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1234Linear functions

Solutions to the exercises

Exercise 1
a

Do it.

b

Cyclist 1: a 1 = 20 t
Cyclist 2: a 2 = -25 t + 150

c

20 t = -25 t + 150 gives you 45 t = 150 and therefore t = 3 1 3 .
The cyclists pass each other after 3 hours and 20 minutes.

Exercise 2
a

3 x - 5 = 0 therefore 3 x = 5 and thus x = 5 3 .The axis intersects are: ( 5 3 , 0 ) and ( 0 , - 5 ) .

b

x - 4 = 0 therefore x = 4 . The axis intersects are: ( 4 , 0 ) and ( 0 , -4 ) .

c

-0.5 x + 4 = 0 therefore 0 . 5 x = 4 and thus x = 8 . The axis intersects are: ( 8 , 0 ) and ( 0 , 4 ) .

d

-2 ( x + 3 ) = 0 therefore x + 3 = 0 and thus x = -3 . The axis intersects are: ( -3 , 0 ) and ( 0 , -6 ) .

Exercise 3
a

y-intersect = ( 0 , -4 ) , and slope 5 .

b

y 2 = 6 + 5 x

c

y 3 = -4 - 5 x

Exercise 4
a

Linear equation y = a x + b through ( 1 , 5 ) gives you 5 = a + b , therefore b = 5 - a . Substitution results in y = a x + 5 - a .

b

If the line goes through A , then a = - 4 and if the line goes through C , then a = - 2 3 . Therefore the line will not intersect the square if a < - 4 a > - 2 3 .

Exercise 5
a

f : 4 - 0.5 x = 0 therefore 0.5 x = 4 and thus x = 8 . The axis intersects are: ( 8 , 0 ) and ( 0 , 4 ) .

g : 2 x - 1 = 0 therefore 2 x = 1 and thus x = 1 2 . The axis intersects are: ( 1 2 , 0 ) and ( 0 , - 1 ) .

b

4 - 0.5 x = 2 x - 1 gives you -2.5 x = -5 and therefore x = 2 . The intersect is ( 2 , 3 ) .

Exercise 6
a

26 km in 45 minutes gives you 34 2 3 km/h.

b

The participant starts the last part of the tour at t = 7 , with a = 174 , and has finished the tour at t = 7 3 4 , with a = 200 .
Therefore a ( t ) = 34 2 3 t + b . Using t = 7 you get 174 = 242 2 3 + b , and thus b = -68 2 3 . The function rule is a ( t ) = 34 2 3 t - 68 2 3 .

c

At t = 9 , a = 174 and at t = 10 , a = 200 . Therefore a ( t ) = 26 t + b and 200 = 260 + b , so b = -60 . The function rule is a ( t ) = 26 t - 60 .

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