The function ​s(t)=t^3-27t+3 gives the distance from a starting point at time t of a particle moving along a line. Find the velocity and acceleration functions. Then find the velocity and acceleration at t=0 and t=3. Assume that time is measured in seconds and distance is measured in centimeters. Velocity will be in centimeters per second​ (cm/sec) and acceleration in centimeters per second per second ​(cm/sec^2​). 1. the velocity function is v(t)=
2.the acceleration function is a(t)=___cm/sec
3.the velocity at t=0 is v(0)=cm/sec
4.the velocity at t=3 is v(3)___cm/sec
5.the acceleration at t=0 is a(0)___cm/sec^2
6.the acceleration at t=3 is a(3)___xm/sec^2

need points to ask question

step-by-step explanation:

option 2 is correct.

step-by-step explanation:

we are given two inequality equation

x+y> 5 and -3x+2y< -2

first we draw the graph of both function using t-table method

equation: x+y> 5

for x=0, y=5

for x=-5, y=0

table:

x   :   0     5

y   :   5     0

test point (0,0) , 0> 5

false

equation: -3x+2y< -2

for x=0, y=-1

for x=2, y=4

table:

x   :   0     2

y   : -1       4

test point (0,0) , 0< -2

true

now we draw the graph. see the attachment for graph.

common region would be solution set.

option 2 is correct.

c

step-by-step explanation:

it would be see because your formula is f(x)=6x, so 6 times 1 is 6 and you go on and the best answer is c.