1. Suppose that F(x) varies directly with x and f(x) = 160 when x = 5.What is F(x) when x = 0.5? A. 16B. 32C. 160D. 1802. Suppose that H(x) varies inversely with x and H(x) = 50 when x = 0.25.What is H(x) when x = 2?A. 0.5B. 6.25 C. 12.5D. 24

1.AnswerA. 16ExplanationWe know that f(x) varies directly with x, so this is a problem involving proportions. We know that when f(x) = 160 when x = 5 and f(x) = ? when x = 0.5; in other words:[tex]\frac{160---->5}{f(x)---->0.5}[/tex]Now, we can express our proportion as a fraction and solve for f(x):[tex]\frac{160}{f(x)} =\frac{5}{0.5}[/tex][tex]f(x)=\frac{(160)(0.5)}{5}[/tex][tex]f(x)=\frac{80}{5}[/tex][tex]f(x)=16[/tex]We can conclude that [tex]f(x)=16[/tex] when [tex]x=0.5[/tex]2.AnswerB. 6.25 ExplanationJust like before, we have a problem involving proportions, but this time is an inverse proportion since h(x) varies inversely with x. First, lets set up our proportion just like before. We know that when h(x) = 50 when x = 0.25 and h(x) = ? when x = 2; in other words:[tex]\frac{50---->0.25}{h(x)----->2}[/tex]Now we can express our proportion as a fraction:[tex]\frac{50}{h(x)} =\frac{0.25}{2}[/tex]But remember that we are dealing with an inverse proportion here, so before solving for h(x) we need to flip the second fraction:[tex]\frac{50}{h(x)} =\frac{2}{0.25}[/tex][tex]h(x)=\frac{(50)(0.25)}{2}[/tex][tex]h(x)=\frac{12.5}{2}[/tex][tex]h(x)=6.25[/tex]We can conclude that [tex]h(x)=6.25[/tex] when [tex]x=2[/tex]