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(B) To find the intersection points,set the equations equal: $10 - x^2 = 2 + x^2$. $2x^2 = 8 \implies x^2 = 4 \implies x = \pm 2$. For $x = 2$,$y = 2

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$\frac{8}{15}$

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(B) To find the intersection points,set the equations equal: $10 - x^2 = 2 + x^2$. $2x^2 = 8 \implies x^2 = 4 \implies x = \pm 2$. For $x = 2$,$y = 2 + (2)^2 = 6$. So,the intersection point is $(2, 6)$. Find the slopes of the tangents at $(2, 6)$: For $y = 10 - x^2$,$\frac{dy}{dx} = -2x$. At $x = 2$,$m_1 = -2(2) = -4$. For $y = 2 + x^2$,$\frac{dy}{dx} = 2x$. At $x = 2$,$m_2 = 2(2) = 4$. The angle $	heta$ between the curves is given by $	an 	heta = |\frac{m_1 - m_2}{1 + m_1 m_2}|$. $|	an 	heta | = |\frac{-4 - 4}{1 + (-4)(4)}| = |\frac{-8}{1 - 16}| = |\frac{-8}{-15}| = \frac{8}{15}$.

If $\theta$ denotes the acute angle between the curves $y = 10 - x^2$ and $y = 2 + x^2$ at a point of their intersection,then $|\tan \theta |$ is equal to

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