Evaluate ∫ e3x cosh 2x dx a. 1/2e5x + 1/2ex + c b. 1/10e5x + 1/5 x + c c. 1/10e5x + 1/2ex + c d. 1/4e3x + 1/2 x + c

∫ e^(3x)*(cosh(2x)dx

= ∫ [e^(3x)*(e^(2x)+e^(-2x))/2]dx

= ∫ [(e^(5x)+e^x)/2]dx

=e^(5x)/10+e^x/2+C

=(1/10)(e^(5x)+5e^x)+C

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.

question 1 is 8, question 2 is 9, question 3 is 24

step-by-step explanation:

now, the numbers they give you at the top are the key. from there, you just have to plug and chug! let's try 1, 2, and 3!

key: a=3, b=5, c=6

1. a+5

remember, a=3, so we plug that in to get 3+5=8

2. 15-c

c=6, so we plug that in to get 15-6=9

3. 4b

b=6, so we plug it in to get 4*6=24

time to try the rest on your own! if you put the answers below in a comment, i'll check them for you!