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jeka94
3 years ago
7

How do you solve 6/15=2/c

Mathematics
2 answers:
madreJ [45]3 years ago
7 0
You cross multiply so you get 15 x 2= 6 x c
this is 30= 6c. if you divide both sides by 6 you get c =5
Anna11 [10]3 years ago
6 0
Solve

\mathsf{\dfrac{6}{15}=\dfrac{2}{c}}


Multiplying both sides by 15:

\mathsf{\diagup\!\!\!\!\! 15\cdot \dfrac{6}{\diagup\!\!\!\!\! 15}=15\cdot \dfrac{2}{c}}\\\\\\ \mathsf{6=\dfrac{30}{c}}


Since \mathsf{c} is a denominator of a fraction, it can't be zero. Then, you can also multiply both sides by \mathsf{c}:

\mathsf{6\cdot c=\dfrac{30}{\diagup\!\!\!\! c}\cdot \diagup\!\!\!\! c}\\\\\\ \mathsf{6\cdot c=30}\\\\\\ \mathsf{c=\dfrac{30}{6}}\\\\\\ \boxed{\begin{array}{l} \mathsf{c=5} \end{array}}\qquad\checkmark


Cross-multiplication works fine as well and you would get to the same result.


I hope this helps. :-)

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ANSWER
{a}^{3}  + {b}^{3} = (a + b)( {a}^{2 }  - ab +  {b}^{2} )


EXPLANATION

To find the expression that is equivalent to
{a}^{3}  + {b}^{3}
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So let's get started.


{(a + b)}^{3}  = (a + b) {(a + b)}^{2}

We expand the parenthesis on the right hand side to get,



{(a + b)}^{3}  = (a + b) ( {a}^{2} + 2ab +  {b}^{2}  )



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Let us group the cubed terms on the right hand side to get,

{(a + b)}^{3}  =  {a}^{3}   +  {b}^{3}  + 3 {a}^{2}b + 3a {b}^{2}




{(a + b)}^{3}  =  {a}^{3}   +  {b}^{3}  + 3ab (a+ b)





We make the cubed terms the subject,

{(a + b)}^{3}  - 3ab (a+ b) =  {a}^{3}   +  {b}^{3}

We factor to get,


(a + b)({(a + b)}^{2}  - 3ab ) =  {a}^{3}   +  {b}^{3}


We expand the bracket on the left hand side to get,

(a + b)( {a}^{2}  + 2ab +  {b}^{2}   - 3ab ) =  {a}^{3}   +  {b}^{3}


We finally simplify to get,

(a + b)( {a}^{2}   - ab +  {b}^{2}  ) =  {a}^{3}   +  {b}^{3}
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