Question

1-58. Mr. Wright was making a table to figure out how much it costs to send a certain numt

Answer 1

A table of values can be used to represent variables that are directly proportional.

The complete table of proportions is:

$\left[\begin{array}{ccccccccc}Letters&10&2&[150 ]&7&1&500&[420] \\Cost&0.45&0.90&6.75&[0.315]&[0.045 ]&[22.5 ] & 18.90\end{array}\right]$

Given that

$\left[\begin{array}{ccccccccc}Letters&10&2&[\ ]&7&1&500&[\ ] \\Cost&0.45&0.90&6.75&[\ ]&[\ ]&[\ ] & 18.90\end{array}\right]$

Let:

$L \to$ Letters

$C \to$ Cost

Using proportional reasoning, we have:

$C = kL$

Where

$k \to$ ratio of proportion

For the first values of C and L, we have:

$C = kL$

$0.45 = k \times 10$

Divide both sides by 10

$k = 0.045$

So, the equation of proportion is:

$C = 0.045L$

When C = 6.75, we have:

$C = 0.045L$

$6.75 = 0.045L$

Solve for L

$L = \frac{6.75}{0.045}$

$L = 150$

When L = 7, we have:

$C = 0.045L$

$C = 0.045 \times 7$

$C = 0.315$

When L = 1, we have:

$C = 0.045L$

$C = 0.045 \times 1$

$C = 0.045$

When L = 500, we have:

$C =0.045L$

$C =0.045 \times 500$

$C =22.5$

When C = 18.90, we have:

$C = 0.045L$

$18.90 = 0.045L$

Solve for L

$L=\frac{18.90}{0.045}$

$L=420$

Hence, the complete table is:

$\left[\begin{array}{ccccccccc}Letters&10&2&[150 ]&7&1&500&[420] \\Cost&0.45&0.90&6.75&[0.315]&[0.045 ]&[22.5 ] & 18.90\end{array}\right]$

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