Question #92b07

It is given that y varies inversely with x .

We are also given that: y equals 0.2 when x = 8.

We must remember that for two quantities with inverse variation, as one quantity increases, the other quantity decreases.

We must determine the equation for the inverse variation .

An inverse variation can be represented by the equation #color(green)(y = k/x)#.

That is, y varies inversely as x if there is some nonzero constant k such that, #x xx y=k or y=k/x# where #x!=0,y!=0#

We can solve Inverse variation problems using the equation #color(green)(y = k/x)#.

We use the information given in the problem to find the value of k .

In our problem, we need to find k when x = 8 and y = 0.2.

#rArr 0.2 = k/8#

We will rewrite the above equation as #rArr 0.2/1 = k/8# to help us with simplification.

When we cross-multiply, we get #k = 8 xx 0.2 = 1.6#

Hence, the required equation that describes the inverse variation is given by:

#color(red)(y = 1.6/x)#

I believe this problem solving process is helpful.

#"the initial statement is "yprop1/x#

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

#rArry=kxx1/x=k/x#

#"to find k use the given condition"#

#y=0.2" when "x=8#

#y=k/xrArrk=yx=0.2xx8=1.6#

#"equation is " color(red)(bar(ul(|color(white)(2/2)color(black)(y=1.6/x)color(white)(2/2)|)))#