Answer:
a) Average velocity at 0.1 s is 696 ft/s.
b) Average velocity at 0.01 s is 7536 ft/s.
c) Average velocity at 0.001 s is 75936 ft/s.
Step-by-step explanation:
Given : If a ball is thrown straight up into the air with an initial velocity of 70 ft/s, its height in feet after t seconds is given by 
.
To find : The average velocity for the time period beginning when t = 2 and lasting. a. 0.1 s.
, b. 0.01 s.
, c. 0.001 s.
Solution :
a) The average velocity for the time period beginning when t = 2 and lasting 0.1 s.
b) The average velocity for the time period beginning when t = 2 and lasting 0.01 s.
c) The average velocity for the time period beginning when t = 2 and lasting 0.001 s.
Answer:
Choose f(x) = 11x + 1
Step-by-step explanation:
Note that we will simply plug the value of 2 into the equation for February:
f(2) = 11(2) + 1 = 23
And plug the value of 6 into the equation for June:
f(6) = 11(6) + 1 = 67
Note how the points on the graph seem to match up with these values. If we evaluate following the same style for each:
f(3) = 11(3) + 1 = 34
f(4) = 11(4) + 1 = 45
f(5) = 11(5) + 1 = 56
Note, these values seems to be very close in approximation to the graph points for each month.
The other three functions return values that are just to far away from what are represented in the graph.
Cheers.
It will be negative 2,cause your equation is already in general form but we can't have a negative on a, so we should changed it on their opposite sign, so 2 will be negative
Answer:
4times1/4(y-(-25/4))=(x-(-3/2)) small 2
4p(y-k)=(x-h)small 2
something like that I belive
or just y>xsmall2+3x-4
Answer:
C
Step-by-step explanation:
This is because a liter is 1000 milliliters, so multiplying the amount of liters to the milliliters you would get 5000, you would then divide that by 500 since a bottle would contain that much, giving you 10 bottles
