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BlackZzzverrR [31]
3 years ago
6

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

Mathematics
2 answers:
WINSTONCH [101]3 years ago
8 0
<h2>1.</h2>

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

<h2>2.</h2>

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



weeeeeb [17]3 years ago
6 0
The answer to question number 1 is 16.
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