The problem requires us to simplify the given expression so that it is in the form $K x^a y^b z^c$.

The expression is:

$z^{-4} \left( \frac{7x^2 y^3}{z y^2 \sqrt{5}} \right)^2 \cdot 12^0$

Step-by-Step Simplification:

1. Handle $12^0$: $12^0 = 1$ So the expression simplifies to: $z^{-4} \left( \frac{7x^2 y^3}{z y^2 \sqrt{5}} \right)^2$

2. Square the fraction inside the parentheses: $\left( \frac{7x^2 y^3}{z y^2 \sqrt{5}} \right)^2 = \frac{(7^2)(x^2)^2(y^3)^2}{(z^2)(y^2)^2(\sqrt{5})^2}$ Simplifying: $= \frac{49 x^4 y^6}{z^2 y^4 \cdot 5}$ So the expression becomes: $z^{-4} \cdot \frac{49 x^4 y^6}{5 z^2 y^4}$

3. Combine terms: The expression now is: $\frac{49 x^4 y^6}{5 z^6 y^4}$ Simplify the powers of $y$ and $z$: $= \frac{49 x^4 y^2}{5 z^6}$

Now, we have the expression in the form $K x^a y^b z^c$. Comparing the simplified expression with $K x^a y^b z^c$, we get:

• $K = \frac{49}{5}$
• $a = 4$
• $b = 2$
• $c = -6$

Final Answers:

• $a = 4$
• $b = 2$
• $c = -6$
• $K = \frac{49}{5}$

Let me know if you need further details or have any questions.

Related Questions:

1. How do we simplify powers of variables in fractions?
2. What is the general rule for exponents when squaring a fraction?
3. Why is $12^0$ equal to 1?
4. How do square roots behave when raised to even powers?
5. How do negative exponents affect variable expressions?

Tip: Always simplify powers and constants separately before combining terms!