DEBES DESARROLLAR CADA EJERCICIO PASO A PASO: NO DEBES ESCRIBIR SÓLO LOS RESULTADOS. z^3 x^4 z^2 x^3=

m^4 n^5 m^2 n^2=

(a^4 b^3 c^2)/(a^2 b^5 c^3 )=

(m^4 n^3 )^5 (m^5 n^2 )^3=

(x^3/y^4 )^2 (y^2/x^3 )^3=

(a^4/b^3 )^5 (b^4/a^6 )^4=

(n^2/m^4 )^5 (m^2/n^3 )^6 (m/n)^8=

[(a/b^8 )^3 (a^(-2)/b^(-5) )^5 ]^2=

[(x^3/y^(-15) y^4/x^7 )^0 ]^4=

(x^(n+2) )(x^(n+1) )=

(a^(n+3) b^(m-2) )^4 (a^(n-2) b^(m+3) )^5

(z^(n+1)/x^(4-n) )^3 ((x^(5-n) y^(m+3))/z^(n+1) )^2

0.02

step-by-step explanation:

time of 9 192 631 770 periods of the radiation corresponding to the transition between the 2 hyperfine levels of the ground state of the caesium 133 atom at 0 k (but other seconds are sometimes used in astronomy)

3/6 chance

step-by-step explanation:

half of the nubers are odd numbers (1,3,5) which in that case if you would be told to simplify it would be 1/2

aloha~! my name is zalgo and i am here to you out on this amazing day. to answer your question, yes, nonzero numbers can have a greatest common factor and that would be 1. still don't get it? let me considering every number is divisible by 1 and hence every number has 1 as a factor of theirs. so, if 2 nonzero numbers didn't share any gcf (greatest common factor), they would still have a factor in common and that would be 1.

i hope that this ! : p

"stay and stay proud! " - zalgo

(by the way, do you mind marking me as brainliest? i'd greatly appreciate it! mahalo! x3)