1.
Answer
A. 16
Explanation
We 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.
Answer
B. 6.25
Explanation
Just 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]
