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Home » \(v = \frac40 \pm \sqrt35202\). - Crosslake

\(v = \frac40 \pm \sqrt35202\). - Crosslake

📅 March 1, 2026 👤 scraface

Mar 01, 2026

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📌 + 12085 + 12087 + 12089 = (12083 + 12089) + (12085 + 12087) = 24172 + 24172 = 48344

📌 Now compute $ 48344 \mod 8 $. Since 1000 is divisible by 8, only the last three digits matter:

📌 \div 8 = 43 \text{ with remainder } 0

📌 Question: What is the sum of all positive divisors of $ 2^3 \cdot 3^2 $?

📌 Solution: The sum of the divisors of $ 2^3 \cdot 3^2 $ is given by:

📌 Thus, the sum of all positive divisors is $ \boxed{195} $.

📌 Question: What is the remainder when $ 7^5 + 7^6 $ is divided by 6?

📌 Solution: First, note that $ 7 \equiv 1 \mod 6 $, so $ 7^k \equiv 1^k = 1 \mod 6 $ for any positive integer $ k $.

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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}.
Since \( x^2 \geq 0 \), we require:
\frac{y + 1}{y - 1} \geq 0.

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