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Question Number 208968 by MWSuSon last updated on 29/Jun/24 $$ \\ $$$$\mathrm{hello}\:\mathrm{everyone}.\:\mathrm{Im}\:\mathrm{writing}\:\mathrm{a}\:\mathrm{project}\:\:\mathrm{on}\:\mathrm{the}\:\mathrm{topic}“\:\mathrm{Solution}\:\mathrm{of} \\ $$$$\mathrm{Nonlinear}\:\mathrm{partial}\:\mathrm{differential}\:\mathrm{equation}\:\mathrm{using}\:\mathrm{charpits}\:\mathrm{methods}'' \\ $$$$\mathrm{curently}\:\mathrm{writing}\:\mathrm{chapter}\:\mathrm{2}\:\mathrm{but}\:\mathrm{I}'\mathrm{m}\:\mathrm{a}\:\mathrm{little}\:\mathrm{bit}\:\mathrm{confused}\:\mathrm{about}\:\mathrm{what}\:\mathrm{ih} \\ $$$$\mathrm{should}\:\mathrm{include}\:\mathrm{in}\:\mathrm{my}\:\mathrm{theoretical}\:\mathrm{framework}.\:\mathrm{Any}\:\mathrm{ideas}? \\ $$ Terms of Service Privacy Policy…
Question Number 208969 by Spillover last updated on 29/Jun/24 Answered by Spillover last updated on 30/Jun/24 Answered by Spillover last updated on 30/Jun/24 Answered by…
Question Number 208970 by Spillover last updated on 29/Jun/24 Answered by Spillover last updated on 30/Jun/24 Answered by Spillover last updated on 30/Jun/24 Answered by…
Question Number 208962 by yaslm last updated on 29/Jun/24 Commented by Rasheed.Sindhi last updated on 29/Jun/24 $${Are}\:{all}\:{the}\:{digits}\:{in}\:{divisor},{dividend} \\ $$$${and}\:{quotient}\:{different}? \\ $$$${Otherwise}\:{many}\:{many}\:{answers}. \\ $$ Commented by…
Question Number 208940 by efronzo1 last updated on 28/Jun/24 Answered by kapoorshah last updated on 28/Jun/24 $${r}\:=\:{OD}\:=\:\sqrt{\mathrm{10}} \\ $$$${BE}\:=\:\sqrt{\mathrm{15}} \\ $$$${BOE}\:\sim\:{BCA} \\ $$$$\frac{{BO}}{{BE}}\:=\:\frac{{BC}}{{AB}} \\ $$$$\frac{\sqrt{\mathrm{10}}}{\:\sqrt{\mathrm{15}}}\:=\:\frac{{BC}}{\mathrm{2}\sqrt{\mathrm{10}}}\:\:\Rightarrow\:{BC}\:=\:\frac{\mathrm{20}}{\:\sqrt{\mathrm{15}}}\:=\:\mathrm{5}.\mathrm{164}…
Question Number 208959 by hardmath last updated on 28/Jun/24 $$\mathrm{If}\:\:\:\boldsymbol{\mathrm{z}}\:=\:−\:\frac{\mathrm{1}}{\mathrm{2}}\:\:+\:\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\boldsymbol{\mathrm{i}} \\ $$$$\mathrm{Find}\:\:\:\left(\mathrm{z}^{\mathrm{4}} \:+\:\mathrm{2z}\right)\centerdot\left(\mathrm{z}^{\mathrm{3}} \:+\:\mathrm{z}\right)\:=\:? \\ $$ Answered by grigoriy last updated on 29/Jun/24 $$ \\…
Question Number 208943 by MWSuSon last updated on 28/Jun/24 Answered by A5T last updated on 28/Jun/24 $${a}.\:\frac{\left[{ADM}\right]}{\left[{ABCD}\right]}=\frac{\frac{\mathrm{1}}{\mathrm{2}}×{AD}×{DM}}{{AD}×\left({DM}×\mathrm{2}\right)}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${b}.\:{Let}\:{AM}\:{and}\:{DN}\:{intersect}\:{at}\:{E} \\ $$$$\frac{\left[{DEM}\right]}{\left[{ABCD}\right]}=\frac{\frac{\mathrm{1}}{\mathrm{2}}×\frac{{AD}}{\mathrm{2}}×{DM}}{{AD}×{DM}×\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{8}} \\ $$$${c}.\:\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{8}}=\frac{\mathrm{5}}{\mathrm{8}} \\ $$…