Question

1. Suppose that F(x) varies directly with x and f(x) = 160 when x = 5.

Answer 1

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:

$\frac{160---->5}{f(x)---->0.5}$

Now, we can express our proportion as a fraction and solve for f(x):

$\frac{160}{f(x)} =\frac{5}{0.5}$

$f(x)=\frac{(160)(0.5)}{5}$

$f(x)=\frac{80}{5}$

$f(x)=16$

We can conclude that $f(x)=16$ when $x=0.5$

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:

$\frac{50---->0.25}{h(x)----->2}$

Now we can express our proportion as a fraction:

$\frac{50}{h(x)} =\frac{0.25}{2}$

But remember that we are dealing with an inverse proportion here, so before solving for h(x) we need to flip the second fraction:

$\frac{50}{h(x)} =\frac{2}{0.25}$

$h(x)=\frac{(50)(0.25)}{2}$

$h(x)=\frac{12.5}{2}$

$h(x)=6.25$

We can conclude that $h(x)=6.25$ when $x=2$

Answer 2

The answer to question number 1 is 16.