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goblinko [34]
4 years ago
13

The length of the boston marathon is 138,435 ft. each stride, or step, by a particular runner is 3ft. long. does the number of s

trides the runner takes fit evenly into the length of the race? explain.
Mathematics
2 answers:
REY [17]4 years ago
6 0
Yes! If you divide 138,435 by 3, the number of strides the runner takes fit evenly into the length of the race. Therefore the answer is 46,145 ft.
inysia [295]4 years ago
5 0

Yes 138,435 can be divided by 3 it equals to 46,145

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What is m∠ABC? Justify your reasoning. Using the addition property of equality, 40 + 15 = 55, so m∠ABC = 55°. Using the subtract
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Answer:

Using the Angle Addition Postulate, 15 + 40 = m∠ABC. So, m∠ABC = 55° after simplifying.

Step-by-step explanation:

The  Angle Addition Postulate is gotten from angles placed side by side to form a resulting angle that is the sum of all the individual angles.

If D is the interior of ABC then, ABC is split into two parts i.e ABD and DBC, then the measure of ABC is the sum of the individual two parts

m∠ABC = m∠ABD + m∠DBC.

Using the Angle Addition Postulate:

m∠ABC = m∠ABD + m∠DBC

m∠ABC = 15 + 40 = 55°

m∠ABC = 55°

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3 years ago
You randomly draw a marble out of a bag that contains 202020 total marbles. 121212 of the marbles in the bag are blue. What is \
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➷ It would be 121212/202020

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3 years ago
Answer the following question below.
mina [271]

30

Explanation:

Angle = 67°

To solve the question, we will use an illustration:

To get the horizontal distance from the base of the skycrapper to the ship, we will apply tangent ratio:

\tan \text{ 67}\degree\text{ = }\frac{opposite}{\text{adjacent}}

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adjacent = horizontal distance from the base of the skycrapper to the ship = x

\begin{gathered} \tan \text{ }67\degree\text{ = }\frac{903}{x} \\ \text{cross multiply:} \\ x(\tan \text{ }67\degree)\text{ = 903} \\ x\text{ = }\frac{903}{\tan 67\degree} \end{gathered}\begin{gathered} x\text{ = }\frac{903}{2.3559} \\ x\text{ = 383.29} \end{gathered}

To the nerest hundredth, the horizontal distance from the base of the skycrapper to the ship is 383.30ft

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