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Home » Thus, \( r = \frac31.42\pi \approx \frac31.46.28 = 5 \) cm. - Crosslake

Thus, \( r = \frac31.42\pi \approx \frac31.46.28 = 5 \) cm. - Crosslake

📅 March 1, 2026 👤 scraface

Mar 01, 2026

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📌 2p + q = 2 \quad ext{(ii)}

📌 + q = 3 \Rightarrow q = 4

📌 Now verify with (3):

📌 h(3) = 27 + 3(-1) + 4 = 27 - 3 + 4 = 28

📌 Contradiction! So our assumption that \( h(x) = x^3 + px + q \) (missing quadratic term) must still hold, but we need to re-express correctly: likely the general form is \( h(x) = ax^3 + bx^2 + px + q \), but the problem states âDefine $ h(x) = x^3 + px + q $â â so it is missing the \( x^2 \) term. But real data suggests discrepancy.

📌 Wait â perhaps \( h(x) = x^3 + px + q \), but we must use all three equations to solve for \( p \) and \( q \), even if third doesn't fit exactly? But the problem gives all three, so maybe it's overdetermined.

📌 But let's suppose the model is \( h(x) = ax^3 + px + q \), as cubic with no \( x^2 \). Use the three values to solve for \( a, p, q \).

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