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3 years ago

The mean amount of money withdrawn from

Mathematics

2 answers:

larisa [96]3 years ago

8 0

Answer:

I really don't know

Step-by-step explanation:

slavikrds [6]3 years ago

7 0

Answer:

i really dont kno ima lil slow so yea i can't help .-.

Step-by-step explanation:

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Michael and 3 friends went to manny's pizza for lunch and their meal cost $32. If they left $4.80 for a tip what percent of thei

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15%

Step-by-step explanation:

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3 years ago

At a restaurant, the bill comes to$90. if you decide to leave a 19% tip, how much is the tip?

vodka [1.7K]

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$17.1

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4 years ago

Read 2 more answers

In the above quadrilateral ABCD,

MakcuM [25]

Answer:

The length of segment $\overline{DO}$ is $\displaystyle \frac{24}{5}$ .

Step-by-step explanation:

Start by verifying that $\triangle DAB \sim \triangle MEB$ , and that $\triangle DOA \sim \triangle BOC$ .

Both $\triangle DAB$ and $\triangle MEB$ contain the angle $\angle DBA$ . Hence, another pair of equal angles will be sufficient to prove $\triangle DAB \sim \triangle MEB$ by the angle-angle similarity postulate.

Notice that because $\overline{AD} \; \| \; \overline{EF}$ , the corresponding angles next to those two lines would be equal: $\angle MEB = \angle DAB$ . Both $\angle MEB$ is in $\rm \triangle MEB$ whereas $\angle DAB$ is in $\triangle DAB$ .

Hence, $\triangle DAB \sim \triangle MEB$ by the angle-angle similarity postulate.

Notice how $\angle DOA$ in $\triangle DOA$ and $\angle BOC$ in $\triangle BOC$ are two vertically opposite angles (and are thus equal to one another.) That is: $\angle DOA = \angle BOC$ .

Because $\overline{AD} \| \overline{BC}$ , the alternate interior angles $\angle DAO$ and $\angle BCO$ are also equal to one another. That is: $\angle DAO = \angle BCO$

Hence, $\triangle DOA \sim \triangle BOC$ , also by the angle-angle similarity postulate.

Notice that $\triangle DAB \sim \triangle MEB$ implies that: $\begin{aligned} & \frac{\left| \overline{BM} \right|}{\left| \overline{BD} \right|} \\ &= \frac{\left|\overline{EB}\right|}{\left|\overline{AB}\right|} && (\triangle MEB \sim \triangle DAB) \\ &= \frac{\left|\overline{EB}\right|}{\left|\overline{AE}\right| + \left|\overline{EB}\right|} \\ &= \frac{\left|\overline{EB}\right|}{2\, \left|\overline{EB}\right| + \left|\overline{EB}\right|} && (\text{given that $\left|\overline{AB}\right| = 2\,\left|\overline{EB}\right|$})\\ &= \frac{1}{3}\end{aligned}$ .

Therefore:

$\begin{aligned}&\left|\overline{BD}\right|\\ &= 3\, \left|\overline{BM}\right| \\ &= 12\end{aligned}$ .

Similarly, $\triangle DOA \sim \triangle BOC$ implies that:

$\begin{aligned} \frac{\left| \overline{DO} \right|}{\left| \overline{BO} \right|} &= \frac{\left|\overline{AD}\right|}{\left|\overline{BC}\right|} && (\triangle DOA \sim \triangle BOC) \\ &= \frac{6}{9} = \frac{2}{3}\end{aligned}$ .

Therefore, $\displaystyle \left|\overline{BO}\right|= \frac{3}{2}\, \left|\overline{DO}\right|$ .

$\begin{aligned} \frac{\left| \overline{DO} \right|}{\left| \overline{BD} \right|} &= \frac{\left| \overline{DO} \right|}{\left| \overline{BO} \right| + \left| \overline{DO} \right|} \\ &= \frac{\left| \overline{DO} \right|}{(3/2)\, \left| \overline{DO} \right| + \left| \overline{DO} \right|} = \frac{2}{5}\end{aligned}$ .