Question and Answers Forum

AllQuestion and Answers: Page 345

Question Number 186544 Answers: 0 Comments: 1

Question Number 186540 Answers: 0 Comments: 0

Complex Numbers Prove that 1. ∣Z_1 +Z_2 ∣≤∣Z_1 ∣+∣Z_2 ∣ 2. ∣Z_1 −Z_2 ∣≥∣Z_1 ∣−∣Z_2 ∣

$${Complex}\:{Numbers} \\ $$$${Prove}\:{that} \\ $$$$\:\mathrm{1}.\:\mid{Z}_{\mathrm{1}} +{Z}_{\mathrm{2}} \mid\leq\mid{Z}_{\mathrm{1}} \mid+\mid{Z}_{\mathrm{2}} \mid \\ $$$$\mathrm{2}.\:\mid{Z}_{\mathrm{1}} −{Z}_{\mathrm{2}} \mid\geq\mid{Z}_{\mathrm{1}} \mid−\mid{Z}_{\mathrm{2}} \mid \\ $$

Question Number 186539 Answers: 0 Comments: 0

(1)∫(lim_(x→x+1) ((x^(x+2) x^(2+x^2 ) )/( (√x))))dx=? (2)∫((2+xcos 2x)/(1−2x+cos 3x))dx=? (3) Σ_(n=5) ^(4n) (3n+2n)=?

$$\left(\mathrm{1}\right)\int\left(\underset{{x}\rightarrow{x}+\mathrm{1}} {\mathrm{lim}}\frac{{x}^{{x}+\mathrm{2}} {x}^{\mathrm{2}+{x}^{\mathrm{2}} } }{\:\sqrt{{x}}}\right){dx}=? \\ $$$$\left(\mathrm{2}\right)\int\frac{\mathrm{2}+{x}\mathrm{cos}\:\mathrm{2}{x}}{\mathrm{1}−\mathrm{2}{x}+\mathrm{cos}\:\mathrm{3}{x}}{dx}=? \\ $$$$\left(\mathrm{3}\right)\:\underset{{n}=\mathrm{5}} {\overset{\mathrm{4}{n}} {\sum}}\left(\mathrm{3}{n}+\mathrm{2}{n}\right)=? \\ $$

Question Number 186531 Answers: 0 Comments: 0

Q.use the parseval relation of hankel transfrom to evaluate the Integral ∫_0 ^∞ ((J_(𝛄+1) (ar)J_(𝛄+1) (br))/r) , for 𝛄>−(1/2) , 0<a<b where J_n (x) are bessel funtions.

$$ \\ $$$$\:\:\mathbb{Q}.\boldsymbol{{use}}\:\boldsymbol{{the}}\:\boldsymbol{{parseval}}\:\boldsymbol{{relation}}\:\boldsymbol{{of}}\:\boldsymbol{{hankel}}\:\boldsymbol{{transfrom}}\:\boldsymbol{{to}}\:\boldsymbol{{evaluate}}\:\boldsymbol{{the}}\:\boldsymbol{{Integral}}\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\overset{\infty} {\int}_{\mathrm{0}} \:\:\frac{\boldsymbol{{J}}_{\boldsymbol{\gamma}+\mathrm{1}} \left(\boldsymbol{{ar}}\right)\boldsymbol{{J}}_{\boldsymbol{\gamma}+\mathrm{1}} \left(\boldsymbol{{br}}\right)}{\boldsymbol{{r}}}\:,\:\:\boldsymbol{{for}}\:\boldsymbol{\gamma}>−\frac{\mathrm{1}}{\mathrm{2}}\:,\:\:\mathrm{0}<\boldsymbol{{a}}<\boldsymbol{{b}} \\ $$$$\:\:\:\boldsymbol{{where}}\:\boldsymbol{{J}}_{\boldsymbol{{n}}} \left(\boldsymbol{{x}}\right)\:\boldsymbol{{are}}\:\boldsymbol{{bessel}}\:\boldsymbol{{funtions}}. \\ $$$$ \\ $$

Question Number 186527 Answers: 2 Comments: 0

Question Number 186517 Answers: 1 Comments: 0

Question Number 186507 Answers: 0 Comments: 0

Question Number 186503 Answers: 0 Comments: 1

function of , f (x) = ax + ∣ x ∣ is one to one .find ” a ” .

$$ \\ $$$$\:\:\:{function}\:{of}\:,\:{f}\:\left({x}\right)\:=\:{ax}\:\:+\:\mid\:{x}\:\mid\:{is}\:\:{one}\:{to}\:{one} \\ $$$$\:\:\:\: \\ $$$$\:\:\:\:\:\:\:.{find}\:\:\:\:''\:\:\:\:{a}\:\:\:\:''\:\:. \\ $$$$\:\: \\ $$

Question Number 186500 Answers: 1 Comments: 0

lim_(x→+∞) (√(x+(√(x+(√(x+(√x)))))))−(√x)=? please solution

$$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}}}}}−\sqrt{\mathrm{x}}=? \\ $$$$\mathrm{please}\:\mathrm{solution} \\ $$

Question Number 186492 Answers: 1 Comments: 1

Question Number 186489 Answers: 1 Comments: 0

Question Number 186486 Answers: 0 Comments: 6

Question propose par Migma −−−−−−−−−−− calcul de X △ABC ∡ACB=40^° AB^2 =AC^2 +BC^2 −2AC×BCcos 40 AC=15 ; AB=X+4 ; BC=10 (X+4)^2 =15^2 +10^2 −300×cos 40 =325−229,81334 (X+4)^2 =95,18666706 posons Z=X+4 Z^2 =9,756^2 ⇒X+4=9,756 X=5,756

$${Question}\:{propose}\:{par} \\ $$$${Migma} \\ $$$$−−−−−−−−−−− \\ $$$${calcul}\:{de}\:{X} \\ $$$$ \\ $$$$\bigtriangleup{ABC}\:\:\:\:\:\:\measuredangle{ACB}=\mathrm{40}^{°} \\ $$$$\mathrm{AB}^{\mathrm{2}} =\mathrm{AC}^{\mathrm{2}} +\mathrm{BC}^{\mathrm{2}} −\mathrm{2AC}×\mathrm{BCcos}\:\mathrm{40}\:\:\:\: \\ $$$$\mathrm{AC}=\mathrm{15}\:\:;\:\:\:\:\mathrm{AB}=\mathrm{X}+\mathrm{4}\:;\:\:\:\mathrm{BC}=\mathrm{10} \\ $$$$\left(\mathrm{X}+\mathrm{4}\right)^{\mathrm{2}} =\mathrm{15}^{\mathrm{2}} +\mathrm{10}^{\mathrm{2}} −\mathrm{300}×\mathrm{cos}\:\mathrm{40} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{325}−\mathrm{229},\mathrm{81334} \\ $$$$\left({X}+\mathrm{4}\right)^{\mathrm{2}} =\mathrm{95},\mathrm{18666706} \\ $$$${posons}\:\:\:{Z}={X}+\mathrm{4} \\ $$$$\:\:\:{Z}^{\mathrm{2}} =\mathrm{9},\mathrm{756}^{\mathrm{2}} \:\:\:\:\Rightarrow\mathrm{X}+\mathrm{4}=\mathrm{9},\mathrm{756} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:{X}=\mathrm{5},\mathrm{756} \\ $$$$ \\ $$

Question Number 186482 Answers: 2 Comments: 0

Question Number 186481 Answers: 0 Comments: 0

Question Number 186476 Answers: 0 Comments: 0

Question Number 186473 Answers: 0 Comments: 0

A metallic cube is subjected to heating such that as the metal expands, the total surface area increases at rate of 6.25 cm^2 s^(−1) . Calculate the rate at which each side of the cube is increasing when the volume is 51.2 cm^3 .

$$\mathrm{A}\:\mathrm{metallic}\:\mathrm{cube}\:\mathrm{is}\:\mathrm{subjected}\:\mathrm{to} \\ $$$$\mathrm{heating}\:\mathrm{such}\:\mathrm{that}\:\mathrm{as}\:\mathrm{the}\:\mathrm{metal} \\ $$$$\mathrm{expands},\:\mathrm{the}\:\mathrm{total}\:\mathrm{surface}\:\mathrm{area} \\ $$$$\mathrm{increases}\:\mathrm{at}\:\mathrm{rate}\:\mathrm{of}\:\mathrm{6}.\mathrm{25}\:\mathrm{cm}^{\mathrm{2}} \mathrm{s}^{−\mathrm{1}} . \\ $$$$\mathrm{Calculate}\:\mathrm{the}\:\mathrm{rate}\:\mathrm{at}\:\mathrm{which}\:\mathrm{each} \\ $$$$\mathrm{side}\:\mathrm{of}\:\mathrm{the}\:\mathrm{cube}\:\mathrm{is}\:\mathrm{increasing}\:\mathrm{when} \\ $$$$\mathrm{the}\:\mathrm{volume}\:\mathrm{is}\:\mathrm{51}.\mathrm{2}\:\mathrm{cm}^{\mathrm{3}} . \\ $$

Question Number 186468 Answers: 1 Comments: 0

Simplify ((1^2 ∙2!+2^2 ∙3!+3^2 ∙4!+∙∙∙+n^2 (n+1)!−2)/((n+1)!)) to n^2 +n−2

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{Simplify} \\ $$$$\frac{\mathrm{1}^{\mathrm{2}} \centerdot\mathrm{2}!+\mathrm{2}^{\mathrm{2}} \centerdot\mathrm{3}!+\mathrm{3}^{\mathrm{2}} \centerdot\mathrm{4}!+\centerdot\centerdot\centerdot+{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)!−\mathrm{2}}{\left({n}+\mathrm{1}\right)!} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{to} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{n}^{\mathrm{2}} +\mathrm{n}−\mathrm{2} \\ $$

Question Number 186464 Answers: 1 Comments: 15