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Home » Total distance: \( 150 + 200 = 350 \text km \) - Portal da AcÃºstica

Total distance: \( 150 + 200 = 350 \text km \) - Portal da AcÃºstica

📅 March 1, 2026 👤 scraface

Mar 01, 2026

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📌 Multiply the second equation by 2: $ 4s + 10m = 370 $. Subtract the first equation: $ 3m = 55 \Rightarrow m = rac{55}{3} $. Substitute back: $ 2s + 5 \cdot rac{55}{3} = 185 \Rightarrow 2s = 185 - rac{275}{3} = rac{280}{3} \Rightarrow s = rac{140}{3} pprox 46.67 $. Convert to cents: $ 4667 $ cents.

📌 oxed{4667}

📌 Question: An ichthyologist tracks two fish species' population trends with lines $ y = -3x + 18 $ (species A) and $ y = 4x - 2 $ (species B). Find their intersection point.

📌 Solution: Set $ -3x + 18 = 4x - 2 $. Solve: $ 18 + 2 = 7x \Rightarrow x = rac{20}{7} $. Substitute into $ y = -3 \cdot rac{20}{7} + 18 = -rac{60}{7} + rac{126}{7} = rac{66}{7} $. Intersection at $ \left( rac{20}{7}, rac{66}{7}

📌 oxed{\left( \dfrac{20}{7}, \dfrac{66}{7}

📌 $$Question: A statistician models a time series using a geometric sequence where the first term is $ 3 $ and the common ratio is $ r $. If the sum of the first five terms exceeds $ 300 $, what is the smallest integer value of $ r $ that satisfies this condition?

📌 The sum of the first five terms of a geometric sequence with first term $ a = 3 $ and common ratio $ r $ is given by:

📌 S_5 = 3(1 + r + r^2 + r^3 + r^4) = 3 \cdot rac{r^5 - 1}{r - 1} \quad ext{(for } r

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