Coordinate Geometry — Page 1 — Tinkutara Math Forum

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Question 229432

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Question 229195

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Question 228474

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Question 228378

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Question 228342

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∣ x+1 ∣ + ∣ x ∣ + ∣ x−3∣ > 8

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Question 227295

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Question 227294

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Question 227047

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Question 226829

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Question 226770

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Question 226588

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Question 226587

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Question 226338

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Question 226337

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Question 226336

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Question 225716

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Question 225497

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Question 225449

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Question 224702

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The foot of the perpendicular from a point of the circle x^2 +y^2 =1,z=0 to the plan 2x+3y+z=6 lie on curve−−−−−−

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Question 221288

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Find the remainder when x^(100) is divided by (x^2 +x+1)

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Question 221184

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Question 221050

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proof that volume of frustum of circular cone is (1/3)h[A1+A2+(√(A1A2)) A_1 and A_2 are areas of base

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Question 219685

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Question 219621

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I_n = ∫_0 ^( 1) (x/((x + 1)^n ))dx find I_0 and I_1 express I_n interms of n for all n ≥ 2

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Question 218384

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Question 218357

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Question 218356

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Question 218345

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Question 218344

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show that ∫_( n) ^( n + 1) ln(t) dt ≤ ln(n + (1/2)) Given u_n = (((4n)^n n!e^(−n) )/((2n)!)), ∀n ≥ 1 prove, using the preceding question that u_n is decreasing and convergent

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Geometrie dans le plan. AB^(→) et CD^(→) sont deux vecteurs du plan. AB^(→) n′est pas nul. Demontre que si AB^(→) et CD^(→) sont colineaires alors il existe un nombre reel k tel que CD^(→) = k AB^(→) .

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Help me please... $\vv{AB}$ and $\vv{CD}$ are two vectors, and $\vv{AB}$ is not the zero vector. Prove that if the vectors $\vv{AB}$ and $\vv{CD}$ are colinear, then there exists a real number $ k $ such that $ \vv{CD} = k \vv{AB} $. (don't use coordinates !)