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Crazy boy [7]
3 years ago
8

Earth travels 30 kilometers each second as it revolves around the Sun. How many miles does Earth travel in 1 second? Round to th

e hundredths if necessary.
Mathematics
2 answers:
Degger [83]3 years ago
8 0

Answer:

18.64 if rounded

Step-by-step explanation:

First you need to know how many miles is 1 kilometer. That would be approximately 0.62137 miles. Multiply this by 3 and you would get 1.86411. Now multiply this by 10 to make it 30 kilometers worth of miles and not 3 kilometers. You would get 18.6411 which rounded to the hundredths place would be 18.64.

artcher [175]3 years ago
7 0

9514 1404 393

Answer:

  18.64 miles

Step-by-step explanation:

There are 1.609344 km per mile, so the distance traveled in 1 second is ...

  (30 km)(1 mi)/(1.609344 km) ≈ 18.64 mi

The Earth travels around the Sun at the rate of about 18.64 miles per second.

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When x = -1/3, the value of y is
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4 years ago
Classify ABC by its sides. Then determine whether it is a right triangle.
m_a_m_a [10]

Answer:

∴Given Δ ABC is not a right-angle triangle

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b = BC = 12

c = AC = √45 = 3√5

Step-by-step explanation:

Given vertices are A(3,3) and B(6,9)

            AB = \sqrt{x_{2}-x_{1} )^{2} +(y_{2}-y_{1} )^{2}  }

            AB = \sqrt{(9-3)^{2}+(6-3)^{2}  } = \sqrt{6^{2}+3^{2}  } =\sqrt{45}

Given vertices are  B(6,9) and C( 6,-3)

       B C = \sqrt{x_{2}-x_{1} )^{2} +(y_{2}-y_{1} )^{2}  }

             =  \sqrt{(-3-9)^{2}+(6-6)^{2}  } =\sqrt{12^{2} } = 12

    BC = 12

Given vertices are  A(3,3) and C( 6,-3)

 AC = \sqrt{(6-3)^{2}+(-3-3)^{2}  } = \sqrt{9+36} = \sqrt{45}

AC² = AB²+BC²

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 45  ≠ 189

∴Given Δ ABC is not a right angle triangle

 

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3 years ago
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Answer:

400 sq units.................

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