** **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