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Home » Also, $\sec x \csc x = \frac1\sin x \cos x$, so: - Nelissen Grade advocaten

Also, $\sec x \csc x = \frac1\sin x \cos x$, so: - Nelissen Grade advocaten

📅 March 1, 2026 👤 scraface

Mar 01, 2026

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📌 Substitute \( w = m \) into \( L(w) \):

📌 L(m) = (m)^2 - 2m(m) + m^2 + 4 = m^2 - 2m^2 + m^2 + 4 = 0 + 4 = 4

📌 But the problem states the minimum loss must be 1, yet we get 4. This suggests either the function is misstated or we must adjust. However, observe:

📌 L(m) = m^2 - 2m \cdot m + m^2 + 4 = 2m^2 - 2m^2 + 4 = 4

📌 So the minimum value is always 4, independent of \( m \). But the problem asks for the minimum loss to be 1, which is impossible under this function form.

📌 Wait — re-examining: perhaps a typo in interpretation. Let’s suppose the function is \( L(w) = (w - m)^2 + 1 \), so minimum is 1. But the given is \( L(w) = w^2 - 2mw + m^2 + 4 = (w - m)^2 + 4 - m^2 + m^2 = (w - m)^2 + 4 \)? No:

📌 Thus minimum value is \( 0 + 4 = 4 \), always. So minimum cannot be 1.

📌 But the problem says "find \( m \) such that the minimum loss is exactly 1". This is impossible unless the constant term is adjusted.

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We seek divisors of \( 360 \) that are perfect squares. A divisor is a...
Let a divisor \( d = 2^a \cdot 3^b \cdot 5^c \) be a perfect square, w...
\( a \in \{0, 2\} \) (since \( a \leq 3 \), and even)
\( b \in \{0, 2\} \) (since \( b \leq 2 \), even)
\( c \in \{0\} \) (since \( c \leq 1 \), and only even choice is 0)
So possible combinations:
\( a = 0 \) or \( 2 \) → 2 choices
\( b = 0 \) or \( 2 \) → 2 choices
\( c = 0 \) → 1 choice
Total: \( 2 \times 2 \times 1 = 4 \) perfect square divisors.

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