Asap: a particle is moving with velocity v(t) = t2 – 9t + 18 with distance, s measured in meters, left or right of zero, and t measured in seconds, with t between 0 and 8 seconds inclusive. the position at time t = 0 sec is 1 meter right of zero, that is, s(0) = 1. the average velocity over the interval 0 to 8 seconds the instantaneous velocity and speed at time 5 secs the time interval(s) when the particle is moving right the time interval(s) when the particle is going faster slowing down find the total distance the particle has traveled between 0 and 8 seconds

step-by-step explanation:

the options are kind of confusing. i can tell you what i see going on here when i look at the graph.

the class and exam medians are closely related, and it looks like the low outlier on the exam did pull the median lower. [tex][/tex]

correct choice is b

step-by-step explanation:

when you calculate the average rate of change of a function, you are finding the slope of the secant line between the two points. for the function the average rate of change from point to point can be calculated as

to find the average rate of change from day 2 to day 10, note that at day 2, 8 houses were sold, so a(2,8) and at day 10, 2 houses were sold, so b(10,2). thus,