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Home » Since $ \phi = 3\theta $, we substitute: - Imagemakers

Since $ \phi = 3\theta $, we substitute: - Imagemakers

📅 April 29, 2026 👤 scraface

Apr 29, 2026

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📌 5Question: An ornithologist tracks a bird's migration, where the daily distance traveled forms an arithmetic sequence. If the bird flies 15 km on day 1 and 35 km on day 5, what is the total distance flown in 10 days?

📌 Solution: Let the arithmetic sequence have first term $ a_1 = 15 $ and fifth term $ a_5 = 35 $. The common difference $ d $ satisfies $ a_5 = a_1 + 4d $. Substituting: $ 35 = 15 + 4d \Rightarrow d = 5 $. The sum of the first $ n = 10 $ terms is $ S_{10} = rac{10}{2} [2(15) + 9(5)] = 5(30 + 45) = 5(75) = 375 $. oxed{375}

📌 Question: A science educator designs a game where each correct challenge awards $ x $ points and a 5-point bonus for every 3 challenges completed. If a student earns 68 points after 10 challenges, what is $ x $?

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So symmetric about x-axis. But both are distinct unless \(y = 0\).
But since 506 ≠ 0, no solution has \(y = 0\) (would require \(m = n\),...
But each \((m,n)\) is distinct, so 16 solutions.
But let’s test a small one: \(m = 1\), \(n = 506\):
\(a = 2\), \(b = 1012\), then \(x = (2+1012)/2 = 507\), \(y = (1012 -...
Check: \(507^2 - 505^2 = (507-505)(507+505) = 2 \cdot 1012 = 2024\), c...
Another: \(m = -1\), \(n = -506\): \(x = -507/2\), not integer — wait!...
But when we list all divisor pairs \((m,n)\) with \(mn = 506\), we inc...
No: for each factorization, we get a point. But since \(m\) and \(n\)...
But each pair \((m,n)\) with \(mn = 506\) gives a unique \((x,y)\)?

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