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Formula
Expand the expression
Answer
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Factorize the expression
Answer
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$a \left( 1- \sqrt{ 2 } \right) - \sqrt{ 2 } \left( 4-3 \sqrt{ 2 } \right)$
$\left ( 1 - \sqrt{ 2 } \right ) a - 4 \sqrt{ 2 } + 6$
Organize polynomials
$\color{#FF6800}{ a } \left ( \color{#FF6800}{ 1 } \color{#FF6800}{ - } \sqrt{ \color{#FF6800}{ 2 } } \right ) - \sqrt{ 2 } \left ( 4 - 3 \sqrt{ 2 } \right )$
$ $ Sort the order of variables in the mononomial expression $ $
$\left ( \color{#FF6800}{ 1 } \color{#FF6800}{ - } \sqrt{ \color{#FF6800}{ 2 } } \right ) \color{#FF6800}{ a } - \sqrt{ 2 } \left ( 4 - 3 \sqrt{ 2 } \right )$
$\left ( 1 - \sqrt{ 2 } \right ) a \color{#FF6800}{ - } \sqrt{ \color{#FF6800}{ 2 } } \left ( \color{#FF6800}{ 4 } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \sqrt{ \color{#FF6800}{ 2 } } \right )$
$ $ Calculate the value $ $
$\left ( 1 - \sqrt{ 2 } \right ) a \color{#FF6800}{ - } \color{#FF6800}{ 4 } \sqrt{ \color{#FF6800}{ 2 } } + \color{#FF6800}{ 6 }$
$\left ( 1 - \sqrt{ 2 } \right ) a - 4 \sqrt{ 2 } + 6$
Arrange the expression in the form of factorization..
$\color{#FF6800}{ a } \left ( \color{#FF6800}{ 1 } \color{#FF6800}{ - } \sqrt{ \color{#FF6800}{ 2 } } \right ) \color{#FF6800}{ - } \sqrt{ \color{#FF6800}{ 2 } } \left ( \color{#FF6800}{ 4 } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \sqrt{ \color{#FF6800}{ 2 } } \right )$
$ $ Expand the expression $ $
$\left ( \color{#FF6800}{ 1 } \color{#FF6800}{ - } \sqrt{ \color{#FF6800}{ 2 } } \right ) \color{#FF6800}{ a } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \sqrt{ \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 6 }$
Solution search results
search-thumbnail-$s|ef\left(-1n$ $\left($ }\right)^{50}\ $\right)$ \ | | is\ equal\ to\ $S$ $s1S$ $S-1S$ $s2S$ $s50s$
7th-9th grade
Other
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search-thumbnail-The is the statement of the Mean Value Theorem from your $te\times tb00k$ $Tneoren$ $m4.5$ $Ne8n$ Value Theorem Let f be continuous over $leclose$ interval $\left(a.b\right)aod$ $i$ $eo$ $xd$ $csnlc$ $\left(ab\right)$ $mmd$ exists at least one point cE $\left(a.b\right)$ $si1dhtba$ $f^{'}\left(c\right)=\dfrac {f\left(b\right)-f\left(a\right)} {b-a}$ What is the geometric interpretation of the conclusion of the theorem? O The tangent line to the graph of \(f\left(x\right)\) at $l\left(c1\right)$ is parallel to the secant line connecting \(\left(a,f\left(a\right)\right)\) and \(\left(b,f\left(b\right)\right)\). O The tangent line to the graph of $\left(f\left(lef\left(x\left(right\right)$ at \(c\) is the secant line connecting \(\left(a,f\left(a\right)\right)\) $ana$ \(\left(b,f\left(b\right)\right)\). O The tangent line to the graph $of$ $\left(f\left(leF\left(x\left(right\right)\left(\right)at1\left(c1\right)is$ perpendicular to the secant line connecting \(\left(a,f\left(a\right)\right)\) and A(\left(b,f\left(b\right)\right)\). O The tangent line to the graph of $\left(fleH\left(x\right)nght\right)\right)\right)$ $at$ $\left(c\right)\right)ishoizonta|$ $Tnere$ is more than one tangent line to the graph $0no$ $\left(flcf\left(x\right)night\right)\left($ $at1\left(c1\right)$
Calculus
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search-thumbnail-What is the inverse of $g\left(x\right)=\sqrt{x-1} $ $Og\left(x\right)=\sqrt{x+1} $ $+1$ $g^{'}\left(x\right)=5-$ $y_{9}$ $g\sqrt{ef\left(x\left(nght\right)} =\left(f_{tac\left(x+1\right)0sq\pi \left(x\right)}$ $Og^{'}\left(x\right)=x^{2}+1$ $○g\left(x\right)=x^{2}-1$ « Previous Ne
Other
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search-thumbnail-What is the inverse of $g\left(x\right)=\sqrt{x-1} $ $Og\left(x\right)=\sqrt{x+1} $ $+1$ $g^{'}\left(x\right)=5-$ $y_{9}$ $g\sqrt{ef\left(x\left(nght\right)} =\left(f_{tac\left(x+1\right)0sq\pi \left(x\right)}$ $Og^{'}\left(x\right)=x^{2}+1$ $○g\left(x\right)=x^{2}-1$ « Previous Ne
Other
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search-thumbnail-$8 \times $ $ = $ In $ \dfrac { E } { 8 } $ $ \left. \begin{array} { l } { \dfrac { 1 } { 3 } } \\ { \dfrac { 11 } { 3 } } \end{array} \right. $ $ \left. \begin{array} { l } { \dfrac { 1 } { 1 } } \\ { \dfrac { 1 } { 1 } } \end{array} \right. $ and $ \left. \begin{array} { l } { δ } \\ { 8 } \end{array} \right. $ Find the length of PR. $ \bar { I } $ $0$ $ \bar { u } $ $2$ $ = $ $ \| = $
7th-9th grade
Other
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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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