hi
I'm confused with this question
it says
Using a trial and improvement method, slove the following equataions giving your answers to 1 decimal place
a) a squared - 28 = 0
b) a squared - 104= 0
c) a cubed - a = 15
how ?
start with some estimation
28 is a bit larger than 5^2, which would have been 5^2
so try 5.1^2 which is 26.01, too small
try 5.2^2 to get 27.04, still too small
5.3^2 = 28.09, almost dead on.
so to one decimal place a = 5.3
do the same with the other questions.
Using a trial and improvement method, slove the following equataions giving your answers to 1 decimal place
a) a squared - 28 = 0
b) a squared - 104= 0
c) a cubed - a = 15
n.........n^2
5.........25
a.........28
6.........36
(6 - a)/(6 - 5) = (36 - 28)/(36 - 25)
6 - a = 8/11
66 - 11a = 8 making a = 5.272
5.272^2 = 27.79
n.......n^2
5.272..27.79
a.......28
6.......36
(6 - a)/(6 - 5.272) = (36 - 28)/(36 - 27.79) making a = 5.2906
5.29062 = 27.99
Carry it as far as you wish.
Try the others now that you have a method.
1 answer
1 answer