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Formula
Solve the equation
Answer
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Graph
$y = 2 x + 1$
$y = \dfrac { 1 } { 2 }$
$x$-intercept
$\left ( - \dfrac { 1 } { 2 } , 0 \right )$
$y$-intercept
$\left ( 0 , 1 \right )$
$2x+1= \left( \dfrac{ 1 }{ 2 } \right)$
$x = - \dfrac { 1 } { 4 }$
$ $ Solve a solution to $ x$
$2 x \color{#FF6800}{ + } \color{#FF6800}{ 1 } = \dfrac { 1 } { 2 }$
$ $ Move the constant to the right side and change the sign $ $
$2 x = \color{#FF6800}{ \dfrac { 1 } { 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 1 }$
$2 x = \color{#FF6800}{ \dfrac { 1 } { 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 1 }$
$ $ Subtract $ 1 $ from $ \dfrac { 1 } { 2 }$
$2 x = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 2 } }$
$\color{#FF6800}{ 2 } \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 2 } }$
$ $ Divide both sides by the same number $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 4 } }$
$ $ 그래프 보기 $ $
Graph
Solution search results
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$s1S$ 
$S-1S$ 
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7th-9th grade
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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
search-thumbnail-Determine Value $f _{ 1 } t$ $ - \quad $ such that 
$ | A | = \left. \begin{array} { cc } { t - 2 } & { 4 } & { 4 } \\ { 1 } & { 0 + 1 } & { - 2 } \\ { 0 } & { 0 } & { - 4 } \end{array} \right| = 0$
1st-6th grade
Algebra
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