1. The problem involves evaluating and simplifying expressions with exponents (potencias). 2. Let's clarify and solve the expressions step-by-step: 3. Expression: $x \mid y = M^3$ - This seems to indicate $M$ raised to the power 3, or $M^3$. 4. Expression: $10^0 = (10^0)^2 = 10^0 * 10^0$ - Recall that any number to the zero power is 1, so $10^0 = 1$. - Then $(10^0)^2 = 1^2 = 1$. - Also, $10^0 * 10^0 = 1 * 1 = 1$. 5. Expression: $2^1$ and $2^2$ - $2^1 = 2$ - $2^2 = 4$ 6. Expression: $2 * 2^2$ - Calculate $2^2 = 4$ - Then multiply: $2 * 4 = 8$ 7. Expression: $( ) ( ) ( )$ - This is unclear without content; possibly multiplication of terms. 8. Expression: $3^2, 5^{1}, 2^5, 5^0, 3^2, 7^2, 10^0$ - Calculate each: - $3^2 = 9$ - $5^1 = 5$ - $2^5 = 32$ - $5^0 = 1$ - $3^2 = 9$ - $7^2 = 49$ - $10^0 = 1$ 9. Expression: $4^1, 2^5, 4^1, 2^5, 6^1, 8^1, 8^1$ - Calculate each: - $4^1 = 4$ - $2^5 = 32$ - $6^1 = 6$ - $8^1 = 8$ 10. Regarding the triangle with vertices $a(0,0)$, $b(0,-6)$, and $c(5,-2)$: - This is a coordinate geometry problem, possibly to find lengths or area. - Length $ab = |0 - (-6)| = 6$ - Length $ac = \sqrt{(5-0)^2 + (-2-0)^2} = \sqrt{25 + 4} = \sqrt{29}$ - Length $bc = \sqrt{(5-0)^2 + (-2 + 6)^2} = \sqrt{25 + 16} = \sqrt{41}$ Final answers: - $M^3$ remains as is. - $10^0 = 1$ - $2^1 = 2$, $2^2 = 4$ - $2 * 2^2 = 8$ - Powers calculated as above. - Triangle side lengths: $ab=6$, $ac=\sqrt{29}$, $bc=\sqrt{41}$. This completes the evaluation and explanation of the given powers and the triangle geometry.