If y varies directly as x and inversely as the square of z and y=1/6 when x=20 and z =6, how do you find y when x = 14 and z=5?

1 Answer
Mar 11, 2018

Answer:

#y=21/125#

Explanation:

#"the initial statement is "ypropx/z^2#

#"to convert to an equation multiply by k the constant"#
#"of variation"#

#rArry=kxxx/z^2=(kx)/z^2#

#"to find k use the given condition"#

#y=1/6" when "x=20" and "z=6#

#y=(kx)/z^2rArrk=(yz^2)/x=(1/6xx36)/20=3/10#

#"equation is " color(red)(bar(ul(|color(white)(2/2)color(black)(y=(3x)/(10z^2))color(white)(2/2)|)))#

#"when "x=14" and "z=5" then"#

#y=(3xx14)/(10xx25)=21/125#