Answer:
[tex] \dfrac{1}{v^3} [/tex]
Step-by-step explanation:
To simplify [tex]v^{-5} \cdot v^2[/tex], we can use the properties of exponents.
Recall the rule:
[tex] \Large\boxed{\boxed{v^m \cdot v^n = v^{m+n}}} [/tex]
Applying this rule to [tex]v^{-5} \cdot v^2[/tex], we add the exponents:
[tex] v^{-5} \cdot v^2 = v^{-5+2} = v^{-3} [/tex]
Now, [tex]v^{-3}[/tex] represents [tex]\dfrac{1}{v^3}[/tex]
(since [tex]v^{-3} = \dfrac{1}{v^3}[/tex])
Therefore,
[tex] v^{-5} \cdot v^2 = \dfrac{1}{v^3} [/tex]
So, the simplified form with positive exponents is:
[tex]\large\boxed{\boxed{ \dfrac{1}{v^3}}} [/tex]
