Prove that: (12^13–12^12+12^11)(11^9–11^8+11^7) is divisible by 3, 7, 19, and 37. The answer should be like: x^b*3*7*19*37. Also this is my first time asking a question on :)

Step-by-step explanation:

Given (12^13–12^12+12^11)(11^9–11^8+11^7),

(12^13–12^12+12^11)(11^9–11^8+11^7) =

[(12^12)12 – 12^12 + 12^11][(11^8)11 – 11^8 + 11^7)

[(12^12)(12 – 1) + 12^11][(11^8)(11 – 1) + 11^7] =

(12^12(11) + 12^11)(11^8(10) + 11^7) =

(12^11(12x11) + 12^11)(11^7(11x10) + 11^7) =

[(12^11)(12x11 + 1)][(11^7)(11x10 + 1)] =

[(12^11)x(11^7)](12x11 + 1)(11x10 + 1) =

[(12^11)x(11^7)](133 x 111) =

[(12^11)x(11^7)](133 x 111) =

[(12^11)x(11^7)](14763) =

[(12^11)x(11^7)](3x7x19x37)

From here, it is clear that the given number is divisible by 3, 7, 19 and 37.

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.

87%

step-by-step explanation:

apex answer