Se y varia direttamente come x e inversamente come il quadrato di z e se # y = 20 # quando # x = 50 # e # z = 5 # come si trova y quando # x = 3 # e # z = 6 #?
Risposta:
#y=5/6#
Spiegazione:
#"the initial statement is "ypropx/z^2#
#"to convert to an equation multiply by k the constant"#
#"of variation"##rArry=kxx x/z^2=(kx)/z^2#
#"to find k use the given condition"#
#y=20" when "x=50" and "z=5#
#y=(kx)/z^2rArrk=(yz^2)/x=(20xx25)/50=10#
#"equation is " color(red)(bar(ul(|color(white)(2/2)color(black)(y=(10x)/z^2)color(white)(2/2)|))#
#"when "x=3" and "z=6" then"#
#y=(10xx3)/36=30/36=5/6#