Respuesta :
Answer:
[tex]L_2 = 2\frac{2}{15}\ ft[/tex]
Step-by-step explanation:
Given
Original
[tex]Length(L_1) = 8\ in[/tex]
[tex]Height (H_1) = 15\ in[/tex]
New
[tex]Height (H_2) = 4\ ft[/tex]
Required
Determine the new length (L2)
Since the original photographs was blown up; the original and the new photographs would have the same ratio
i.e.
[tex]Ratio = L_1 : H_1[/tex]
and
[tex]Ratio = L_2 : H_2[/tex]
Equate both ratios
[tex]L_1 : H_1 = L_2 : H_2[/tex]
Rewrite as fraction
[tex]\frac{L_1}{H_1} = \frac{L_2}{H_2}[/tex]
Substitute values for L1, H1 and H2
[tex]\frac{8}{15} = \frac{L_2}{4}[/tex]
Multiply both sides by 4
[tex]4 * \frac{8}{15} = \frac{L_2}{4} * 4[/tex]
[tex]4 * \frac{8}{15} = L_2[/tex]
[tex]\frac{4 * 8}{15} = L_2[/tex]
[tex]\frac{32}{15} = L_2[/tex]
[tex]2\frac{2}{15} = L_2[/tex]
[tex]L_2 = 2\frac{2}{15}\ ft[/tex]
The new length must be [tex]2\frac{2}{15}\ ft[/tex]
The new picture must be [tex]2\frac{2}{15}[/tex] feet or 2.1334 feet in length.
We know, [tex]1\ inch= \frac{1}{12}\ feet[/tex]
Given to us:
Initial length, [tex]L_i =8 inches[/tex];
Initial height, [tex]H_i =15\ inches[/tex];
Final height, [tex]H_f = 4\ feet[/tex];
We know that a when a picture is blown up(enlarged), it is always done in a specific ratio, therefore
Ratio, [tex]r=\dfrac{Length}{Height}[/tex]
So,
[tex]\dfrac{Initial\ Length}{Initial\ Height} =\dfrac{Final\ Length}{Final\ Height}[/tex],
[tex]\dfrac{8}{15} =\dfrac{Final\ Length}{4}\\\\{Final\ Length}=\dfrac{8}{15} \times 4 \\\\{Final\ Length}=\dfrac{32}{15}\\\\{Final\ Length}=2\frac{2}{15}\ feet[/tex]
Hence, the new picture must be [tex]2\frac{2}{15}[/tex] feet or 2.1334 feet in length.
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