Respuesta :
Answer:
[tex]f(d)=\left\{ \begin{matrix} 11 & \text{for} & 0< d\leq 4 \\-3+3.5d & \text{for}& d > 4 \end[/tex]
Step-by-step explanation:
Cost of first 4 kilometers ride = 11 pesos.
So, for the first 4 km, the fair is constant i.e.
for 1 km the fare is 11 pesos,
for 2 km the fair is also 11 pesos,
similarly, for 3 km of 4 km, the fare is 11 pesos.
Hence, for the distance [tex]0\geq d\geq 4[/tex], the fair function,
[tex]f(d)=11\cdots(i)[/tex]
After 4 km, there is an increment of $ 3.50 for each kilometer.
So, the fare function up to 5 kilometers,
[tex]f(d)=11+3.5=14.5[/tex]
So, the fare function up to 6 kilometers, i.e for the distance [tex]5<d\leq6[/tex],
[tex]f(d)=11+2\times3.5=18[/tex]
This can be arranged as the fare function up to 6 km,
[tex]f(d)=11+(6-4)\times3.5=18[/tex]
Similarly, the fare function up to [tex]d[/tex] kilometer [tex](n>4)[/tex],
[tex]f(d)=11+(d-4)\times3.5[/tex]
[tex]\Rightarrrow f(d)=-3+3.5d\cdots(ii)[/tex]
Hence, from equations (i) and (ii),
[tex]f(d)=\left\{ \begin{matrix} 11 & \text{for}\; 0< d \leq 4 \\-3+3.5d & \text{for}\; d > 4 \end[/tex]
