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Coordinate GeometryQuestion and Answers: Page 1

Question Number 222531 Answers: 1 Comments: 0

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−−−−−−

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

Question Number 221288 Answers: 1 Comments: 0

Question Number 221271 Answers: 1 Comments: 0

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

$$\:\:{Find}\:{the}\:{remainder}\:{when}\:{x}^{\mathrm{100}} \: \\ $$$$\:\:{is}\:{divided}\:{by}\:\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right) \\ $$

Question Number 221184 Answers: 3 Comments: 2

Question Number 221050 Answers: 0 Comments: 1

Question Number 220257 Answers: 2 Comments: 0

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

$${proof}\:{that}\:{volume}\:{of}\:{frustum}\:{of} \\ $$$$\:{circular}\:{cone}\:{is}\:\frac{\mathrm{1}}{\mathrm{3}}{h}\left[{A}\mathrm{1}+{A}\mathrm{2}+\sqrt{{A}\mathrm{1}{A}\mathrm{2}}\right. \\ $$$${A}_{\mathrm{1}} {and}\:{A}_{\mathrm{2}} \:{are}\:\:{areas}\:{of}\:{base} \\ $$

Question Number 219685 Answers: 2 Comments: 0

Question Number 219621 Answers: 2 Comments: 1

Question Number 219461 Answers: 1 Comments: 0

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

$${I}_{{n}} \:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{x}}{\left({x}\:+\:\mathrm{1}\right)^{{n}} }{dx} \\ $$$${find}\:{I}_{\mathrm{0}} \:{and}\:{I}_{\mathrm{1}} \\ $$$${express}\:{I}_{{n}} \:{interms}\:{of}\:{n}\:{for}\:{all}\:{n}\:\geqslant\:\mathrm{2} \\ $$

Question Number 218384 Answers: 3 Comments: 0

Question Number 218356 Answers: 3 Comments: 0

Question Number 218357 Answers: 3 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

$${show}\:{that}\:\int_{\:{n}} ^{\:{n}\:+\:\mathrm{1}} {ln}\left({t}\right)\:{dt}\:\leqslant\:{ln}\left({n}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$${Given}\:{u}_{{n}} \:=\:\frac{\left(\mathrm{4}{n}\right)^{{n}} {n}!{e}^{−{n}} }{\left(\mathrm{2}{n}\right)!},\:\forall{n}\:\geqslant\:\mathrm{1} \\ $$$${prove},\:{using}\:{the}\:{preceding}\:{question}\:{that} \\ $$$${u}_{{n}} \:{is}\:{decreasing}\:{and}\:{convergent} \\ $$

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^(→) .

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

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 $ k $ such that $ \vv{CD} = k \vv{AB} $. (don't use coordinates !)

$$ \\ $$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 !)

Question Number 216859 Answers: 1 Comments: 3