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Aleks [24]
3 years ago
9

a town park has 2 biking trails. Marsh Trail is 5 and three over ten miles long. Woodland trail is 2 and 4 over 5 miles long. Ho

w much longer is Marsh Trail than Woodland trail?
Mathematics
1 answer:
Allisa [31]3 years ago
5 0

Marsh Trail is longer than Woodland trail by 2.5 miles

Step-by-step explanation:

A town park has 2 biking trails

1. Marsh Trail is 5 and three over ten miles long

2. Woodland trail is 2 and 4 over 5 miles long

We need to find that Marsh Trail is how much longer than

Woodland trail

To do that we will subtract the length of the Woodland trail from the

length of the Marsh Trail

∵ The length of Marsh Trail is 5 and three over ten miles

- 5 and three over ten = 5\frac{3}{10}

∵ \frac{3}{10} = 0.3

∴ 5\frac{3}{10} = 5.3 miles

∵ The length of Woodland trail is 2 and 4 over 5 miles

- 2 and 4 over 5 = 2\frac{4}{5}

∵ \frac{4}{5}=\frac{4*2}{5*2}=\frac{8}{10} = 0.8

∴ 2\frac{4}{5} = 2.8

∴ The difference = 5.3 - 2.8

∴ The difference = 2.5 miles

Marsh Trail is longer than Woodland trail by 2.5 miles

Learn more:

You can learn more about fractions and decimals in brainly.com/question/1648978

#LearnwithBrainly

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8 0
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Find f.<br><br> A) 7.4<br><br> B) 8.2<br><br> C) 10.5<br><br> D) 11.1
tatyana61 [14]
F=72

g=6

------------

\cos { \left( F \right)  } =\frac { { e }^{ 2 }+{ g }^{ 2 }-{ f }^{ 2 } }{ 2eg }

Therefore:

\cos { \left( 72 \right)  } =\frac { { e }^{ 2 }+{ 6 }^{ 2 }-{ f }^{ 2 } }{ 2\cdot e\cdot 6 } \\ \\ \cos { \left( 72 \right)  } =\frac { { e }^{ 2 }+36-{ f }^{ 2 } }{ 12e }

\\ \\ 12e\cdot \cos { \left( 72 \right)  } ={ e }^{ 2 }+36-{ f }^{ 2 }\\ \\ \therefore \quad { f }^{ 2 }={ e }^{ 2 }-12e\cdot \cos { \left( 72 \right)  } +36\\ \\ \therefore \quad f=\sqrt { { e }^{ 2 }-12e\cdot \cos { \left( 72 \right) +36 }  } \\ \\ \therefore \quad f=\sqrt { e\left( e-12\cos { \left( 72 \right)  }  \right) +36 }

But what is e?

E=76

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And:

\frac { e }{ \sin { \left( E \right)  }  } =\frac { g }{ \sin { \left( G \right)  }  }

Which means that:

\frac { e }{ \sin { \left( 76 \right)  }  } =\frac { 6 }{ \sin { \left( 32 \right)  }  } \\ \\ \therefore \quad e=\frac { 6\cdot \sin { \left( 76 \right)  }  }{ \sin { \left( 32 \right)  }  }

If you take this value into account, you will discover that f is...

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So I would have to say that the answer is approximately (c).
6 0
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