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Find the difference
Answer
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$\dfrac{ 5 }{ 3 } - \dfrac{ 1 }{ 2 }$
$\dfrac { 7 } { 6 }$
Find the difference
$\dfrac { 5 } { \color{#FF6800}{ 3 } } - \dfrac { 1 } { \color{#FF6800}{ 2 } }$
$ $ The smallest common multiple in denominator is $ 6$
$\dfrac { 5 } { \color{#FF6800}{ 3 } } - \dfrac { 1 } { \color{#FF6800}{ 2 } }$
$\dfrac { 5 } { 3 } - \dfrac { 1 } { 2 }$
$ $ Multiply the denominator and the numerator so that the denominator is the smallest common multiple $ $
$\dfrac { 5 \times \color{#FF6800}{ 2 } } { 3 \times \color{#FF6800}{ 2 } } - \dfrac { 1 \times \color{#FF6800}{ 3 } } { 2 \times \color{#FF6800}{ 3 } }$
$\color{#FF6800}{ \dfrac { 5 \times 2 } { 3 \times 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 \times 3 } { 2 \times 3 } }$
$ $ Organize the expression $ $
$\color{#FF6800}{ \dfrac { 10 } { 6 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 6 } }$
$\color{#FF6800}{ \dfrac { 10 } { 6 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 6 } }$
$ $ Since the denominator is the same as $ 6 $ , combine the fractions into one $ $
$\color{#FF6800}{ \dfrac { 10 - 3 } { 6 } }$
$\dfrac { \color{#FF6800}{ 10 } \color{#FF6800}{ - } \color{#FF6800}{ 3 } } { 6 }$
$ $ Subtract $ 3 $ from $ 10$
$\dfrac { \color{#FF6800}{ 7 } } { 6 }$
Solution search results
search-thumbnail-Question $4$ 
If $x$ is $6$ what $|s$ 
$\dfrac {1} {3}x$ 
$1frac\left(1\right)\left(3\right)x$
1st-6th grade
Other
search-thumbnail-$\left(iV\right)$ $\dfrac {19} {21}+\left(\dfrac {2} {3}- \begin{cases} - \\ \dfrac {5} {7}-\left(4\dfrac {1} {2}-\dfrac {1} {3}+\dfrac {1} {4}-\dfrac {1} {6}\right)\right) \end{cases} $
1st-6th grade
Algebra
search-thumbnail-Can you answer this? 
$20$ $25$ 

$18$ 

$\left($ 
$\left(A\right)$ $A\right)2$ 
$21frac\left(5\right)\left(9\right)$ \) 
$\left(B\right)$ $B\right)$ $1\left(211$ 
$\left(C\right)$ $1\left(21$ $21+rac\left(7\right)+9\right)$ \) 
$\left(D\right)$ $1\left(2\right)$ 2\frac{8}{9} 
$ac\left(8\right)\left(9\right)$ \) $9:18PM\sqrt{} $
1st-6th grade
Algebra
search-thumbnail-$11.$ Question $11$ 
Solve the $:$ $folloMlng'$ $0<θ<90^{°}$ 
$\left(1\right)$ $2sin^{2}θ=1\right)$ $\left(rac\left(3\right)\left(2\right)\right)$ 
$\left(11\right)$ $3tan^{2}θ+2=3$ 
$\left(111\right)cos^{2}θ$ $11rac\left(1\right)\left(4\right)\right)=$ 
$c\left(1\right)\left(4\right)\right)=11113c\left(1\right)\left(2\right)\right)$
10th-13th grade
Trigonometry
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