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Find the difference
Answer
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$\dfrac{ 5 }{ 4 } - \dfrac{ 3 }{ 5 }$
$\dfrac { 13 } { 20 }$
Find the difference
$\dfrac { 5 } { \color{#FF6800}{ 4 } } - \dfrac { 3 } { \color{#FF6800}{ 5 } }$
$ $ The smallest common multiple in denominator is $ 20$
$\dfrac { 5 } { \color{#FF6800}{ 4 } } - \dfrac { 3 } { \color{#FF6800}{ 5 } }$
$\dfrac { 5 } { 4 } - \dfrac { 3 } { 5 }$
$ $ Multiply the denominator and the numerator so that the denominator is the smallest common multiple $ $
$\dfrac { 5 \times \color{#FF6800}{ 5 } } { 4 \times \color{#FF6800}{ 5 } } - \dfrac { 3 \times \color{#FF6800}{ 4 } } { 5 \times \color{#FF6800}{ 4 } }$
$\color{#FF6800}{ \dfrac { 5 \times 5 } { 4 \times 5 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 \times 4 } { 5 \times 4 } }$
$ $ Organize the expression $ $
$\color{#FF6800}{ \dfrac { 25 } { 20 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 12 } { 20 } }$
$\color{#FF6800}{ \dfrac { 25 } { 20 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 12 } { 20 } }$
$ $ Since the denominator is the same as $ 20 $ , combine the fractions into one $ $
$\color{#FF6800}{ \dfrac { 25 - 12 } { 20 } }$
$\dfrac { \color{#FF6800}{ 25 } \color{#FF6800}{ - } \color{#FF6800}{ 12 } } { 20 }$
$ $ Subtract $ 12 $ from $ 25$
$\dfrac { \color{#FF6800}{ 13 } } { 20 }$
Solution search results
search-thumbnail-$\dfrac {10} {3}+\dfrac {3} {5}+\dfrac {11} {4}-\dfrac {1} {g}$ $-\dfrac {5} {4}+\dfrac {2} {5}$ $2$
7th-9th grade
Other
search-thumbnail-$\dfrac {10} {3}+\dfrac {3} {5}+\dfrac {11} {4}-\dfrac {1} {g}$ $-\dfrac {5} {4}+\dfrac {2} {5}$ $2$
7th-9th grade
Other
search-thumbnail-Perform the operation on the following fractions. 
$1$ $\dfrac {1} {2}+\dfrac {3} {2}$ $4$ $\dfrac {10} {13}-\dfrac {5} {13}$ 
$2.$ $\dfrac {5} {4}+\dfrac {9} {4}$ $5.$ $\dfrac {5} {4}-\dfrac {1} {4}$ 
$3.$ $\dfrac {9} {5}+\dfrac {3} {5}$
7th-9th grade
Algebra
search-thumbnail-$\dfrac {2} {9}\times -\dfrac {5} {4}$ $-\dfrac {1} {2y}-\dfrac {3} {4}\times \dfrac {3} {5}$
7th-9th grade
Other
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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