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3 years ago

select the domain and range, then determine if the relation is a function R= {(7,5), (2,0), (1,-1), (-5,-1), (2,6)}

Mathematics

1 answer:

-BARSIC- [3]3 years ago

7 0

Answer:

range =5,0,-1,-1,6

domain=5,2,1,-5,2

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$2\frac { 1 }{ 2 } \quad divided\quad by\quad 5\frac { 1 }{ 3 } \\ \\ 2\frac { 1 }{ 2 } =2+\frac { 1 }{ 2 } =\frac { 5 }{ 2 } \\ \\ 5\frac { 1 }{ 3 } =5+\frac { 1 }{ 3 } =\frac { 15+1 }{ 3 } =\frac { 16 }{ 3 } \\ \\ \\ \frac { \frac { 5 }{ 2 } }{ \frac { 16 }{ 3 } }$

We're dividing fractions, so we will flip the one in the denominator upside down and we will multiply it with the numerator.

$\frac { \frac { 5 }{ 2 } }{ \frac { 16 }{ 3 } } \quad =\quad \frac { 5 }{ 2 } \cdot \frac { 3 }{ 16 } =\frac { 5\cdot 3 }{ 2\cdot 16 } =\frac { 15 }{ 32 }$

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3 years ago

select the domain and range, then determine if the relation is a function R= {(7,5), (2,0), (1,-1), (-5,-1), (2,6)}​

select the domain and range, then determine if the relation is a function R= {(7,5), (2,0), (1,-1), (-5,-1), (2,6)}