Linear relations

Exercise 1

$f (x) = \frac{1}{3} x + 3 \frac{1}{3}$ ; $g (x) = \frac{2}{3} x + 1$ ; $h (x) = -2 x + 4$ ; $k (x) = -5 x - 2$

Exercise 2

a

$y = -3 x + 158$

b

$y = 100$

c

$y = 0.5 x + 5$

d

$y = 0$

e

$x = 0$

Exercise 3

Line $P Q$ has equation $y = \frac{4}{3} x - \frac{28}{3}$ and line $R S$ has equation $y = - \frac{1}{2} x + \frac{99}{2}$ .
You can find the point of intersection by solving $\frac{4}{3} x - \frac{28}{3} = - \frac{1}{2} x + \frac{99}{2}$ . This gives you $x = \frac{353}{11} = 32 \frac{1}{11}$ and the point of intersection is therefore $(\frac{353}{11}, \frac{368}{11})$ .

Exercise 4

a

$V (t) = \frac{V (0)}{273} \cdot (t + 273) = V (0) \cdot \frac{1}{273} \cdot (t + 273) = V (0) \cdot (\frac{t}{273} + 1) = V (0) \cdot (1 + \frac{1}{273} t)$

b

$V (0)$ is a constant, so the formula can be written as $V (t) = a \cdot t + b$ .
You have to assume that the pressure remains constant at all temperatures.

The domain is $D = [- 273, \to ⟩$

c

Enter: Y1=1+1/273X. Window: $-300 \leq x \leq 300$ and $-1 \leq y \leq 3$ .

d

$V (20) = 1 + \frac{20}{273} = 1.073$ m³

e

$1.5 = 1 + \frac{1}{273} t$ gives you $t = 136.5$ . Thus the temperature is $136.5$ °C.

Exercise 5

a

$s (0)$ is the distance covered at time $t = 0$ , and $v$ is the speed in m/s.

b

$s (t) = 20 t$ . Enter: Y1=20X. Window: $0 \leq x \leq 50$ and $0 \leq y \leq 1000$ .

c

$s (t) = 400 + 15 t$ , therefore use Y2=400+15X.

d

$20 t = 400 + 15 t$ gives you $t = 80$

Exercise 6

a

The speed at $t = 0$ (initial speed).

b

$v (t) = 40 + 10 t$ and $v (t) = 350$ , therefore $40 + 10 t = 350$ . This gives you $t = 31$ .

c

$v (8) = 40 + a \cdot 8 = 0$ gives you $a = - 5$ m/s².
Therefore $F = m \cdot a = 1000 \cdot -5 = -5000$ Newton. (It is a negative force because it works in the opposite direction of the movement of the object.)

Solutions to the exercises