Question and Answers Forum

All Questions Topic List

Coordinate GeometryQuestion and Answers: Page 1

Question Number 218384 Answers: 3 Comments: 0

Question Number 218356 Answers: 3 Comments: 0

Question Number 218357 Answers: 1 Comments: 0

Question Number 218345 Answers: 1 Comments: 0

Question Number 218344 Answers: 0 Comments: 1

Question Number 217088 Answers: 1 Comments: 0

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

$s h o w t h a t \int_{n}^{n + 1} l n (t) d t ⩽ l n (n + \frac{1}{2})$ $G i v e n u_{n} = \frac{{(4 n)}^{n} n! e^{- n}}{(2 n)!}, \forall n ⩾ 1$ $p r o v e, u s i n g t h e p r e c e d i n g q u e s t i o n t h a t$ $u_{n} i s d e c r e a s i n g a n d c o n v e r g e n t$

Question Number 217084 Answers: 0 Comments: 0

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^(→) .

$Geometr i e dans le plan .$ $\vec{AB} et \vec{CD} sont deux vecteurs du plan .$ $\vec{AB} n^{'} est pas nul .$ $Demontre que si \vec{AB} et \vec{CD} sont colineaires$ $alors il existe un nombre reel k tel que$ $\vec{CD} = k \vec{AB} .$

Question Number 217075 Answers: 0 Comments: 3

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