Question Number 140138 by henderson last updated on 04/May/21 Answered by mr W last updated on 04/May/21 Commented by mr W last updated on 04/May/21…
Question Number 74601 by ~blr237~ last updated on 27/Nov/19 $${solve}\:\:\:{y}''+\:{a}\left({x}\right){y}={b}\left({x}\right)\: \\ $$$${the}\:\:{general}\:\:{form}\:{of}\:\:{the}\:{solution}\:{if}\:\:{possible} \\ $$$${or}\:\:{juzt}\:{a}\:{solving}\:{metbod} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 74599 by mind is power last updated on 27/Nov/19 $$\mathrm{Hello},\mathrm{verry}\:\mathrm{Nice}\:\mathrm{day}\: \\ $$$$\mathrm{let}\:\mathrm{U}_{\mathrm{n}} =\mathrm{E}\left(\left(\frac{\mathrm{3}+\sqrt{\mathrm{17}}}{\mathrm{2}}\right)^{\mathrm{n}} \right),\mathrm{n}\in\mathbb{N}^{\ast} \\ $$$$\mathrm{show}\:\mathrm{that}\:\mathrm{U}_{\mathrm{n}} \equiv\mathrm{n}\left(\mathrm{2}\right) \\ $$ Terms of Service Privacy…
Question Number 9061 by tawakalitu last updated on 16/Nov/16 Commented by tawakalitu last updated on 16/Nov/16 $$\mathrm{please}\:\mathrm{help}. \\ $$ Answered by mrW last updated on…
Question Number 9060 by tawakalitu last updated on 16/Nov/16 Answered by mrW last updated on 17/Nov/16 $$\left.{b}\right) \\ $$$$\mathrm{60}^{\mathrm{2}} =\mathrm{40}^{\mathrm{2}} +\mathrm{92}^{\mathrm{2}} −\mathrm{2}×\mathrm{40}×\mathrm{92}×\mathrm{cos}\:\alpha \\ $$$$\mathrm{cos}\:\alpha=\frac{\mathrm{40}^{\mathrm{2}} +\mathrm{92}^{\mathrm{2}}…
Question Number 140129 by mathdanisur last updated on 04/May/21 $${x};{y}\in\mathbb{R}^{+} \:;\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{2} \\ $$$${proof}:\:\mathrm{3}−{xy}\geqslant\left({x}+{y}\right)\sqrt{{xy}}+\left({x}−{y}\right)^{\mathrm{2}} \geqslant\mathrm{2}{xy} \\ $$ Answered by mr W last updated on…
Question Number 74594 by ajfour last updated on 27/Nov/19 Commented by ajfour last updated on 27/Nov/19 $${Reposted}:\:{Q}.\mathrm{74557} \\ $$$${A}\:{particle}\:{of}\:{mass}\:{m}\:{tied}\:{to}\:{O} \\ $$$${with}\:{a}\:{string}\:{of}\:{length}\:{b}\:{and} \\ $$$${released}\:{at}\:{t}=\mathrm{0}\:{at}\:{the}\:{rim}\:{of} \\ $$$${a}\:{hollow}\:{hemisphere}\:{crucible}.…
Question Number 9057 by sandipkd@ last updated on 16/Nov/16 Answered by aydnmustafa1976 last updated on 16/Nov/16 $${nsin}\frac{\mathrm{1}}{{n}}={lim}\frac{{sin}\frac{\mathrm{1}}{{n}}}{\frac{\mathrm{1}}{{n}}}={lim}\frac{{sint}}{{t}}=\mathrm{1}\:{therefore}\:\:\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}{dx}=\mathrm{4}.{arctgx}\mid_{\mathrm{0}} ^{\mathrm{1}} =\mathrm{4}\left(\frac{\Pi}{\mathrm{4}}−\mathrm{0}\right)=\Pi \\ $$ Commented…
Question Number 74590 by Aditya789 last updated on 27/Nov/19 Answered by MJS last updated on 27/Nov/19 $${n}=\mathrm{7} \\ $$$$\mathrm{the}\:\mathrm{coefficients}\:\mathrm{then}\:\mathrm{are}\:\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}\:\mathrm{with}\:{n}=\mathrm{7};\:\mathrm{0}\leqslant{k}\leqslant\mathrm{7} \\ $$$$\begin{pmatrix}{\mathrm{7}}\\{\mathrm{0}}\end{pmatrix};\:\begin{pmatrix}{\mathrm{7}}\\{\mathrm{1}}\end{pmatrix};\:\begin{pmatrix}{\mathrm{7}}\\{\mathrm{2}}\end{pmatrix};\:\begin{pmatrix}{\mathrm{7}}\\{\mathrm{3}}\end{pmatrix};\:\begin{pmatrix}{\mathrm{7}}\\{\mathrm{4}}\end{pmatrix};\:\begin{pmatrix}{\mathrm{7}}\\{\mathrm{5}}\end{pmatrix};\:\begin{pmatrix}{\mathrm{7}}\\{\mathrm{6}}\end{pmatrix};\:\begin{pmatrix}{\mathrm{7}}\\{\mathrm{7}}\end{pmatrix} \\ $$$$\mathrm{1}\:\mathrm{7}\:\mathrm{21}\:\mathrm{35}\:\mathrm{35}\:\mathrm{21}\:\mathrm{7}\:\mathrm{1} \\ $$…
Question Number 74591 by Aditya789 last updated on 27/Nov/19 Terms of Service Privacy Policy Contact: info@tinkutara.com