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evansville school board land sale - NBX Soluciones

📅 May 5, 2026 👤 scraface

May 05, 2026

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Solving gives:
v_1 = 1, \quad v_2 \text{ is free}, \quad v_3 = 1.
Thus, \(\mathbf{v} = \langle 1, t, 1 \rangle\) for any real \(t\).
\boxed{\langle 1, t, 1 \rangle \text{ for any } t \in \mathbb{R}}
Find the range of the function \( f(x) = \frac{x^2 + 1}{x^2 - 1} \) as...
Let \( y = \frac{x^2 + 1}{x^2 - 1} \). Rearranging gives:
y(x^2 - 1) = x^2 + 1 \quad \Rightarrow \quad yx^2 - y = x^2 + 1.
Rearranging terms:
If \( y \neq 1 \),
x^2 = \frac{y + 1}{y - 1}.

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