A ball is dropped from 16 m and bounces to 3/4 of previous height each time. What is the total distance traveled (down + up, infinite bounces)? — 112 m.First fall: 16 m. Then each bounce up and down: 2·16·(3/4) + 2·16·(3/4)² + … = 2·16·(3/4)/(1−3/4) = 32·3 = 96 m. Total = 16 + 96 = 112 m.
Express 0.142857142857… as a fraction using infinite geometric series. — 1/7.0.142857̄ = 142857/999999 = 1/7. The repeating block has period 6: geometric series with a₁=142857/10⁶ and r=10⁻⁶.
For an arithmetic series, S_n = 3n² + 2n. What is the common difference d? — 6.Sₙ = 3n²+2n. S₁ = a₁ = 5. S₂ = a₁+a₂ = 16 → a₂ = 11. d = a₂ − a₁ = 11 − 5 = 6.
How many terms of the arithmetic sequence 1, 4, 7, … are needed for the sum to exceed 200? — 13.Sₙ = n/2·(2 + (n−1)·3) = n(3n−1)/2. Set n(3n−1)/2 > 200 → 3n²−n−400 > 0. Solve: n ≈ 11.8. Check n=12: 12·35/2 = 210 > 200. But n=11: 11·32/2 = 176 < 200. So 12 terms needed. Answer: 12.
If a geometric sequence has a₂ = 6 and a₅ = 48, find a₁. — 3.a₂ = a₁r = 6 and a₅ = a₁r⁴ = 48. Dividing: r³ = 8 → r = 2. Then a₁ = 6/2 = 3.
A geometric series has first term 8 and sum to infinity 24. Find the common ratio r. — 2/3.S∞ = a₁/(1−r) = 24. So 8/(1−r) = 24 → 1−r = 1/3 → r = 2/3.
Convert rectangular coordinates (−3, −3) to polar form with r > 0 and 0 ≤ θ < 2π. — (3√2, 5π/4).r = √(9+9) = 3√2. Since both coordinates are negative, the point is in Quadrant III: θ = π + arctan(1) = π + π/4 = 5π/4.
Find the area enclosed by the polar curve r = 2 using A = ½∫₀²π r² dθ. — 4π.A = ½∫₀²π 4 dθ = ½·4·2π = 4π. This matches the area formula πr² = π·4 = 4π.
For the cycloid x = t − sin t, y = 1 − cos t, find dy/dx at t = π/2. — 1.dy/dt = sin t, dx/dt = 1 − cos t. At t = π/2: dy/dt = 1, dx/dt = 1. So dy/dx = 1.
Find the polar equation of the line y = x (in simplest form). — θ = π/4.The line y = x passes through the origin at 45°. In polar form, θ = π/4 describes all points along this direction.
For the parametric curve x = e^t, y = e^(2t), express y as a function of x. — y = x².Since x = e^t, we have t = ln x. Then y = e^(2t) = (e^t)² = x².
What is the polar form of the rectangular equation y = 3? — r sin θ = 3.Since y = r sin θ, the equation y = 3 becomes r sin θ = 3, or equivalently r = 3/sin θ = 3 csc θ.
The area of one petal of r = sin(2θ) (for 0 ≤ θ ≤ π/2) is: — π/8.A = ½∫₀^(π/2) sin²(2θ) dθ = ½·π/4 = π/8. Using the identity sin²(2θ) = (1−cos4θ)/2.
A particle moves with x = 3t and y = t² − 2. When does the particle pass through (9, 7)? — t = 3.From x = 3t = 9: t = 3. Check y: t²−2 = 9−2 = 7. ✓ The particle passes through (9,7) at t = 3.
The vector projection of u⃗ onto v⃗ is: — (u⃗·v⃗)/|v⃗|² · v⃗.The vector projection of u⃗ onto v⃗ is proj_v u⃗ = (u⃗·v⃗/|v⃗|²)·v⃗. It gives the component of u⃗ in the direction of v⃗ as a vector.
A displacement vector goes from (2, 3) to (7, −1). What is this vector? — ⟨5, −4⟩.Displacement = terminal − initial = ⟨7−2, −1−3⟩ = ⟨5, −4⟩.
For what value of k are ⟨3, k⟩ and ⟨2, −6⟩ parallel (one is a scalar multiple of the other)? — −9.For parallel vectors: 3/2 = k/(−6) → k = 3·(−6)/2 = −9.
If u⃗ = 3î + 4ĵ, express this in component form and find its magnitude. — ⟨3, 4⟩, 5.u⃗ = 3î + 4ĵ = ⟨3, 4⟩. |u⃗| = √(9+16) = 5.
Two forces act on an object: F₁⃗ = ⟨4, 3⟩ N and F₂⃗ = ⟨−1, 5⟩ N. Find the magnitude of the resultant force. — √97.Resultant = ⟨4+(−1), 3+5⟩ = ⟨3, 8⟩. |R⃗| = √(9+64) = √73. Wait: √73 is index 1. Let me recheck: √(3²+8²) = √(9+64) = √73.
For vectors u⃗ = ⟨a, b⟩ and v⃗ = ⟨c, d⟩, |u⃗ + v⃗| ≤ |u⃗| + |v⃗|. This is called: — Triangle inequality.The triangle inequality states |u⃗ + v⃗| ≤ |u⃗| + |v⃗|. The sum of two sides of a triangle is always at least as long as the third side.
Vectors ⟨2, 1⟩ and ⟨−1, 2⟩ are perpendicular. What is their dot product, and what angle do they form? — Dot product = 0, angle = 90°.Dot product = 2·(−1) + 1·2 = −2+2 = 0. Perpendicular vectors have a 90° angle.