Answer :
To determine the dimensions of the original photo, we need to understand that Sylvia used a dilation factor to enlarge the original dimensions to create the final poster dimensions of 24 inches by 32 inches. The dilation factor provided is 0.4, which means the final dimensions are scaled by this factor to obtain the original sizes.
Let's denote the original width and height as [tex]\(w_{\text{original}}\)[/tex] and [tex]\(h_{\text{original}}\)[/tex], respectively.
The relationship between the final and original dimensions with the dilation factor is given by:
[tex]\[ w_{\text{final}} = w_{\text{original}} \times \text{dilation factor} \][/tex]
[tex]\[ h_{\text{final}} = h_{\text{original}} \times \text{dilation factor} \][/tex]
Given:
[tex]\[ w_{\text{final}} = 24 \, \text{inches} \][/tex]
[tex]\[ h_{\text{final}} = 32 \, \text{inches} \][/tex]
[tex]\[ \text{dilation factor} = 0.4 \][/tex]
We can find the original dimensions by rearranging the above equations to solve for [tex]\(w_{\text{original}}\)[/tex] and [tex]\(h_{\text{original}}\)[/tex]:
[tex]\[ w_{\text{original}} = \frac{w_{\text{final}}}{\text{dilation factor}} = \frac{24}{0.4} \][/tex]
[tex]\[ h_{\text{original}} = \frac{h_{\text{final}}}{\text{dilation factor}} = \frac{32}{0.4} \][/tex]
Perform the division:
[tex]\[ w_{\text{original}} = 24 \times \frac{1}{0.4} = 24 \times 2.5 = 9.6 \, \text{inches} \][/tex]
[tex]\[ h_{\text{original}} = 32 \times \frac{1}{0.4} = 32 \times 2.5 = 12.8 \, \text{inches} \][/tex]
Therefore, the dimensions of the original photo are:
[tex]\[ w_{\text{original}} = 9.6 \, \text{inches} \][/tex]
[tex]\[ h_{\text{original}} = 12.8 \, \text{inches} \][/tex]
We can conclude that none of the given multiple-choice options correspond to the dimensions calculated. Hence, the original dimensions are [tex]\(9.6 \times 12.8\)[/tex] inches.
Let's denote the original width and height as [tex]\(w_{\text{original}}\)[/tex] and [tex]\(h_{\text{original}}\)[/tex], respectively.
The relationship between the final and original dimensions with the dilation factor is given by:
[tex]\[ w_{\text{final}} = w_{\text{original}} \times \text{dilation factor} \][/tex]
[tex]\[ h_{\text{final}} = h_{\text{original}} \times \text{dilation factor} \][/tex]
Given:
[tex]\[ w_{\text{final}} = 24 \, \text{inches} \][/tex]
[tex]\[ h_{\text{final}} = 32 \, \text{inches} \][/tex]
[tex]\[ \text{dilation factor} = 0.4 \][/tex]
We can find the original dimensions by rearranging the above equations to solve for [tex]\(w_{\text{original}}\)[/tex] and [tex]\(h_{\text{original}}\)[/tex]:
[tex]\[ w_{\text{original}} = \frac{w_{\text{final}}}{\text{dilation factor}} = \frac{24}{0.4} \][/tex]
[tex]\[ h_{\text{original}} = \frac{h_{\text{final}}}{\text{dilation factor}} = \frac{32}{0.4} \][/tex]
Perform the division:
[tex]\[ w_{\text{original}} = 24 \times \frac{1}{0.4} = 24 \times 2.5 = 9.6 \, \text{inches} \][/tex]
[tex]\[ h_{\text{original}} = 32 \times \frac{1}{0.4} = 32 \times 2.5 = 12.8 \, \text{inches} \][/tex]
Therefore, the dimensions of the original photo are:
[tex]\[ w_{\text{original}} = 9.6 \, \text{inches} \][/tex]
[tex]\[ h_{\text{original}} = 12.8 \, \text{inches} \][/tex]
We can conclude that none of the given multiple-choice options correspond to the dimensions calculated. Hence, the original dimensions are [tex]\(9.6 \times 12.8\)[/tex] inches.
