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cestrela7 [59]
3 years ago
11

Describe the end behavior of the given function f(x)=2/3x-2

Mathematics
1 answer:
Ipatiy [6.2K]3 years ago
4 0

Answer:

The end behavior of f(x)=2/3x-2 is: as x->+ infinity, f(x)->+ infinity

as x->- infinity, f(x)->- infinity

Step-by-step explanation:

When you are asked about the end behavior of a function, look to see where the function is traveling on the graph. For instance, this graph is linear, so you should look to see if the slope is positive or negative. This linear function is positive, so as x is reaching positive infinity the f(x) would also be reaching positive infinity. As x is reaching negative infinity, f(x) would also be reaching negative infinity. The end behavior of a function describes the trend of the graph on the left and right side of the x- axis. (As x approaches negative infinity and as x approaches positive infinity).  

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P+q/2; use p=1, and q=-1
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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

G=32

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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...

f=\sqrt { \frac { 6\cdot \sin { \left( 76 \right)  }  }{ \sin { \left( 32 \right)  }  } \left( \frac { 6\cdot \sin { \left( 76 \right)  }  }{ \sin { \left( 32 \right)  }  } -12\cos { \left( 72 \right)  }  \right) +36 } \\ \\ \therefore \quad f=10.8\quad \left( 1\quad d.p \right)

So I would have to say that the answer is approximately (c).
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Step-by-step explanation:

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