

The y intercept doesn't matter at all when you are finding the gradient. If you are trying to find the equation of the line theny=mx+c,m=−125,c=10y = mx + c, m = \frac{-1}{25}, c = 10

therefore y=−125x+10 y = \frac{-1}{25}x + 10


Ohh damn, thanks I see what i done wrong
I didn't see the 50 i thought it was 0 so i put 100 instead of 50.
Thanks
Reply 6
Avatar for AdeptDz
AdeptDz
OP
21
Original post by Shipreck
change in y=−2,y = -2, , change in x=50x = 50, m=−250=−125m = \frac{-2}{50} = \frac{-1}{25}

The y intercept doesn't matter at all when you are finding the gradient. If you are trying to find the equation of the line theny=mx+c,m=−125,c=10y = mx + c, m = \frac{-1}{25}, c = 10

therefore y=−125x+10 y = \frac{-1}{25}x + 10


It asks for the y intercept and I did put 10 but then as a hint it said
'remember the graph doesn't start from 0'
Reply 7
Avatar for JN17
JN17
10
Original post by AdeptDz
It asks for the y intercept and I did put 10 but then as a hint it said
'remember the graph doesn't start from 0'


If 10 wasn't the answer I'd assume they mean to continue the line backwards to when the current drawn is 0, then find the voltage at that point.
Reply 8
Avatar for AdeptDz
AdeptDz
OP
21
Original post by JN17
If 10 wasn't the answer I'd assume they mean to continue the line backwards to when the current drawn is 0, then find the voltage at that point.


Ahh yh true, thanks

Spoiler

Original post by AdeptDz
It asks for the y intercept and I did put 10 but then as a hint it said
'remember the graph doesn't start from 0'

Woops, didn't see that xD.y−10=−125(x−50),y=−x25+2+10,y=−x25+12y - 10 = -\frac{1}{25}(x - 50), y = -\frac{x}{25} + 2 + 10, y = -\frac{x}{25} +12
(edited 7 years ago)
Reply 10
Avatar for AdeptDz
AdeptDz
OP
21
Original post by Shipreck
Woops, didn't see that xD.y−10=−125(x−50),y=−x25+2+10,y=−x25+12y - 10 = -\frac{1}{25}(x - 50), y = -\frac{x}{25} + 2 + 10, y = -\frac{x}{25} +12


sorry, this is correct
Thanks!
so it is 12
(edited 7 years ago)

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you