Question

Which expression is equivalent to StartFraction 28 p Superscript 9 Baseline q Superscript negative 5 Baseline Over 12 p Superscript negative 6 Baseline q Superscript 7 Baseline EndFraction? Assume p not-equals 0, q not-equals 0.

Answer 1

The expression \(\frac{28p^{9}q^{-5}}{12p^{-6}q^{7}}\) simplifies to \(\frac{7p^{3}}{3q^{12}}\) by dividing the coefficients and applying the rule of adding exponents for \(p\) and subtracting them for \(q\).

Step-by-step explanation:

The expression given is \(\frac{28p^{9}q^{-5}}{12p^{-6}q^{7}}\). To simplify this expression, we apply the properties of exponents. For the numerical coefficients (28 and 12), we divide 28 by 12 to get \(\frac{7}{3}\) after simplifying. Next, for the powers of \(p\), we use the rule \(x^{p}x^{q} = x^{(p+q)}\), adding the exponents for \(p\) which are 9 and \(-6\) to get \(p^{3}\). Similarly, for the powers of \(q\), we subtract the exponents for \(q\) which are \(-5\) and 7 to get \(q^{-12}\). Combining these results, we get the simplified expression \(\frac{7p^{3}}{3q^{12}}\).

Answer 2

`(7p^(15))/(3q^(12))`

Step-by-step explanation:

We want to determine which expression is equivalent to:

`(28p^(9)q^(-5) )/(12p^(-6)q^7) , p\\eq0, q\\eq 0`

To simplify, we first let us separate the fraction in terms of the terms.

`(28p^(9)q^(-5) )/(12p^(-6)q^7) =(28 )/(12) X(p^(9) )/(p^(-6)) X(q^(-5))/(q^7)`

Next, we apply the division law of indices:
`(a^x)/(a^y)=a^(x-y)`

`(28 )/(12) Xp^(9-(-6))Xq^(-5-7)\\=(7)/(3)Xp^(15)Xq^(-12)\\$Using negative index law of indices: a^(-n)=(1)/(a^n) \\=(7p^(15))/(3q^(12))`

Therefore:

`(28p^(9)q^(-5) )/(12p^(-6)q^7)=(7p^(15))/(3q^(12))`