How do you solve for #x# in #20/27 = 6/x#?
1 Answer
Mar 23, 2018
Explanation:
#"one way is to use the method of "color(blue)"cross-multiplication"#
#•color(white)(x)a/b=c/drArrad=bc#
#rArr20/27=6/x#
#rArr20x=27xx6#
#rArrx=(27xx6)/20=81/10#
