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.

yes

step-by-step explanation:

so of course you want to start by plugging the 5 in which goes to 7(5)+3(5)=50 because 7*5=35 and 3*5=15 and then add them together, then plug in the 5 to the other equation as well 9(5)+5=50 because 9*5= 45 and then add 5 and so yes they are equivalent.

the absolute value function