Question

How do you simplify $-\dfrac{5{{c}^{2}}{{d}^{5}}}{8c{{d}^{5}}{{f}^{0}}}$ ?

Answer 1

Hint: Here in this question we have been asked to simplify the given algebraic expression $-\dfrac{5{{c}^{2}}{{d}^{5}}}{8c{{d}^{5}}{{f}^{0}}}$ . For doing that we will perform simple basic algebraic operations and transfer all the variables to the numerator and simplify them.

Complete step-by-step answer:
Now considering from the question we have been asked to simplify the given algebraic expression $-\dfrac{5{{c}^{2}}{{d}^{5}}}{8c{{d}^{5}}{{f}^{0}}}$ .
For doing that we will perform simple basic algebraic operations and transfer all the variables to the numerator and simplify them.
Firstly we will transfer variable $c$ to the numerator from the denominator. By doing that we will have $-\dfrac{5{{c}^{2}}{{d}^{5}}\left( {{c}^{-1}} \right)}{8{{d}^{5}}{{f}^{0}}}$ .
Now we will transfer variable $d$ to the numerator from the denominator. By doing that we will have $-\dfrac{5{{c}^{2}}{{d}^{5}}\left( {{c}^{-1}} \right)\left( {{d}^{-5}} \right)}{8{{f}^{0}}}$ .
We know that any variable having zero as a power will have one as its value. So in this case ${{f}^{0}}=1$ .
By using this we will have $-\dfrac{5{{c}^{2}}{{d}^{5}}\left( {{c}^{-1}} \right)\left( {{d}^{-5}} \right)}{8}$ .
Now by further simplifying this expression we will have $-\dfrac{5{{c}^{2-1}}{{d}^{5-5}}}{8}$ .
Therefore we can conclude that the simplified form of the given algebraic expression $-\dfrac{5{{c}^{2}}{{d}^{5}}}{8c{{d}^{5}}{{f}^{0}}}$ will be given as $-\dfrac{5c}{8}$ .

Note: While answering questions of this type we should be sure with the concepts that we are going to apply and the calculations that we are going to perform in between the steps. Alternatively the given expression can be simplified by considering it as $-\dfrac{5{{c}^{2}}{{d}^{5}}}{8c{{d}^{5}}{{f}^{0}}}=-\dfrac{5{{c}^{2}}{{d}^{5}}}{8c{{d}^{5}}}$ since ${{f}^{0}}=1$ and then $-\dfrac{5{{c}^{2}}}{8c}$ since we have same power for the denominator and numerator of $d$ now we can simplify for $c$ after that we will have $\dfrac{-5c}{8}$ which is same.