A table of values can be used to represent variables that are directly proportional.
The complete table of proportions is:
[tex]\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][/tex]
Given that
[tex]\left[\begin{array}{ccccccccc}Letters&10&2&[\ ]&7&1&500&[\ ] \\Cost&0.45&0.90&6.75&[\ ]&[\ ]&[\ ] & 18.90\end{array}\right][/tex]
Let:
[tex]L \to[/tex] Letters
[tex]C \to[/tex] Cost
Using proportional reasoning, we have:
[tex]C = kL[/tex]
Where
[tex]k \to[/tex] ratio of proportion
For the first values of C and L, we have:
[tex]C = kL[/tex]
[tex]0.45 = k \times 10[/tex]
Divide both sides by 10
[tex]k = 0.045[/tex]
So, the equation of proportion is:
[tex]C = 0.045L[/tex]
When C = 6.75, we have:
[tex]C = 0.045L[/tex]
[tex]6.75 = 0.045L[/tex]
Solve for L
[tex]L = \frac{6.75}{0.045}[/tex]
[tex]L = 150[/tex]
When L = 7, we have:
[tex]C = 0.045L[/tex]
[tex]C = 0.045 \times 7[/tex]
[tex]C = 0.315[/tex]
When L = 1, we have:
[tex]C = 0.045L[/tex]
[tex]C = 0.045 \times 1[/tex]
[tex]C = 0.045[/tex]
When L = 500, we have:
[tex]C =0.045L\\[/tex]
[tex]C =0.045 \times 500[/tex]
[tex]C =22.5[/tex]
When C = 18.90, we have:
[tex]C = 0.045L[/tex]
[tex]18.90 = 0.045L[/tex]
Solve for L
[tex]L=\frac{18.90}{0.045}[/tex]
[tex]L=420[/tex]
Hence, the complete table is:
[tex]\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][/tex]
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