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Question-96542

Question Number 96542 by Hamida last updated on 02/Jun/20 Commented by bobhans last updated on 02/Jun/20 $$\mathrm{average}\:\mathrm{rate}\:=\:\frac{\mathrm{1}}{\mathrm{3}−\left(−\mathrm{3}\right)}\:\underset{−\mathrm{3}} {\overset{\mathrm{3}} {\int}}\left(\mathrm{x}^{\mathrm{2}} −\mathrm{2}\right)\:\mathrm{dx} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{6}}\:\left[\frac{\mathrm{1}}{\mathrm{3}}\mathrm{x}^{\mathrm{3}} −\mathrm{2x}\right]_{−\mathrm{3}} ^{\mathrm{3}} \\…

Question-96538

Question Number 96538 by Hamida last updated on 02/Jun/20 Answered by john santu last updated on 02/Jun/20 $$\mathrm{y}'=\frac{\mathrm{2}.\mathrm{7}−\mathrm{2}.\mathrm{8}}{\left(\mathrm{8x}+\mathrm{7}\right)^{\mathrm{2}} }\:=\:−\frac{\mathrm{2}}{\left(\mathrm{8x}+\mathrm{7}\right)^{\mathrm{2}} } \\ $$ Commented by Hamida…

If-1-m-3-3m-3-2-0-m-1-Find-1-m-2-3m-2-

Question Number 96530 by abony1303 last updated on 02/Jun/20 $$\boldsymbol{\mathrm{If}}:\:\left(\mathrm{1}+{m}\right)^{\mathrm{3}} −\mathrm{3}{m}^{\mathrm{3}} =\mathrm{2}\:\:\:\:\:\:\:\:\mathrm{0}\leqslant{m}\leqslant\mathrm{1} \\ $$$$\boldsymbol{\mathrm{Find}}:\:\left(\mathrm{1}+{m}\right)^{\mathrm{2}} −\mathrm{3}{m}^{\mathrm{2}} \\ $$ Commented by MJS last updated on 02/Jun/20 $$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{think}\:\mathrm{there}'\mathrm{s}\:\mathrm{a}\:“\mathrm{nice}''\:\mathrm{solution}…

Question-30986

Question Number 30986 by mondodotto@gmail.com last updated on 01/Mar/18 Answered by rahul 19 last updated on 01/Mar/18 $${y}={mx}+{c}\:…\left(\mathrm{1}\right) \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={a}^{\mathrm{2}} …\left(\mathrm{2}\right)\: \\ $$$${put}\:{the}\:{value}\:{of}\:{y}\:{in}\:{eq}^{{n}}…

0-1-1-1-1-x-dx-

Question Number 96514 by Farruxjano last updated on 02/Jun/20 $$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\mathrm{1}}{\mathrm{1}+\left[\frac{\mathrm{1}}{\boldsymbol{{x}}}\right]}\boldsymbol{{dx}}=? \\ $$ Commented by Farruxjano last updated on 02/Jun/20 $$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{know}\:\mathrm{english}\:\mathrm{perfect},\:\mathrm{but}\:\mathrm{i}\:\mathrm{can}\:\mathrm{say} \\ $$$$\left[\mathrm{x}\right]−\mathrm{the}\:\mathrm{whole}\:\mathrm{part}\:\mathrm{of}\:\mathrm{x}.\:\:\:\mathrm{x}=\left[\mathrm{x}\right]+\left\{\mathrm{x}\right\} \\…

1-2014-2-2014-3-2014-1007-2014-

Question Number 162035 by naka3546 last updated on 25/Dec/21 $$\left(\underset{\mathrm{1}} {\overset{\mathrm{2014}} {\:}}\right)\:+\:\left(\underset{\mathrm{2}} {\overset{\mathrm{2014}} {\:}}\right)\:+\:\left(\underset{\mathrm{3}} {\overset{\mathrm{2014}} {\:}}\right)\:+\:\ldots+\:\left(\underset{\mathrm{1007}} {\overset{\mathrm{2014}} {\:}}\right)\:=\:? \\ $$ Answered by Rasheed.Sindhi last updated…