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iogann1982 [59]
3 years ago
8

Mark all the statements that are true a) the domain for this function is the set {3}

Mathematics
1 answer:
telo118 [61]3 years ago
8 0

Answer:

the answer is 6 statements

Step-by-step explanation:

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Hoochie [10]
1) B
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3) A
8 0
3 years ago
Conjugate/Rational Number?
hram777 [196]

Answer:

1)  \dfrac{2}{\sqrt{5} }  = \dfrac{2 \cdot \sqrt{5} }{5}

2)  -\dfrac{5}{\sqrt{3} } = -\dfrac{5 \cdot \sqrt{3} }{3}

3)  \dfrac{\sqrt{2} + \sqrt{5}  }{\sqrt{10} } =\dfrac{\sqrt{5}  }{5} + \dfrac{ \sqrt{2}  }{2}

4)  \dfrac{3 + \sqrt{2} }{\sqrt{3} } \times \dfrac{\sqrt{3} }{\sqrt{3} } = \sqrt{3} + \dfrac{\sqrt{6} }{3}

5)  \dfrac{\sqrt{3} }{\sqrt{5} + \sqrt{2}  }= \dfrac{\sqrt{15} - \sqrt{6} }{3}

Step-by-step explanation:

The rationalization of the denominator of the surds are found as follows;

1) \dfrac{2}{\sqrt{5} }

\dfrac{2}{\sqrt{5} } \times \dfrac{\sqrt{5} }{\sqrt{5} } = \dfrac{2 \cdot \sqrt{5} }{5}

\dfrac{2}{\sqrt{5} }  = \dfrac{2 \cdot \sqrt{5} }{5}

2) -\dfrac{5}{\sqrt{3} }

-\dfrac{5}{\sqrt{3} } \times \dfrac{\sqrt{3} }{\sqrt{3} } = -\dfrac{5 \cdot \sqrt{3} }{3}

-\dfrac{5}{\sqrt{3} } = -\dfrac{5 \cdot \sqrt{3} }{3}

3) \dfrac{\sqrt{2} + \sqrt{5}  }{\sqrt{10} }

\dfrac{\sqrt{2} + \sqrt{5}  }{\sqrt{10} } \times \dfrac{ \sqrt{10}  }{\sqrt{10} } = \dfrac{\sqrt{20} + \sqrt{50}  }{10 } = \dfrac{2\cdot \sqrt{5} + 5 \cdot \sqrt{2}  }{10} = \dfrac{\sqrt{5}  }{5} + \dfrac{ \sqrt{2}  }{2}

\dfrac{\sqrt{2} + \sqrt{5}  }{\sqrt{10} } =\dfrac{\sqrt{5}  }{5} + \dfrac{ \sqrt{2}  }{2}

4) \dfrac{3 + \sqrt{2} }{\sqrt{3} }

\dfrac{3 + \sqrt{2} }{\sqrt{3} } \times \dfrac{\sqrt{3} }{\sqrt{3} } = \dfrac{3 \cdot \sqrt{3}+\sqrt{6}  }{3 } = \sqrt{3} + \dfrac{\sqrt{6} }{3}

\dfrac{3 + \sqrt{2} }{\sqrt{3} } \times \dfrac{\sqrt{3} }{\sqrt{3} } = \sqrt{3} + \dfrac{\sqrt{6} }{3}

5) \dfrac{\sqrt{3} }{\sqrt{5} + \sqrt{2}  }

\dfrac{\sqrt{3} }{\sqrt{5} + \sqrt{2}  } = \dfrac{\sqrt{5} - \sqrt{2} }{\sqrt{5} - \sqrt{2} }  = \dfrac{\sqrt{15} -\sqrt{6} }{5 - 2} = \dfrac{\sqrt{15} - \sqrt{6} }{3}

\dfrac{\sqrt{3} }{\sqrt{5} + \sqrt{2}  }= \dfrac{\sqrt{15} - \sqrt{6} }{3}

6) \dfrac{\sqrt{7} }{\sqrt{3} - \sqrt{5}  }

\dfrac{\sqrt{7} }{\sqrt{3} - \sqrt{5}  } \times \dfrac{\sqrt{3} + \sqrt{5}}{\sqrt{3} + \sqrt{5}}  = \dfrac{\sqrt{21} + \sqrt{35}}{{3} + {5}} = \dfrac{\sqrt{21} + \sqrt{35}}{8}

\dfrac{\sqrt{7} }{\sqrt{3} - \sqrt{5}  } \times \dfrac{\sqrt{3} + \sqrt{5}}{\sqrt{3} + \sqrt{5}}  =\dfrac{\sqrt{21} + \sqrt{35}}{8}

8 0
3 years ago
Factor 9x^2 -5 . using difference of squares
lawyer [7]
\bf \textit{difference of squares}
\\ \quad \\
(a-b)(a+b) = a^2-b^2\qquad \qquad 
a^2-b^2 = (a-b)(a+b)\\\\
-------------------------------\\\\
9x^2-5\qquad 
\begin{cases}
9x^2=3^2x^2\\
\qquad (3x)^2\\
5=\sqrt{5^2}\\
\qquad (\sqrt{5})^2
\end{cases}\implies
\begin{array}{llll}
(3x)^2-(\sqrt{5})^2
\\\\\\
(3x-\sqrt{5})~(3x+\sqrt{5})
\end{array}
5 0
3 years ago
Which equation could be used to find the cost of each Apple? 4x+3(0.89)=$7.15 , 3x+4(0.89)=$7.15 , $7.15-3x=0.89 , or 4x-3(0.89)
Dmitry_Shevchenko [17]

Step-by-step explanation:

wkskwkkwkwiwwwwsisiiqoqnznqiqw

5 0
3 years ago
The surface area of a cube is 96 in2. What is the volume of the cube?​
Yakvenalex [24]

Answer:

64

Step-by-step explanation:

6 0
3 years ago
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