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📅 May 3, 2026 👤 scraface

May 03, 2026

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📌 Multiply the first equation by 3: $ 21l + 15h = 361.50 $. Multiply the second equation by 5: $ 20l + 15h = 425.00 $. Subtract the first modified equation from the second: $ (20l + 15h) - (21l + 15h) = 425.00 - 361.50 \Rightarrow -l = 63.50 \Rightarrow l = -63.50 $. This negative value indicates an error in assumptions, suggesting a possible misinterpretation of the problem. However, if the equations are correct, the solution is $ \boxed{-63.50} $, which is invalid in context. Rechecking the problem setup is advised.

📌 Question: If $ f(x) = 3x - 7 $ and $ g(x) = x^2 + 2x $, what is $ g(f(4)) $?

📌 Solution: First compute $ f(4) = 3(4) - 7 = 12 - 7 = 5 $. Then substitute into $ g(x) $: $ g(5) = 5^2 + 2(5) = 25 + 10 = 35 $. The value of $ g(f(4)) $ is $ \boxed{35} $.

📌 Question: Solve for $ x $: $ \frac{2}{3}(x + 4) = \frac{5}{6}(x - 1) $.

📌 Solution: Multiply both sides by 6 to eliminate denominators: $ 4(x + 4) = 5(x - 1) $. Expand: $ 4x + 16 = 5x - 5 $. Rearrange: $ 16 + 5 = 5x - 4x \Rightarrow x = 21 $. The solution is $ \boxed{21} $.

📌 Question: If $ 2a + 3b = 14 $ and $ a - b = 1 $, what is the value of $ 5a + 2b $?

📌 Solution: From $ a - b = 1 $, express $ a = b + 1 $. Substitute into the first equation: $ 2(b + 1) + 3b = 14 \Rightarrow 2b + 2 + 3b = 14 \Rightarrow 5b = 12 \Rightarrow b = \frac{12}{5} $. Then $ a = \frac{12}{5} + 1 = \frac{17}{5} $. Compute $ 5a + 2b = 5\left(\frac{17}{5}\right) + 2\left(\frac{12}{5}\right) = 17 + \frac{24}{5} = \frac{109}{5} $. The value is $ \boxed{\dfrac{109}{5}} $.Question: A hydrologist is modeling the cross-section of a riverbed, approximated as an isosceles trapezoid with bases of 120 m and 180 m, and legs of 50 m each. What is the height of the trapezoid, in meters?

📌 Solution: Let the trapezoid have bases of lengths $ a = 120 $ m and $ b = 180 $ m, and equal legs of length $ l = 50 $ m. The height $ h $ can be found by dropping perpendiculars from the ends of the shorter base to the longer base, forming two right triangles on either side and a rectangle in the middle. The difference in the bases is $ b - a = 60 $ m, so each of the right triangles has a base of $ \frac{60}{2} = 30 $ m. Using the Pythagorean theorem:

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