A block is set in motion hanging from a spring and oscillates about its resting position
x = 0
according to the function
x = 0.6sin(2t)+0.4cos(2t),
where x is in centimeters and t is in seconds. For what values of t in the interval [0,3] is the block at its resting position
x =0?

t = 1.277 sec and t = 2.848 sec

Step-by-step explanation:

This problem is much more easily done by graphing it than by computing it using algebra.

The values of t we're looking for are the ones that make x = 0, so we want the solutions of on the interval [0, 3].

According to the graph, this is true when t = 1.277 seconds and t = 2.848 seconds.

t = 1.28, 2.85

Step-by-step explanation:

x = 0.6sin(2t)+0.4cos(2t)

x = 0, so

0.6sin(2t)+0.4cos(2t) = 0

0.6sin(2t) = -0.4cos(2t)

Convert into tan(2t) by dividing both sides by 0.6cos(2t):

tan(2t) = -0.4/0.6

tan(2t) =-2/3

Since t lies in [0,3] ,

2t lies in [0,6]

tan(2t) =-2/3

Basic angle:

0.5880026035

Tan is negative is second and fourth quadrants

2t = 2.553590005, 5.695182704

t = 1.276795025, 2.847591352

a) one student is chosen from classroom a, then a student is chosen from classroom b.  

step-by-step explanation:

independent events are ones in which the probability of one event does not affect the probability of the second event.

one student is chosen from classroom a; this does not affect the chances of choosing a student from classroom b.   this makes these independent events.

when one card is chosen from a standard deck and then set aside before a second card is chosen, the probability of the second card being chosen changes.   this makes them dependent events.

when one student is chosen from classroom a and then a second student is chosen from classroom a, the probability of choosing the second student changes.   this makes them dependent events.

when a letter is chosen from the alphabet and then a second letter is chosen, the probability of choosing the second letter changes.   this makes them dependent events.

because they could drop even lower..