Sure, let's figure out the dimensions of the original photo step-by-step.
1. Sylvia enlarged her photo to a poster with the dimensions [tex]\( 24 \)[/tex] inches in width and [tex]\( 32 \)[/tex] inches in height. This means that the enlarged poster has dimensions [tex]\( 24 \times 32 \)[/tex] inches.
2. The poster was created using the dilation [tex]\( D_{Q, 4} \)[/tex]. This indicates that every dimension of the original photo was scaled up by a factor of 4. To find the original dimensions, we need to reverse this dilation (essentially scaling down by the same factor).
3. Let's denote the original width of the photo as [tex]\( w \)[/tex] and the original height as [tex]\( h \)[/tex].
4. According to the dilation factor:
[tex]\[ \text{New width} = \text{Original width} \times 4 \][/tex]
[tex]\[ \text{New height} = \text{Original height} \times 4 \][/tex]
5. We know the new dimensions (enlarged poster dimensions), so we can substitute those in:
[tex]\[ 24 = w \times 4 \][/tex]
[tex]\[ 32 = h \times 4 \][/tex]
6. To find the original width [tex]\( w \)[/tex], divide the new width by 4:
[tex]\[ w = \frac{24}{4} = 6 \][/tex]
7. To find the original height [tex]\( h \)[/tex], divide the new height by 4:
[tex]\[ h = \frac{32}{4} = 8 \][/tex]
So, the dimensions of the original photo are [tex]\( 6 \times 8 \)[/tex] inches.
Thus, the answer is:
[tex]\[ \boxed{6 \times 8} \][/tex]
