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Question Number 88977    Answers: 0   Comments: 1

x^3 + 3^x = 17 x =?

$$\:{x}^{\mathrm{3}} \:+\:\mathrm{3}^{{x}} \:=\:\mathrm{17} \\ $$$${x}\:=? \\ $$

Question Number 88973    Answers: 0   Comments: 0

∫((3tan(x))/(2sin^2 (x)+5cos^2 (x)+sec(x)))dx

$$\int\frac{\mathrm{3}{tan}\left({x}\right)}{\mathrm{2}{sin}^{\mathrm{2}} \left({x}\right)+\mathrm{5}{cos}^{\mathrm{2}} \left({x}\right)+{sec}\left({x}\right)}{dx} \\ $$

Question Number 88970    Answers: 0   Comments: 0

Question Number 88967    Answers: 1   Comments: 1

Question Number 88963    Answers: 1   Comments: 1

How far can a cyclist travel in 4 h if his average speed is 11.5 km/h ?

$$\mathrm{How}\:\mathrm{far}\:\mathrm{can}\:\mathrm{a}\:\mathrm{cyclist}\:\mathrm{travel}\:\mathrm{in}\:\mathrm{4}\:\mathrm{h} \\ $$$$\mathrm{if}\:\mathrm{his}\:\mathrm{average}\:\mathrm{speed}\:\mathrm{is}\:\mathrm{11}.\mathrm{5}\:\mathrm{km}/\mathrm{h}\:? \\ $$

Question Number 88961    Answers: 0   Comments: 3

One care travels due east at 40 km/h. and second car travels north at 40 km/h. are their velocities equal?

$$\mathrm{One}\:\mathrm{care}\:\mathrm{travels}\:\mathrm{due}\:\mathrm{east}\:\mathrm{at}\:\mathrm{40}\:\mathrm{km}/\mathrm{h}. \\ $$$$\mathrm{and}\:\mathrm{second}\:\mathrm{car}\:\mathrm{travels}\:\mathrm{north}\:\mathrm{at}\: \\ $$$$\mathrm{40}\:\mathrm{km}/\mathrm{h}.\:\mathrm{are}\:\mathrm{their}\:\mathrm{velocities}\:\mathrm{equal}? \\ $$

Question Number 88955    Answers: 0   Comments: 10

hello floor function ∫_a ^b ⌊x⌋ dx a,b∈z and b>a =∫_0 ^b ⌊x⌋ dx −∫_0 ^a ⌊x⌋ dx =((b^2 −b)/2)−((a^2 −a)/2) ....(1) now ∫_m ^k ⌊x⌋ dx when m,k∉z when m<a<b<k b=[k] and a=[m] ∫_m ^a ⌊x⌋dx +∫_a ^b ⌊x⌋dx +∫_b ^k ⌊x⌋dx =(a−m)⌊m⌋+((b^2 −b)/2)−((a^2 −a)/2)+(k−b)⌊k⌋ =(⌊m⌋−m)⌊m⌋+((⌊k⌋^2 −⌊k⌋)/2)−((⌊m⌋^2 −⌊m⌋)/2)+(k−⌊k⌋)⌊k⌋ ⌊m⌋^2 −m⌊m⌋ +(1/2)⌊k⌋^2 −(1/2)⌊k⌋−(1/2)⌊m⌋^2 −(1/2)⌊m⌋+k⌊k⌋−⌊k⌋^2 =k⌊k⌋−m⌊m⌋+(1/2)⌊m⌋^2 −(1/2)⌊k⌋^2 +(1/2)⌊m⌋−(1/2)⌊k⌋ ∴∫_m ^k ⌊x⌋dx=k⌊k⌋−m⌊m⌋+(1/2)(⌊m⌋−⌊k⌋)(⌊m⌋+⌊k⌋)+1 example ∫_(−1.5) ^(3.7) ⌊x⌋dx=(3.7)3−((−1.5)(−2))+(1/2)(−2−3)(−2+3+1) =11.1−3−5=3.1

$${hello}\: \\ $$$${floor}\:{function} \\ $$$$\int_{{a}} ^{{b}} \lfloor{x}\rfloor\:\:{dx}\:\:\:\:\:\:\:\:\:\:{a},{b}\in{z}\:\:\:{and}\:{b}>{a} \\ $$$$=\int_{\mathrm{0}} ^{{b}} \lfloor{x}\rfloor\:{dx}\:−\int_{\mathrm{0}} ^{{a}} \lfloor{x}\rfloor\:{dx}\:=\frac{{b}^{\mathrm{2}} −{b}}{\mathrm{2}}−\frac{{a}^{\mathrm{2}} −{a}}{\mathrm{2}}\:....\left(\mathrm{1}\right) \\ $$$${now} \\ $$$$\int_{{m}} ^{{k}} \lfloor{x}\rfloor\:{dx}\:\:\:{when}\:{m},{k}\notin{z}\:\:\:\:{when}\:{m}<{a}<{b}<{k} \\ $$$${b}=\left[{k}\right]\:\:\:\:{and}\:{a}=\left[{m}\right] \\ $$$$\int_{{m}} ^{{a}} \lfloor{x}\rfloor{dx}\:+\int_{{a}} ^{{b}} \lfloor{x}\rfloor{dx}\:+\int_{{b}} ^{{k}} \lfloor{x}\rfloor{dx} \\ $$$$=\left({a}−{m}\right)\lfloor{m}\rfloor+\frac{{b}^{\mathrm{2}} −{b}}{\mathrm{2}}−\frac{{a}^{\mathrm{2}} −{a}}{\mathrm{2}}+\left({k}−{b}\right)\lfloor{k}\rfloor \\ $$$$=\left(\lfloor{m}\rfloor−{m}\right)\lfloor{m}\rfloor+\frac{\lfloor{k}\rfloor^{\mathrm{2}} −\lfloor{k}\rfloor}{\mathrm{2}}−\frac{\lfloor{m}\rfloor^{\mathrm{2}} −\lfloor{m}\rfloor}{\mathrm{2}}+\left({k}−\lfloor{k}\rfloor\right)\lfloor{k}\rfloor \\ $$$$\lfloor{m}\rfloor^{\mathrm{2}} −{m}\lfloor{m}\rfloor\:+\frac{\mathrm{1}}{\mathrm{2}}\lfloor{k}\rfloor^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\lfloor{k}\rfloor−\frac{\mathrm{1}}{\mathrm{2}}\lfloor{m}\rfloor^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\lfloor{m}\rfloor+{k}\lfloor{k}\rfloor−\lfloor{k}\rfloor^{\mathrm{2}} \\ $$$$={k}\lfloor{k}\rfloor−{m}\lfloor{m}\rfloor+\frac{\mathrm{1}}{\mathrm{2}}\lfloor{m}\rfloor^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\lfloor{k}\rfloor^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\lfloor{m}\rfloor−\frac{\mathrm{1}}{\mathrm{2}}\lfloor{k}\rfloor \\ $$$$\therefore\int_{{m}} ^{{k}} \lfloor{x}\rfloor{dx}={k}\lfloor{k}\rfloor−{m}\lfloor{m}\rfloor+\frac{\mathrm{1}}{\mathrm{2}}\left(\lfloor{m}\rfloor−\lfloor{k}\rfloor\right)\left(\lfloor{m}\rfloor+\lfloor{k}\rfloor\right)+\mathrm{1} \\ $$$${example} \\ $$$$\int_{−\mathrm{1}.\mathrm{5}} ^{\mathrm{3}.\mathrm{7}} \lfloor{x}\rfloor{dx}=\left(\mathrm{3}.\mathrm{7}\right)\mathrm{3}−\left(\left(−\mathrm{1}.\mathrm{5}\right)\left(−\mathrm{2}\right)\right)+\frac{\mathrm{1}}{\mathrm{2}}\left(−\mathrm{2}−\mathrm{3}\right)\left(−\mathrm{2}+\mathrm{3}+\mathrm{1}\right) \\ $$$$=\mathrm{11}.\mathrm{1}−\mathrm{3}−\mathrm{5}=\mathrm{3}.\mathrm{1} \\ $$$$ \\ $$$$ \\ $$

Question Number 88951    Answers: 1   Comments: 0

∫_a ^b ⌈x⌉ dx=? a,b∈R ⌈..⌉ is ceil

$$\int_{{a}} ^{{b}} \lceil{x}\rceil\:{dx}=? \\ $$$${a},{b}\in{R}\:\:\:\:\:\:\: \\ $$$$ \\ $$$$\lceil..\rceil\:{is}\:{ceil}\: \\ $$

Question Number 88943    Answers: 0   Comments: 2

Question Number 88940    Answers: 1   Comments: 0

Question Number 88937    Answers: 0   Comments: 0

Question Number 88936    Answers: 1   Comments: 0

Question Number 88933    Answers: 0   Comments: 5

Question Number 88930    Answers: 0   Comments: 2

find A_λ =∫_0 ^∞ ((cos(λx))/((x^2 −x+1)^2 ))dx with λ>0 2)find the value of ∫_0 ^∞ ((cos(3x))/((x^2 −x+1)^2 ))dx

$${find}\:{A}_{\lambda} =\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cos}\left(\lambda{x}\right)}{\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)^{\mathrm{2}} }{dx}\:{with}\:\lambda>\mathrm{0} \\ $$$$\left.\mathrm{2}\right){find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cos}\left(\mathrm{3}{x}\right)}{\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$

Question Number 88929    Answers: 0   Comments: 1

cakculate ∫_0 ^∞ ((arctan(ch(x)))/(4+x^2 ))dx

$${cakculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left({ch}\left({x}\right)\right)}{\mathrm{4}+{x}^{\mathrm{2}} }{dx} \\ $$

Question Number 88928    Answers: 0   Comments: 4

calculate ∫_0 ^∞ (dx/((x^4 +x^2 +3)^2 ))

$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\left({x}^{\mathrm{4}} \:\:+{x}^{\mathrm{2}} \:\:+\mathrm{3}\right)^{\mathrm{2}} } \\ $$

Question Number 88927    Answers: 0   Comments: 0

calculate ∫_0 ^∞ ((sin(∣arctanx∣))/(x^2 +1))dx

$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{sin}\left(\mid{arctanx}\mid\right)}{{x}^{\mathrm{2}} \:+\mathrm{1}}{dx} \\ $$

Question Number 88921    Answers: 2   Comments: 0

find x,y x−2y−(√(xy))=0 (√(x−1))−(√(2y−1))=1

$${find}\:{x},{y} \\ $$$${x}−\mathrm{2}{y}−\sqrt{{xy}}=\mathrm{0} \\ $$$$\sqrt{{x}−\mathrm{1}}−\sqrt{\mathrm{2}{y}−\mathrm{1}}=\mathrm{1} \\ $$

Question Number 88918    Answers: 0   Comments: 10

Question Number 88902    Answers: 1   Comments: 4

∫_(1/e) ^e ln∣x∣ dx

$$\int_{\frac{\mathrm{1}}{{e}}} ^{{e}} {ln}\mid{x}\mid\:{dx} \\ $$

Question Number 88899    Answers: 0   Comments: 14

Simplify ((((35+18i(√3)))^(1/3) +((35−18i(√3)))^(1/3) −4)/3)

$${Simplify} \\ $$$$\frac{\sqrt[{\mathrm{3}}]{\mathrm{35}+\mathrm{18}{i}\sqrt{\mathrm{3}}}+\sqrt[{\mathrm{3}}]{\mathrm{35}−\mathrm{18}{i}\sqrt{\mathrm{3}}}−\mathrm{4}}{\mathrm{3}} \\ $$

Question Number 88896    Answers: 1   Comments: 2

find lim_(x→0) ((ln(sin(3x)+cos(3x)))/(ln(sin(x)+cos(x))))

$${find}\: \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{{ln}\left({sin}\left(\mathrm{3}{x}\right)+{cos}\left(\mathrm{3}{x}\right)\right)}{{ln}\left({sin}\left({x}\right)+{cos}\left({x}\right)\right)} \\ $$

Question Number 88892    Answers: 0   Comments: 2

Question Number 88894    Answers: 0   Comments: 1

Question Number 88889    Answers: 0   Comments: 1

Question Number 88886    Answers: 1   Comments: 5

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