Step-by-step explanation:
m(t) = 11t + 55
the slope is always the factor of the variable, so here the slope is 11.
since t is the number of hours riding (after the noon rest stop), the slope (11) is actually the mean speed the rider is going (after the noon rest stop) : with every hour riding the cyclist goes another 11 miles, so this speed is 11 mph.
the y-intercept is the m value when t = 0, because here in our example y (the result variable) is renamed m.
it is here 55.
t = 0 means no time riding yet, so 55 is the "starting value" of the function. that means that the cyclist traveled already 55 miles earlier that day before the noon rest stop.
the cyclist can only ride until 6pm. but we don't know how long the noon rest stop is (i.e. when he continues riding after the rest stop).
so I need to make some assumptions :
the mean speed (11 mph) after lunch break, and the y-intercept (55) suggest to me that the cyclist was riding 5 hours before the stop (5×11 = 55), and will also ride 5 hours after the stop (= from 1pm to 6pm).
if this is wrong, and the cyclist starts at e.g. 12pm, then please just adjust the following numbers accordingly.
but here again, I am assuming the cyclist will go from 1pm to 6pm (for 5 hours).
the domain is the interval or set of all valid values of the input variable (here t) : all real numbers in [0 .. 5].
the "timer" starts at 0 right at the end of the noon test stop. and then the cyclist can go max. 5 hours. and I assume any part of an hour is valid, so we use real numbers instead of e.g. whole numbers.
the range is the interval or set of all valid values of the result variable (here m) : all real numbers in [55 .. 110].
for t = 0 we have m = 55. and this goes continuously up with t until t = 5, leading to 11×5 + 55 = 55+55 = 110.
