Let S=left{(x, y) in N times N : 9(x-3)^{2}+1 | vedclass

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(B) The set $S$ represents points $(x, y)$ with $x, y \in N$ (natural numbers) inside or on the ellipse $\frac{(x-3)^2}{16} + \frac{(y-4)^2}{9} \leq 1

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$26$

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(B) The set $S$ represents points $(x, y)$ with $x, y \in N$ (natural numbers) inside or on the ellipse $\frac{(x-3)^2}{16} + \frac{(y-4)^2}{9} \leq 1$.Since $x, y \geq 1$,we test integer values for $x$ and $y$:For $x=1: 9(-2)^2 + 16(y-4)^2 \leq 144$ $\Rightarrow 36 + 16(y-4)^2 \leq 144$ $\Rightarrow 16(y-4)^2 \leq 108$ $\Rightarrow (y-4)^2 \leq 6.75$. Possible $y \in \{1, 2, 3, 4, 5, 6\}$.For $x=2: 9(-1)^2 + 16(y-4)^2 \leq 144$ $\Rightarrow 9 + 16(y-4)^2 \leq 144$ $\Rightarrow 16(y-4)^2 \leq 135$ $\Rightarrow (y-4)^2 \leq 8.43$. Possible $y \in \{1, 2, 3, 4, 5, 6\}$.For $x=3: 9(0)^2 + 16(y-4)^2 \leq 144$ $\Rightarrow 16(y-4)^2 \leq 144$ $\Rightarrow (y-4)^2 \leq 9$. Possible $y \in \{1, 2, 3, 4, 5, 6, 7\}$.For $x=4: 9(1)^2 + 16(y-4)^2 \leq 144$ $\Rightarrow 9 + 16(y-4)^2 \leq 144$ $\Rightarrow (y-4)^2 \leq 8.43$. Possible $y \in \{1, 2, 3, 4, 5, 6\}$.For $x=5: 9(2)^2 + 16(y-4)^2 \leq 144$ $\Rightarrow 36 + 16(y-4)^2 \leq 144$ $\Rightarrow (y-4)^2 \leq 6.75$. Possible $y \in \{1, 2, 3, 4, 5, 6\}$.Now check which of these points satisfy $T: (x-7)^2 + (y-4)^2 \leq 36$.Points $(x, y)$ in $S$ are: $(1, 1..6), (2, 1..6), (3, 1..7), (4, 1..6), (5, 1..6)$.Checking $(x-7)^2 + (y-4)^2 \leq 36$ for these points:For $x=1: (-6)^2 + (y-4)^2 \leq 36$ $\Rightarrow 36 + (y-4)^2 \leq 36$ $\Rightarrow (y-4)^2 \leq 0$ $\Rightarrow y=4$. Point $(1, 4)$.For $x=2: (-5)^2 + (y-4)^2 \leq 36$ $\Rightarrow 25 + (y-4)^2 \leq 36$ $\Rightarrow (y-4)^2 \leq 11$ $\Rightarrow y \in \{1, 2, 3, 4, 5, 6, 7\}$. Intersection with $S$ gives $y \in \{1, 2, 3, 4, 5, 6\}$. ($6$ points)For $x=3: (-4)^2 + (y-4)^2 \leq 36$ $\Rightarrow 16 + (y-4)^2 \leq 36$ $\Rightarrow (y-4)^2 \leq 20$ $\Rightarrow y \in \{0..8\}$. Intersection with $S$ gives $y \in \{1, 2, 3, 4, 5, 6, 7\}$. ($7$ points)For $x=4: (-3)^2 + (y-4)^2 \leq 36$ $\Rightarrow 9 + (y-4)^2 \leq 36$ $\Rightarrow (y-4)^2 \leq 27$ $\Rightarrow y \in \{-1..9\}$. Intersection with $S$ gives $y \in \{1, 2, 3, 4, 5, 6\}$. ($6$ points)For $x=5: (-2)^2 + (y-4)^2 \leq 36$ $\Rightarrow 4 + (y-4)^2 \leq 36$ $\Rightarrow (y-4)^2 \leq 32$ $\Rightarrow y \in \{-1..9\}$. Intersection with $S$ gives $y \in \{1, 2, 3, 4, 5, 6\}$. ($6$ points)Total points $= 1 + 6 + 7 + 6 + 6 = 26$.

Let $S=\left\{(x, y) \in N \times N : 9(x-3)^{2}+16(y-4)^{2} \leq 144\right\}$ and $T=\left\{(x, y) \in R \times R :(x-7)^{2}+(y-4)^{2} \leq 36\right\}$. Then $n(S \cap T)$ is equal to $......$

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