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Question Number 89063    Answers: 0   Comments: 0

let f(x) = x^3 + 2x^2 + 3x + 4 find the region enclosed by f ′, f ′′ and f ′′′

$$\:\mathrm{let}\:{f}\left({x}\right)\:=\:{x}^{\mathrm{3}} \:+\:\mathrm{2}{x}^{\mathrm{2}} \:+\:\mathrm{3}{x}\:+\:\mathrm{4} \\ $$$$\:\mathrm{find}\:\mathrm{the}\:\mathrm{region}\:\mathrm{enclosed}\:\mathrm{by}\:{f}\:',\:{f}\:''\:\mathrm{and}\:{f}\:''' \\ $$

Question Number 89089    Answers: 1   Comments: 0

16^(3/4) +2

$$\mathrm{16}^{\mathrm{3}/\mathrm{4}} \:+\mathrm{2} \\ $$

Question Number 89052    Answers: 0   Comments: 4

∫_0 ^(π/2) (x/(tan(x)))dx

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{x}}{{tan}\left({x}\right)}{dx} \\ $$

Question Number 89047    Answers: 1   Comments: 1

cos x+sin x=(4/5) 5sin x = ?

$$\mathrm{cos}\:{x}+\mathrm{sin}\:{x}=\frac{\mathrm{4}}{\mathrm{5}} \\ $$$$\mathrm{5sin}\:{x}\:=\:? \\ $$

Question Number 89086    Answers: 0   Comments: 1

show that (((2/3)−((5(√(33)))/(18))))^(1/3) +(((2/3)+((5(√(33)))/(18))))^(1/3) =1

$${show}\:{that} \\ $$$$\sqrt[{\mathrm{3}}]{\frac{\mathrm{2}}{\mathrm{3}}−\frac{\mathrm{5}\sqrt{\mathrm{33}}}{\mathrm{18}}}+\sqrt[{\mathrm{3}}]{\frac{\mathrm{2}}{\mathrm{3}}+\frac{\mathrm{5}\sqrt{\mathrm{33}}}{\mathrm{18}}}=\mathrm{1} \\ $$

Question Number 89070    Answers: 0   Comments: 0

(((√(a^2 +(b^2 /x^2 )))+(√(a^2 +b^2 x^2 )))/((√(a^2 +(b^2 /x^2 )))−(√(a^2 +b^2 x^2 )))) = ((a^2 /b^2 ))x x∈R^+ ; Express x in terms of a^2 , b^2 .

$$\frac{\sqrt{{a}^{\mathrm{2}} +\frac{{b}^{\mathrm{2}} }{{x}^{\mathrm{2}} }}+\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} {x}^{\mathrm{2}} }}{\sqrt{{a}^{\mathrm{2}} +\frac{{b}^{\mathrm{2}} }{{x}^{\mathrm{2}} }}−\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} {x}^{\mathrm{2}} }}\:=\:\left(\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\right){x} \\ $$$${x}\in\mathbb{R}^{+} \:\:;\:\:{Express}\:{x}\:{in}\:{terms}\:{of} \\ $$$${a}^{\mathrm{2}} ,\:{b}^{\mathrm{2}} \:. \\ $$

Question Number 89029    Answers: 0   Comments: 1

Question Number 89038    Answers: 1   Comments: 0

solve log_x (x−3)=log_x (5−x)

$${solve} \\ $$$${log}_{{x}} \left({x}−\mathrm{3}\right)={log}_{{x}} \left(\mathrm{5}−{x}\right) \\ $$

Question Number 89033    Answers: 1   Comments: 0

Question Number 89032    Answers: 0   Comments: 0

find by using de moivre′s formula cos(2°)=?

$${find}\:{by}\:{using}\:{de}\:{moivre}'{s}\:{formula} \\ $$$${cos}\left(\mathrm{2}°\right)=? \\ $$

Question Number 89010    Answers: 2   Comments: 4

Question Number 89025    Answers: 1   Comments: 0

∫((3^x +4^x )/5^x )dx

$$\int\frac{\mathrm{3}^{{x}} +\mathrm{4}^{{x}} }{\mathrm{5}^{{x}} }{dx} \\ $$

Question Number 88997    Answers: 0   Comments: 5

Question Number 89007    Answers: 1   Comments: 0

Question Number 88985    Answers: 0   Comments: 9

lim_(x→3) ln ∣x−3∣ exist or no?

$$\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\:\mathrm{ln}\:\mid{x}−\mathrm{3}\mid\: \\ $$$${exist}\:{or}\:{no}? \\ $$

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

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