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Author: Tinku Tara

Solve-for-all-integer-x-y-Z-x-2-y-2-19-

Question Number 22561 by Bruce Lee last updated on 20/Oct/17 $$\boldsymbol{\mathrm{Solve}}\:\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{all}}\:\boldsymbol{\mathrm{integer}}\:\boldsymbol{{x}},\boldsymbol{{y}}\:\in\boldsymbol{{Z}} \\ $$$$\:\:\:\:\:\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{2}} =\mathrm{19} \\ $$ Answered by $@ty@m last updated on 20/Oct/17 $${Case}−\mathrm{1}.\:{Let}\:{x}=\mathrm{0}…

dx-2x-3-2-3-

Question Number 88097 by sahnaz last updated on 08/Apr/20 $$\int\frac{\mathrm{dx}}{\left(\mathrm{2x}−\mathrm{3}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} } \\ $$ Answered by john santu last updated on 08/Apr/20 $$\int\:{t}^{−\mathrm{2}/\mathrm{3}} ×\frac{\mathrm{1}}{\mathrm{2}}{dt}\:\:\:\left[\:{t}\:=\:\mathrm{2}{x}−\mathrm{3}\:\right] \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{3}\:×\:\sqrt[{\mathrm{3}\:\:}]{{t}}\:+\:{c}\:…

Question-153629

Question Number 153629 by mathdanisur last updated on 08/Sep/21 Commented by benhamimed last updated on 09/Sep/21 $$\mathrm{1}+{x}+…+{x}^{{n}} =\frac{\mathrm{1}−{x}^{{n}} }{\mathrm{1}−{x}}=\frac{\frac{{t}^{{n}} −\mathrm{1}}{{t}^{{n}} }}{\frac{{t}−\mathrm{1}}{{t}}}=\frac{\left({t}^{{n}} −\mathrm{1}\right)}{{t}^{{n}−\mathrm{1}} \left({t}−\mathrm{1}\right)}\:\:\:\:{tq}\:{x}=\frac{\mathrm{1}}{{t}} \\ $$$${si}\:{x}=\frac{\mathrm{1}}{\mathrm{2021}}\:\:;{t}=\mathrm{2021}…

A-ball-is-thrown-vertically-upwards-Its-hight-h-meters-at-time-t-seconds-is-given-by-h-5-30t-5t-2-Find-the-maximum-hight-reached-by-the-ball-

Question Number 153628 by otchereabdullai@gmail.com last updated on 08/Sep/21 $$\mathrm{A}\:\mathrm{ball}\:\mathrm{is}\:\mathrm{thrown}\:\mathrm{vertically}\:\mathrm{upwards}. \\ $$$$\mathrm{Its}\:\mathrm{hight}\left(\mathrm{h}\right)\mathrm{meters}\:\mathrm{at}\:\mathrm{time}\left(\:\mathrm{t}\right)\mathrm{seconds} \\ $$$$\mathrm{is}\:\mathrm{given}\:\mathrm{by}\:\mathrm{h}=\:\mathrm{5}+\mathrm{30t}−\mathrm{5t}^{\mathrm{2}} .\:\mathrm{Find}\:\mathrm{the} \\ $$$$\mathrm{maximum}\:\mathrm{hight}\:\mathrm{reached}\:\mathrm{by}\:\mathrm{the}\:\mathrm{ball}. \\ $$ Answered by mr W last updated…