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zepelin [54]
3 years ago
11

Pls help due today need ASAP

Mathematics
1 answer:
nevsk [136]3 years ago
3 0
30s dirjdididiekekekekeieiwiwiwkej
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2(c + 3) = c - 13<br> Solve
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A rectangle is reduced by a scale factor of One-fourth.
prohojiy [21]

Answer:

(\frac{4}{16})^2

\frac{12}{192}

(\frac{3}{12})^2

Step-by-step explanation:

we know that

If two figures are similar, the the ratio of its areas is equal to the scale factor squared

In this problem

The scale factor is 1/4

Let

z ---> the scale factor

x ---> the area of the smaller rectangle

y ---> the area of the large rectangle

so

z^2=\frac{x}{y}

we have

z=\frac{1}{4}

substitute

z^2=(\frac{1}{4})^2 =\frac{1}{16}

<u><em>Verify each option</em></u>

a) we have

\frac{4}{16}

Compare with  \frac{1}{16}

so

\frac{4}{16} \neq \frac{1}{16}

This option no show the ratio of the area of the smaller rectangle to the area of the larger rectangle

b) we have

(\frac{4}{16})^2=\frac{16}{256}=\frac{1}{16}

Compare with  \frac{1}{16}

so

\frac{1}{16} = \frac{1}{16}

This option show the ratio of the area of the smaller rectangle to the area of the larger rectangle

c) we have

\frac{12}{192}=\frac{1}{16}

Compare with  \frac{1}{16}

so

\frac{1}{16} = \frac{1}{16}

This option show the ratio of the area of the smaller rectangle to the area of the larger rectangle

d) we have

(\frac{4}{12})^2=\frac{16}{144}=\frac{1}{9}

Compare with  \frac{1}{16}

so

\frac{1}{9} \neq \frac{1}{16}

This option no show the ratio of the area of the smaller rectangle to the area of the larger rectangle

e) we have

(\frac{3}{12})^2=\frac{9}{144}=\frac{1}{16}

Compare with  \frac{1}{16}

so

\frac{1}{16} = \frac{1}{16}

This option show the ratio of the area of the smaller rectangle to the area of the larger rectangle

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