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Formula
Find the approximate value
Answer
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Square root
Answer
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$15 \sqrt{ 2 }$
$\approx 21.2132$
Find the approximate value
$\color{#FF6800}{ 15 } \sqrt{ \color{#FF6800}{ 2 } }$
$ $ Find the approximate value of the square root $ $
$\approx \color{#FF6800}{ 21.2132 }$
$\sqrt{ 450 }$
Change the expression
$\color{#FF6800}{ 15 } \sqrt{ \color{#FF6800}{ 2 } }$
$ $ Put the coefficient into the square root $ $
$\sqrt{ \color{#FF6800}{ 15 } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } }$
$\sqrt{ \color{#FF6800}{ 15 } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } }$
$ $ Simplify the expression $ $
$\sqrt{ \color{#FF6800}{ 225 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } }$
$\sqrt{ \color{#FF6800}{ 225 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } }$
$ $ Multiply $ 225 $ and $ 2$
$\sqrt{ \color{#FF6800}{ 450 } }$
Solution search results
search-thumbnail-$2.$ $cos^{2}45^{°}-sin^{2}15^{°}=$ 
$3.$ $cos^{2}15^{°}-sin^{2}45^{°}=$ 
$4.$ $sin^{2}75^{°}-sin^{2}15^{°}=$ 
$5.$ $cos^{2}15^{°}-cos^{2}75^{°}=$
1st-6th grade
Algebra
search-thumbnail-$2$ $cos^{2}45^{°}-sin^{2}15^{°}=$ 
$3.$ $cos^{2}15^{°}-sin^{2}45^{°}=$ 
$4.$ $sin^{2}75^{°}-sin^{2}15^{°}=$ 
$5.$ $cos^{2}15^{°}-cos^{2}75^{°}=$
7th-9th grade
Trigonometry
search-thumbnail-$cos^{2}15^{°}$ $-sin^{2}15^{°}=7$
10th-13th grade
Trigonometry
search-thumbnail-$\dfrac {1-tan^{2}15^{0}} {1+tan^{2}150}$
10th-13th grade
Other
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