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Question Number 75822 Answers: 0 Comments: 0
Question Number 75821 Answers: 0 Comments: 1
Question Number 75818 Answers: 1 Comments: 3
$${Given}\:\:{the}\:\:{increasing}\:\:{sequence}\:: \\ $$$$\mathrm{1},\:\mathrm{4},\:\mathrm{8},\:\mathrm{13},\:... \\ $$$${a}.\:{Find}\:\:{U}_{\mathrm{2019}} \\ $$$${b}.\:{Find}\:\:{S}_{\mathrm{2019}} \\ $$$${U}_{{n}} \:\:{is}\:\:{nth}−{term}\:\:{of}\:\:{the}\:\:{sequence} \\ $$$${S}_{{n}} \:\:{is}\:\:{sum}\:\:{of}\:\:{n}\:−\:{term}\:\:{of}\:\:{the}\:\:{sequence} \\ $$$${Arithmetic}\:\:{Sequence}\:\:{Degree}\:\:{Two} \\ $$
Question Number 75814 Answers: 1 Comments: 1
Question Number 75809 Answers: 0 Comments: 2
Question Number 75802 Answers: 1 Comments: 0
$$\mathrm{A}\:\mathrm{fair}\:\mathrm{die}\:\mathrm{is}\:\mathrm{thrown}\:\mathrm{4}\:\mathrm{times}.\:\mathrm{What}\:\mathrm{is}\:\mathrm{the} \\ $$$$\mathrm{probability}\:\mathrm{of}\:\mathrm{obtaining}\:\mathrm{a}\:\mathrm{6}\:\mathrm{twice}? \\ $$
Question Number 75796 Answers: 1 Comments: 0
$$\mathrm{let}\:\mathrm{be}\:\mathrm{a},\mathrm{b}\:\mathrm{such}\:\mathrm{as}\:\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} =\mathrm{ab} \\ $$$$\mathrm{find}\:\:\mathrm{out}\:\mathrm{Z}=\frac{\mathrm{a}^{\mathrm{n}} +\mathrm{b}^{\mathrm{n}} }{\mathrm{a}^{\mathrm{n}} −\mathrm{b}^{\mathrm{n}} }\:\:\mathrm{when}\:\:\mathrm{a}\neq\mathrm{0} \\ $$$$ \\ $$
Question Number 75795 Answers: 0 Comments: 1
$$\mathrm{Find}\:\mathrm{out}\:\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{argsh}\left(\mathrm{x}\right)}{\mathrm{x}}\mathrm{dx} \\ $$
Question Number 75793 Answers: 1 Comments: 1
$$\mathrm{Prove}\:\mathrm{that}\:\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{arctanx}}{\mathrm{x}\sqrt{\mathrm{log2}}}\right)^{\mathrm{2}} \mathrm{dx}=\:\pi \\ $$
Question Number 75783 Answers: 1 Comments: 1
Question Number 75778 Answers: 1 Comments: 0
$${if}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={p},\:{x}^{\mathrm{3}} +{y}^{\mathrm{3}} ={q}, \\ $$$${find}\:{x}^{{n}} +{y}^{{n}} \:{in}\:{terms}\:{of}\:{p},\:{q}\:{and}\:{n}. \\ $$$$\left({n}\geqslant\mathrm{4}\right) \\ $$
Question Number 75773 Answers: 1 Comments: 0
Question Number 75772 Answers: 0 Comments: 0
Question Number 75771 Answers: 0 Comments: 2
Question Number 75770 Answers: 3 Comments: 6
Question Number 75769 Answers: 0 Comments: 0
Question Number 75768 Answers: 1 Comments: 0
Question Number 75763 Answers: 2 Comments: 0
$$\mathrm{please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{it}\:\mathrm{in}\:\mathbb{R} \\ $$$$\mathrm{3cosx}−\sqrt{\mathrm{3}}\mathrm{sinx}+\sqrt{\mathrm{6}}=\mathrm{0} \\ $$
Question Number 75762 Answers: 0 Comments: 0
$${if}\:\:{y}\:{cos}\left({x}\right)+{x}\:{cos}\left({y}\right)\:=\:\pi \\ $$$$ \\ $$$${find}\:\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:\:\:,\:{x}=\mathrm{0} \\ $$
Question Number 75756 Answers: 0 Comments: 1
$$\underset{{x}=\mathrm{1}} {\overset{\mathrm{99}} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{2}{x}+\mathrm{1}}\right) \\ $$
Question Number 75753 Answers: 1 Comments: 0
$$\left(\frac{\left(\mathrm{x}^{\mathrm{1}/\mathrm{3}} +\mathrm{x}^{−\mathrm{1}/\mathrm{3}} \right)^{\mathrm{2}} −\mathrm{2}}{\left(\mathrm{x}^{\mathrm{1}/\mathrm{3}} +\mathrm{x}^{−\mathrm{1}/\mathrm{3}} \right)^{\mathrm{2}} +\mathrm{2}}−\mathrm{x}\right)^{\mathrm{3}/\mathrm{4}} \\ $$
Question Number 75751 Answers: 0 Comments: 0
Question Number 75750 Answers: 0 Comments: 2
Question Number 75743 Answers: 2 Comments: 3
Question Number 75742 Answers: 0 Comments: 0
$$\mathrm{plz}\:\mathrm{solve}\:\mathrm{this}\:\mathrm{my}\:\mathrm{handsome}\:\mathrm{guys}\:\mathrm{and}\:\mathrm{sisters}... \\ $$$$\mathrm{complex}\:\mathrm{integral}\:\int_{−\infty} ^{+\infty} \:\frac{{e}^{\boldsymbol{{i}}{t}} }{\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}\mathrm{d}{t} \\ $$
Question Number 75741 Answers: 0 Comments: 0
$$\mathrm{solve}\:\mathrm{complex}\:\mathrm{integral} \\ $$$$\int_{−\infty} ^{+\infty} \:\frac{{e}^{\boldsymbol{{i}}{t}} }{\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}\mathrm{d}{t}=???? \\ $$$$\mathrm{plz}.......\mathrm{help}\:\mathrm{me}...\mathrm{T}\frown\mathrm{T}\:\:\: \\ $$$$\mathrm{my}\:\mathrm{handsome}\:\mathrm{brothers}\:\mathrm{and}\:\mathrm{sisters}... \\ $$
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