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Question Number 196972 Answers: 1 Comments: 0
Question Number 196971 Answers: 1 Comments: 0
$$\:\:{solve}\:\begin{cases}{\mathrm{3}{x}^{\mathrm{2}} −\mathrm{9}{y}=\mathrm{1}}\\{\mathrm{3}{y}^{\mathrm{2}} −\mathrm{9}{x}=\mathrm{0}}\end{cases} \\ $$
Question Number 196968 Answers: 2 Comments: 0
Question Number 196965 Answers: 1 Comments: 0
Question Number 196964 Answers: 2 Comments: 0
Question Number 196959 Answers: 2 Comments: 0
$$\:\:\:\:\: \\ $$$$ \underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{sin}\:\mathrm{2x}}{\mathrm{x}^{\mathrm{3}} }\:+\:\frac{\mathrm{a}}{\mathrm{x}^{\mathrm{2}} }\:+\:\mathrm{b}\right)\:=\:\mathrm{1} \\ $$
Question Number 196957 Answers: 1 Comments: 0
Question Number 196955 Answers: 0 Comments: 1
Question Number 196954 Answers: 1 Comments: 0
Question Number 196950 Answers: 1 Comments: 0
$$\mathrm{Prove}\:\mathrm{that}\:\underset{\:\mathrm{0}} {\int}^{\:\frac{\pi}{\mathrm{2}}} \frac{\mathrm{ln}\left(\mathrm{1}+\alpha\mathrm{sin}{t}\right)}{\mathrm{sin}{t}}{dt}=\:\frac{\pi^{\mathrm{2}} }{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{arccos}\alpha\right)^{\mathrm{2}} \\ $$
Question Number 196946 Answers: 2 Comments: 0
Question Number 196940 Answers: 1 Comments: 0
Question Number 196938 Answers: 1 Comments: 0
Question Number 196953 Answers: 1 Comments: 0
Question Number 196934 Answers: 1 Comments: 0
Question Number 196929 Answers: 1 Comments: 0
Question Number 196928 Answers: 0 Comments: 1
$${I}_{{n}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{{n}} {x}\:{dx} \\ $$
Question Number 196918 Answers: 1 Comments: 0
Question Number 196917 Answers: 3 Comments: 0
Question Number 196916 Answers: 0 Comments: 0
Question Number 196915 Answers: 0 Comments: 1
Question Number 196914 Answers: 2 Comments: 0
Question Number 196913 Answers: 1 Comments: 0
$${soit}\:\left\{_{{r}_{{n}+\mathrm{1}} ={r}_{{n}} /\left(\mathrm{2}+{r}_{{n}} ^{\mathrm{2}} \right)} ^{{r}_{\mathrm{0}} =\mathrm{1}} \right. \\ $$$${demontrer}\:{sans}\:{recurrence}\:{que}\:{r}_{{n}} >\mathrm{0} \\ $$$${demontrer}\:{par}\:{recurrence}\:{que}\:{r}_{{n}+\mathrm{1}} \leq\frac{\mathrm{1}}{\mathrm{2}}{r}_{{n}} \\ $$$${demontrer}\:{sans}\:{recurrence}\:{que}\:{r}_{{n}} \leq\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}} \\ $$$$\bullet{erly}\:{rolvinst}\bullet \\ $$
Question Number 196910 Answers: 0 Comments: 0
Question Number 196902 Answers: 1 Comments: 0
Question Number 196893 Answers: 1 Comments: 0
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