What is the scale factor of figure ABCD?

Mathematics

1 answer:

ivann1987 [24]3 years ago

6 0

We know that
measure ABCD=scale factor*[measure A1B1C1D1]

<span>in the axis of the y coordinate
we have
scale factor=measure ABCD/measure A1B1C1D1
</span>scale factor=3/1----->3

in the axis of the x coordinate
we have
scale factor=measure ABCD/measure A1B1C1D1
scale factor=8/6----->4/3

therefore

the answer is the option
<span>d. the figure isn't dilated properly</span>

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A machine at a bottling company fills water bottles. the number of bottles filled is proportional to the amount of time the mach

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In the table shown below:

Let N be the number of bottles filled,

Let T be the time in hours.

Given that the number of bottles filled is proportional to the amount of time the machine runs, we have

$\begin{gathered} N\propto T \\ \text{Introducing a proportionality constant, we have } \\ N\text{ = kT} \\ \Rightarrow k\text{ = }\frac{N}{T} \\ \end{gathered}$

Let's evaluate the value of k for each day.

Thus, on monday,

$\begin{gathered} N\text{ = 9900} \\ T\text{ = 5.5} \\ \text{thus,} \\ k\text{ = }\frac{9900}{5.5} \\ \Rightarrow k\text{ = 1800} \end{gathered}$

Tuesday:

$\begin{gathered} N\text{ = }11160 \\ T\text{ = }6.2 \\ \text{thus,} \\ k\text{ = }\frac{11160}{6.2} \\ \Rightarrow k\text{ = 1800} \end{gathered}$

Wednesday:

$\begin{gathered} N\text{ = }12330 \\ T\text{ = }6.25 \\ \text{thus,} \\ k\text{ = }\frac{12330}{6.25} \\ \Rightarrow k\text{ = 1972.8} \end{gathered}$

Thursday:

$\begin{gathered} N\text{ = }10440 \\ T\text{ = }5.80 \\ \text{thus,} \\ k\text{ = }\frac{10440}{5.8} \\ \Rightarrow k=1800 \end{gathered}$

It is observed that all exept wednesday have the same value of k.

Thus, the amount of time required for the number of bottles filled on wednesday is evaluated as

$\begin{gathered} N\text{ = }12330 \\ k\text{ = 1800} \\ T\text{ = ?} \\ \text{but} \\ k\text{ = }\frac{N}{T} \\ 1800\text{ = }\frac{12330}{T} \\ \Rightarrow T\text{ = }\frac{\text{12330}}{1800} \\ T\text{ = 6.85} \end{gathered}$

Hence, the incorrect day is Wednesday. The amount of time for that many bottles should be 6.85 hours.