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

Answer:

$\dfrac{7p^{15}}{3q^{12}}$

Step-by-step explanation:

We want to determine which expression is equivalent to:

$\dfrac{28p^{9}q^{-5} }{12p^{-6}q^7} , p\neq0, q\neq 0$

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

$\dfrac{28p^{9}q^{-5} }{12p^{-6}q^7} =\dfrac{28 }{12} X\dfrac{p^{9} }{p^{-6}} X\dfrac{q^{-5}}{q^7}$

Next, we apply the division law of indices: $\dfrac{a^x}{a^y}=a^{x-y}$

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

Therefore:

$\dfrac{28p^{9}q^{-5} }{12p^{-6}q^7}=\dfrac{7p^{15}}{3q^{12}}$

Answer 2

The expression that is equivalent to (28p^(9)•q^(-5))/(12p^(6)•q^(7)) is; 7p³/q^(12)

We want to find the fraction that is equivalent to;

(28p^(9)•q^(-5))/(12p^(6)•q^(7))

Applying laws of exponents we know that;

x⁴/x³ = x^(4 - 3)

Thus in our question;

(28p^(9)•q^(-5))/(12p^(6)•q^(7)) gives;

(28/12) × p^(9 - 6) × q^(-5 - 7)

>> (7/3) × p³ × q^(-12)

>> 7p³/q^(12)

Read more about rules of exponents at; https://brainly.com/question/11388301