Respuesta :

From our answer, he raced 1 1/3 (4/3) hours at 90 mph and 2/3 hours at 60 mph

Here, we want to fill the table based on the information presented in the question

We proceed as follows;

We will fill the table, line by line

First line;

a) Rate mph is 90

b) Time at this rate is f hours

Second line;

a) Time is l hours

b) Distance is the product of the rate and 60 mph and the time l which is 60l

So let us set up the system of equations;

a)

[tex]f\text{ + l = 2}[/tex]

b)

[tex]90\text{ F + 60 l = 160}[/tex]

Now, we need to solve this system of equations simultaneously

We have this as;

[tex]\begin{gathered} f\text{ + l = 2} \\ 90f\text{ + 60l = 160} \\ \\ \text{From i;} \\ f\text{ = 2-l} \\ \\ 90\text{ (2-l) + 60l = 160} \\ 180-90l\text{ + 60l = 160} \\ 30l\text{ = 180-60} \\ 30l\text{ = 20} \\ l\text{ = }\frac{20}{30}\text{ = }\frac{2}{3} \\ \\ f\text{ = 2-}\frac{2}{3}\text{ = }\frac{4}{3} \end{gathered}[/tex]

From our answer, he raced 1 1/3 (4/3) hours at 90 mph and 2/3 hours at 60 mph

ACCESS MORE
EDU ACCESS
Universidad de Mexico