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zalisa [80]
4 years ago
15

david ran on a straight track. every time he ran 480m, he took a break. After 2 breaks, he still had to run 120 m to complete th

e track. If the track was twice as long as the playground, what was the distance of the playground?
Mathematics
1 answer:
kipiarov [429]4 years ago
4 0

Answer:

The distance of the playground is 540m

Step-by-step explanation:

Here, we are interested in calculating the length of the playground given the information in the question.

Where to attack the question from is the segment that states that he took a break after 480m and also had 2 breaks. Thus, the distance traveled would be 480 * 2 = 960 m

Now, to find the length of the track, we add the distance covered plus the distance uncovered. Mathematically that would be 960m + 120m = 1,080m

In the last part of the question, we are told that the track is twice as long as the playground. This means that the length of the playground or the distance of the playground is 1080/2 = 540 m

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Find the volume of the figure. Round your answer to the nearest tenth if necessary. Use 3.14 for T.
seraphim [82]

Answer:

\boxed{ \rm \: Volume_{(Cylinder)} \approx \: 461.6  \:  {m}^{3} } \rm (rounded \: to \: nearest \: tenth)

Step-by-step explanation:

Given dimensions:

  • Radius of the cylinder = 7 metres
  • Height of the cylinder = 3 metres

Given value of π :

  • π = 3.14

To find:

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Solution:

Here, we'll need to use the formulae of the volume of cylinder,to find it's volume.Its actually like a savior while solving these type of questions.

\pink{\star}\boxed{\rm \: Volume_{(Cylinder)} = \pi{r} {}^{2} h}\pink{\star}

where,

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  • r² = (radius)²
  • h = height

Plug/substitute them onto the formulae,then simplify it using PEMDAS.

  • [We'll substitute the value of π later]

\rm \: Volume_{(Cylinder)} = \pi(7) {}^{2}  \times 3

\rm \: Volume_{(Cylinder)} = \pi(49)(3)

\rm \: Volume_{(Cylinder)} = 147\pi \:

  • Now substitute the value of π.

\rm \: Volume_{(Cylinder)} = 147 \times 3.14

\rm \: Volume_{(Cylinder)} = 461.58 \:  {m}^{3}

\boxed{\rm \: Volume_{(Cylinder)} \approx \: 461.6  \:  {m}^{3}} \rm (rounded \: to \: nearest \: tenth) \:

<u>Hence, we can conclude that:</u>

The volume of the cylinder is approximately

461.6 m³.

\rule{225pt}{2pt}

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2 years ago
What's the answer to 3.75 ✖️4
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Softa [21]
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I hope that this helps you! =D
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3 years ago
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