Let $A = \{(\alpha, \beta) \varepsilon R\times R: |\alpha-1|\leq 4$ and $|\beta-5|\leq6\}$ and $B = \{(\alpha, \beta) \in R \times R : 16(\alpha-2)^2+9(\beta-6)^2 \leq 144 \}$. Then
B ⊂ A
Set A is defined as $A = \{(\alpha, \beta) \varepsilon R\times R: |\alpha-1|\leq 4 \text{ and } |\beta-5|\leq6\}$.
Set B is defined as $B = \{(\alpha, \beta) \in R \times R : 16(\alpha-2)^2+9(\beta-6)^2 \leq 144 \}$.
To understand the shape of Set B, we can rewrite the inequality:
$ \frac{16(\alpha-2)^2}{144} + \frac{9(\beta-6)^2}{144} \leq 1 $ $ \frac{(\alpha-2)^2}{9} + \frac{(\beta-6)^2}{16} \leq 1 $This is the standard inequality for an ellipse centered at $(2, 6)$.
We need to determine the relationship between Set A (a rectangle) and Set B (an ellipse).
The rectangle A extends from $\alpha=-3$ to $\alpha=5$ and $\beta=-1$ to $\beta=11$. The ellipse B covers $\alpha$ from -1 to 5 and $\beta$ from 2 to 10. Points in A but outside these ranges of B (e.g., $\alpha=-3$) are not in B. Therefore, A is not a subset of B.
Based on the analysis, Set B is entirely contained within Set A.
Final Answer: The final answer is $\boxed{B \subset A}$
Let S be the set of the first 11 natural numbers. Then the number of elements in $A = \{B \subseteq S : n(B) \geq 2$ and the product of all elements of B is even is ________.
If $ \left(\frac{1}{^{15}C_0} + \frac{1}{^{15}C_1}\right)$ $\left(\frac{1}{^{15}C_1} + \frac{1}{^{15}C_2}\right)$ $\dots$ $ \left(\frac{1}{^{15}C_{12}} + \frac{1}{^{15}C_{13}}\right)$ = $\frac{\alpha^{13}}{^{14}C_0 ^{14}C_1 \dots ^{14}C_{12}} $, then $30\alpha$ is equal to __________.
Let S be the set of the first 11 natural numbers. Then the number of elements in $A = \{B \subseteq S : n(B) \geq 2$ and the product of all elements of B is even is ________.