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Formula
Calculate the value
Answer
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$6 \sqrt{ 2 } + \sqrt{ 75 } - \dfrac{ 6 }{ \sqrt{ 2 } } + \sqrt{ 27 }$
$3 \sqrt{ 2 } + 8 \sqrt{ 3 }$
Calculate the value
$6 \sqrt{ 2 } + \sqrt{ \color{#FF6800}{ 75 } } - \dfrac { 6 } { \sqrt{ 2 } } + \sqrt{ 27 }$
$ $ Organize the part that can be taken out of the radical sign inside the square root symbol $ $
$6 \sqrt{ 2 } + \color{#FF6800}{ 5 } \sqrt{ \color{#FF6800}{ 3 } } - \dfrac { 6 } { \sqrt{ 2 } } + \sqrt{ 27 }$
$6 \sqrt{ 2 } + 5 \sqrt{ 3 } - \color{#FF6800}{ \dfrac { 6 } { \sqrt{ 2 } } } + \sqrt{ 27 }$
$ $ Calculate the expression $ $
$6 \sqrt{ 2 } + 5 \sqrt{ 3 } - \left ( \color{#FF6800}{ 3 } \sqrt{ \color{#FF6800}{ 2 } } \right ) + \sqrt{ 27 }$
$6 \sqrt{ 2 } + 5 \sqrt{ 3 } \color{#FF6800}{ - } \left ( \color{#FF6800}{ 3 } \sqrt{ \color{#FF6800}{ 2 } } \right ) + \sqrt{ 27 }$
$ $ Get rid of unnecessary parentheses $ $
$6 \sqrt{ 2 } + 5 \sqrt{ 3 } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \sqrt{ \color{#FF6800}{ 2 } } + \sqrt{ 27 }$
$6 \sqrt{ 2 } + 5 \sqrt{ 3 } - 3 \sqrt{ 2 } + \sqrt{ \color{#FF6800}{ 27 } }$
$ $ Organize the part that can be taken out of the radical sign inside the square root symbol $ $
$6 \sqrt{ 2 } + 5 \sqrt{ 3 } - 3 \sqrt{ 2 } + \color{#FF6800}{ 3 } \sqrt{ \color{#FF6800}{ 3 } }$
$\color{#FF6800}{ 6 } \sqrt{ \color{#FF6800}{ 2 } } + 5 \sqrt{ 3 } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \sqrt{ \color{#FF6800}{ 2 } } + 3 \sqrt{ 3 }$
$ $ Calculate between similar terms $ $
$\color{#FF6800}{ 3 } \sqrt{ \color{#FF6800}{ 2 } } + 5 \sqrt{ 3 } + 3 \sqrt{ 3 }$
$3 \sqrt{ 2 } + \color{#FF6800}{ 5 } \sqrt{ \color{#FF6800}{ 3 } } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \sqrt{ \color{#FF6800}{ 3 } }$
$ $ Calculate between similar terms $ $
$3 \sqrt{ 2 } + \color{#FF6800}{ 8 } \sqrt{ \color{#FF6800}{ 3 } }$
Solution search results
search-thumbnail-
$\sqrt{75} $ $-\sqrt{2} +\sqrt{27} $
7th-9th grade
Algebra
search-thumbnail-$∠$ 
$5.\sqrt{27} +\sqrt{75} =$ 
$6.\sqrt{75} -\sqrt{27} =$ 
$7$ $7.\sqrt{27} $ $k$ $\sqrt{75} =$ 
$\square $ $\sqrt{75} $ $C$
7th-9th grade
Algebra
search-thumbnail-The rationalizing factor of \sqrt{23} is 
$°$ $Options^{°}$ $0$ 
A 24 
23 
C \sqrt{23} 
D None of these
7th-9th grade
Other
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