# Can anyone solve this please?

Variables t and N are such that when 1g N is plotted against 1g t, a straight line graph passing

through the points (0.45, 1.2) and (1, 3.4) is obtained.

(i) Express the equation of the straight line graph in the form 1g N = m 1g t + 1g c, where m and c

are constants to be found.                 (4)

(ii) Hence express N in terms of t. [1]

Posted Thu 7th March, 2013 @ 14:38 by ishrar afrida

a) you just put the coordinates into the equation and get simultanious equations which you can solve for m and c

So you get:

1.2 = 0.45m + 1g c

3.4 = 1m + 1g c

Then we can work out that 1g c by substitution

1g c = 1.2 - 0.45m

3.4 = m + 1.2 - 0.45m

2.2 = 0.55m

m=4

So 1g c =  1.2 - (0.45x4) = -0.6

Working out c is where i get stuck when we dont know what g is :/

b) So you know divide the whole equation by 1g gives you -

N = mt + c

N = 4t + (what ever you find out for c)

Sorry I couldn't help completely!

Answered Sat 9th March, 2013 @ 16:07 by Katie Howgate