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Question Number 157054    Answers: 0   Comments: 1

suppose you drop a tennis ball from a hieght of 15 feet.after the ballhits the floor it rebounds to85% of its previous height.how high will the ball rebound after its ghird bounce round tl the nearest tenth

$${suppose}\:{you}\:{drop}\:{a}\:{tennis}\:{ball}\:{from}\:{a}\:{hieght}\:{of}\:\mathrm{15}\:{feet}.{after}\:{the}\:{ballhits}\:{the}\:{floor}\:{it}\:{rebounds}\:\:{to}\mathrm{85\%}\:{of}\:{its}\:{previous}\:{height}.{how}\:{high}\:{will}\:{the}\:{ball}\:{rebound}\:{after}\:{its}\:{ghird}\:{bounce}\:{round}\:{tl}\:{the}\:{nearest}\:{tenth} \\ $$$$ \\ $$

Question Number 156911    Answers: 0   Comments: 0

Find: 𝛀 =lim_(nβ†’βˆž) ∫_( 0) ^( 1) ([nx] βˆ™ ∣x - [x + (1/2)∣])dx [βˆ—] - GIF

$$\mathrm{Find}: \\ $$$$\boldsymbol{\Omega}\:\:=\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\underset{\:\mathrm{0}} {\overset{\:\mathrm{1}} {\int}}\left(\left[\mathrm{nx}\right]\:\centerdot\:\mid\mathrm{x}\:-\:\left[\mathrm{x}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\mid\right]\right)\mathrm{dx} \\ $$$$\left[\ast\right]\:-\:\mathrm{GIF} \\ $$

Question Number 156910    Answers: 0   Comments: 2

Question Number 156909    Answers: 0   Comments: 0

Find: 𝛀 =lim_(nβ†’βˆž) (n - Ξ£_(k=1) ^n (((e - 1)βˆ™n)/(n + (e - 1)βˆ™k))) = ?

$$\mathrm{Find}: \\ $$$$\boldsymbol{\Omega}\:\:=\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{n}\:-\:\underset{\boldsymbol{\mathrm{k}}=\mathrm{1}} {\overset{\boldsymbol{\mathrm{n}}} {\sum}}\:\frac{\left(\mathrm{e}\:-\:\mathrm{1}\right)\centerdot\mathrm{n}}{\mathrm{n}\:+\:\left(\mathrm{e}\:-\:\mathrm{1}\right)\centerdot\mathrm{k}}\right)\:=\:? \\ $$

Question Number 156905    Answers: 2   Comments: 0

find the value of x and y, x:3:5=8:y:9

$${find}\:{the}\:{value}\:{of}\:{x}\:{and}\:{y},\:{x}:\mathrm{3}:\mathrm{5}=\mathrm{8}:{y}:\mathrm{9} \\ $$

Question Number 156904    Answers: 1   Comments: 0

find the value of x and y , x:3:5=2:y:10

$${find}\:{the}\:{value}\:{of}\:{x}\:{and}\:{y}\:,\:{x}:\mathrm{3}:\mathrm{5}=\mathrm{2}:{y}:\mathrm{10} \\ $$

Question Number 156900    Answers: 0   Comments: 0

Ξ©_1 = 1 - (Ο€/2) +Ξ£_(n=2) ^∞ (- (1/𝛑))^n βˆ™ (1/(n+1)) Ξ©_2 = 1 - (Ο€/2) + Ξ£_(n=2) ^∞ (- (1/e))^n βˆ™ (1/(n+1)) A) Ξ©_1 < Ξ©_2 B) Ξ©_1 = Ξ©_2 C) Ξ©_1 > Ξ©_2

$$\Omega_{\mathrm{1}} \:=\:\mathrm{1}\:-\:\frac{\pi}{\mathrm{2}}\:+\underset{\boldsymbol{\mathrm{n}}=\mathrm{2}} {\overset{\infty} {\sum}}\left(-\:\frac{\mathrm{1}}{\boldsymbol{\pi}}\right)^{\boldsymbol{\mathrm{n}}} \centerdot\:\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}} \\ $$$$\Omega_{\mathrm{2}} \:=\:\mathrm{1}\:-\:\frac{\pi}{\mathrm{2}}\:+\:\underset{\boldsymbol{\mathrm{n}}=\mathrm{2}} {\overset{\infty} {\sum}}\left(-\:\frac{\mathrm{1}}{\boldsymbol{\mathrm{e}}}\right)^{\boldsymbol{\mathrm{n}}} \centerdot\:\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}} \\ $$$$\left.\mathrm{A}\left.\right)\left.\:\Omega_{\mathrm{1}} \:<\:\Omega_{\mathrm{2}} \:\:\:\mathrm{B}\right)\:\Omega_{\mathrm{1}} \:=\:\Omega_{\mathrm{2}} \:\:\:\mathrm{C}\right)\:\Omega_{\mathrm{1}} \:>\:\Omega_{\mathrm{2}} \\ $$

Question Number 156895    Answers: 0   Comments: 0

(dy/dx)βˆ’(x/y)+x^3 cos y = 0

$$\:\frac{{dy}}{{dx}}βˆ’\frac{{x}}{{y}}+{x}^{\mathrm{3}} \:\mathrm{cos}\:{y}\:=\:\mathrm{0} \\ $$

Question Number 156915    Answers: 1   Comments: 0

Question Number 156914    Answers: 2   Comments: 0

Question Number 156891    Answers: 1   Comments: 0

Question Number 156887    Answers: 1   Comments: 0

tan 2x tan 3x tan 5x =1

$$\:\mathrm{tan}\:\mathrm{2}{x}\:\mathrm{tan}\:\mathrm{3}{x}\:\mathrm{tan}\:\mathrm{5}{x}\:=\mathrm{1} \\ $$

Question Number 156869    Answers: 2   Comments: 0

Question Number 156867    Answers: 1   Comments: 0

Question Number 156864    Answers: 0   Comments: 4

Ο† := ∫_0 ^( 1) (( ln (1βˆ’x^( 2) ))/(1+ x^( 2) )) dx = proof : Ο† = ∫_0 ^( 1) (( ln(1βˆ’x ))/(1+x^( 2) ))dx + (Ο€/8)ln(2) .... I= ∫_0 ^( 1) ((ln ( 1βˆ’x ))/(1+x^( 2) ))dx =^(x=tan(t)) ∫_0 ^( (Ο€/4)) ln( cos(t)βˆ’sin(t))dtβˆ’βˆ«_0 ^( (Ο€/4)) ln(cos(t))dt = ∫_0 ^( (Ο€/4)) ln((√2) )dt +∫_0 ^( (Ο€/4)) ln(sin((Ο€/4) βˆ’t))dtβˆ’(G/2) +(Ο€/4)ln(2) =((3Ο€)/8) ln(2)βˆ’(G/2) βˆ’(G/2) βˆ’(Ο€/4) ln(2)=(Ο€/8)ln(2)βˆ’G Ο† = (Ο€/4)ln(2) βˆ’ G β–  m.n

$$ \\ $$$$\:\:\:\:\phi\::=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:{ln}\:\left(\mathrm{1}βˆ’{x}^{\:\mathrm{2}} \right)}{\mathrm{1}+\:{x}^{\:\mathrm{2}} }\:{dx}\:= \\ $$$$\:\:{proof}\:: \\ $$$$\:\:\:\:\phi\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:{ln}\left(\mathrm{1}βˆ’{x}\:\right)}{\mathrm{1}+{x}^{\:\mathrm{2}} }{dx}\:+\:\frac{\pi}{\mathrm{8}}{ln}\left(\mathrm{2}\right) \\ $$$$\:\:\:\:....\:\mathrm{I}=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\:\left(\:\mathrm{1}βˆ’{x}\:\right)}{\mathrm{1}+{x}^{\:\mathrm{2}} }{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\overset{{x}={tan}\left({t}\right)} {=}\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} {ln}\left(\:{cos}\left({t}\right)βˆ’{sin}\left({t}\right)\right){dt}βˆ’\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} {ln}\left({cos}\left({t}\right)\right){dt} \\ $$$$\:\:\:\:\:\:\:=\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} {ln}\left(\sqrt{\mathrm{2}}\:\right){dt}\:+\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} {ln}\left({sin}\left(\frac{\pi}{\mathrm{4}}\:βˆ’{t}\right)\right){dt}βˆ’\frac{\mathrm{G}}{\mathrm{2}}\:+\frac{\pi}{\mathrm{4}}{ln}\left(\mathrm{2}\right) \\ $$$$\:\:\:=\frac{\mathrm{3}\pi}{\mathrm{8}}\:{ln}\left(\mathrm{2}\right)βˆ’\frac{\mathrm{G}}{\mathrm{2}}\:βˆ’\frac{\mathrm{G}}{\mathrm{2}}\:βˆ’\frac{\pi}{\mathrm{4}}\:{ln}\left(\mathrm{2}\right)=\frac{\pi}{\mathrm{8}}{ln}\left(\mathrm{2}\right)βˆ’\mathrm{G} \\ $$$$\:\:\:\:\phi\:=\:\frac{\pi}{\mathrm{4}}{ln}\left(\mathrm{2}\right)\:βˆ’\:\mathrm{G}\:\:\:\:\:\:\blacksquare\:{m}.{n} \\ $$

Question Number 156860    Answers: 1   Comments: 0

∫_0 ^∞ ((x)^(1/n) /(x^3 +x^2 +x+1))dx=?

$$\int_{\mathrm{0}} ^{\infty} \frac{\sqrt[{{n}}]{{x}}}{{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}+\mathrm{1}}{dx}=? \\ $$

Question Number 156855    Answers: 1   Comments: 0

(dy/dx) = ((y^3 βˆ’xy^2 βˆ’x^2 yβˆ’5x^3 )/(xy^2 βˆ’x^2 yβˆ’2x^3 ))

$$\:\:\frac{{dy}}{{dx}}\:=\:\frac{{y}^{\mathrm{3}} βˆ’{xy}^{\mathrm{2}} βˆ’{x}^{\mathrm{2}} {y}βˆ’\mathrm{5}{x}^{\mathrm{3}} }{{xy}^{\mathrm{2}} βˆ’{x}^{\mathrm{2}} {y}βˆ’\mathrm{2}{x}^{\mathrm{3}} }\: \\ $$$$\: \\ $$

Question Number 156881    Answers: 0   Comments: 0

𝛀 =∫_( 0) ^( 1) ∫_( 0) ^( 1) ((log(1 - x) log(1 - y))/(1 - xy)) dxdy = ?

$$\boldsymbol{\Omega}\:=\underset{\:\mathrm{0}} {\overset{\:\mathrm{1}} {\int}}\underset{\:\mathrm{0}} {\overset{\:\mathrm{1}} {\int}}\:\frac{\mathrm{log}\left(\mathrm{1}\:-\:\mathrm{x}\right)\:\mathrm{log}\left(\mathrm{1}\:-\:\mathrm{y}\right)}{\mathrm{1}\:-\:\mathrm{xy}}\:\mathrm{dxdy}\:=\:? \\ $$

Question Number 156880    Answers: 2   Comments: 0

𝛀 =Ξ£_(n=0) ^∞ Ξ£_(k=0) ^n (1/𝛑^n ) βˆ™ ((Ο€/e))^k = ?

$$\boldsymbol{\Omega}\:=\underset{\boldsymbol{\mathrm{n}}=\mathrm{0}} {\overset{\infty} {\sum}}\:\underset{\boldsymbol{\mathrm{k}}=\mathrm{0}} {\overset{\boldsymbol{\mathrm{n}}} {\sum}}\:\frac{\mathrm{1}}{\boldsymbol{\pi}^{\boldsymbol{\mathrm{n}}} }\:\centerdot\:\left(\frac{\pi}{\mathrm{e}}\right)^{\boldsymbol{\mathrm{k}}} =\:? \\ $$

Question Number 156841    Answers: 1   Comments: 0

Question Number 156849    Answers: 0   Comments: 0

∫_0 ^1 ((ln(e+(1/(1βˆ’t))))/( (√t)))dt=?

$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({e}+\frac{\mathrm{1}}{\mathrm{1}βˆ’{t}}\right)}{\:\sqrt{{t}}}{dt}=? \\ $$

Question Number 156835    Answers: 0   Comments: 0

Question Number 156834    Answers: 1   Comments: 0

Question Number 156828    Answers: 0   Comments: 0

Question Number 156824    Answers: 1   Comments: 0

If x βˆ’ z = tan^(βˆ’ 1) (yz) and z = z(x, y), find ((Ξ΄z)/(Ξ΄x)) , ((Ξ΄z)/(Ξ΄y))

$$\mathrm{If}\:\:\:\:\:\:\mathrm{x}\:\:\:βˆ’\:\:\:\mathrm{z}\:\:\:\:=\:\:\:\:\mathrm{tan}^{βˆ’\:\mathrm{1}} \left(\mathrm{yz}\right)\:\:\:\:\:\mathrm{and}\:\:\:\:\:\:\mathrm{z}\:\:\:=\:\:\:\mathrm{z}\left(\mathrm{x},\:\:\mathrm{y}\right),\:\:\:\:\:\:\mathrm{find}\:\:\:\:\frac{\delta\mathrm{z}}{\delta\mathrm{x}}\:,\:\:\:\frac{\delta\mathrm{z}}{\delta\mathrm{y}} \\ $$

Question Number 156820    Answers: 1   Comments: 3

∫ (dx/((x^2 βˆ’x+1)((√(x^2 +x+1)))))

$$\int\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} βˆ’{x}+\mathrm{1}\right)\left(\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}\right)} \\ $$

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