Respuesta :
-3p^8/4q^3
Here's why:
We started with 15p^-4q^-6/-20p^-12q^-3
Knowing that p^-2=1/p^2 we get to this: -15p^12q^3/20p^4q^6
Then simplifying both the numerator and the denominator, we get -15x^8/20y^3 which equals to our final solution mentioned above :)
Here's why:
We started with 15p^-4q^-6/-20p^-12q^-3
Knowing that p^-2=1/p^2 we get to this: -15p^12q^3/20p^4q^6
Then simplifying both the numerator and the denominator, we get -15x^8/20y^3 which equals to our final solution mentioned above :)
Answer: The required quotient is [tex]-\dfrac{3}{4}p^8q^{-3}.[/tex]
Step-by-step explanation: We are given to find the following quotient :
[tex]Q=\dfrac{15p^{-4}q^{-6}}{-20p^{-12}q^{-3}}~~~~~~~~~~~~~~~~~~~~~~~~~~(i)[/tex]
We will be using the following property of exponents :
[tex]\dfrac{x^a}{x^b}=x^{a-b}.[/tex]
From (i), we have
[tex]Q\\\\\\=\dfrac{15p^{-4}q^{-6}}{-20p^{-12}q^{-3}}\\\\\\=-\dfrac{3}{4}p^{-4-(-12)}q^{-6-(-3)}\\\\\\=-\dfrac{3}{4}p^{-4+12}q^{-6+3}\\\\\\=-\dfrac{3}{4}p^8q^{-3}.[/tex]
Thus, the required quotient is [tex]-\dfrac{3}{4}p^8q^{-3}.[/tex]
