- Messages
- 63
- Reaction score
- 87
- Points
- 28
4 markscan u tell me hw much we lose for if we had not plot the graphs..i hurry i forgot to see marks ..
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4 markscan u tell me hw much we lose for if we had not plot the graphs..i hurry i forgot to see marks ..
have you written the scale and marked the both axis with units??r u sure we only lose 2marks if we had not drawn the graph
then you'll lose only 2 marks maximum..just chill man..yeah my point on the graph was not in straight line nd rest all things was done ..only points were lying here nd there on the graph
just chill it varies according to the position allocated to the student...e.g mine was different as i was standing under an air conditioner..what about k mine was near 1
same lol the chill of the AC gave me an idea for the errorjust chill it varies according to the position allocated to the student...e.g mine was different as i was standing under an air conditioner..
what thats so high mine was 0.09k value was 478.22 sumthing
yeahk value was 478.22 sumthing
Mine was negative too and i am 100 percent sure its righy!!!Mine was negative too!
yea the gradient was negative because of 1/Myes the gradient is negative
I am also 100% sureMine was negative too and i am 100 percent sure its righy!!!
the height was decreasing with increasing weight :Oso when the mass increases the h increases so how can the gradient be negative
yeah it was decreasingthe height was decreasing with increasing weight :O
then don't worry beacuse even here at xtreme papers u'll note that most ppl got a negative gradientmine was negative too but every other kid in the school had a postive gradient =[
p was equal to the gradient you calculated while q was equal to ur value of y-interceptI even have my values and the table and I keep redrawing it.. it doesn't make sense how people got a positive gradient.. anyway what was p and q?
p was equal to the gradient you calculated while q was equal to ur value of y-intercept
and btw ppl, anyone who calculated the y-intercept from the equation of y=mx + c, then its really not possible for ur gradients to be positive since a positive gradient would give a negative y-intercept and the y-intercept obviously wasn't negative since there were no negative values used in the graph![]()
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