D is partly constant and partly varies with V. when V=40, D=150 and when V=54, D=192.
(i)find the formula connecting D and V
(ii) find D when V =73.
D=av+b
solving for a and b
when v=40,D=150 and when v=54,D=192
150=40a+b.........(1)
192=54a+b.........(2)
Subtract (2) from (1)
42=14a
divide both sides by 14
a=3
Substitute a =3 into (1)
150=40Γ3+b
150=120+b
b=30.
a) find the relationship
D=av+b
D=3v+b
b) find D when v=73
D=3Γ73+30
D=249
D = mV + C , ( compare this to your familiar y = mx + b)
when V=40, D=150
150 = 40m + C **
when V=54, D=192
192 = 54m + C ***
subtract ** from ***
42 = 14m
m = 3
sub into ** , 150 = 120 + C
C = 30
so you have D = 3V + 30
sub in your given value of V
D=av+b
Solution for a and b
When v=40,D=150 and when v=54, D=192
150=40a+b..........(1)
192=54a+b..........(2)
Subtract (1) from (2)
(192=54a+b - 150=40a+b)
42=14a+0
Therefore:- 42=14a
Divide both sides by the co-efficient of (a) which is 14
a=3
Substitute a=3 into (1)
150=40Γ3+b
150=120+b
Collect like terms
b=150-120
b=30
i. The formula connecting D and V
D=av+b
D=3v+b
ii. D when V=73
D=av+b
D=3v+b
D=(3Γ73)+30
D=219+30
D=249