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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 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 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 208943 by MWSuSon last updated on 28/Jun/24 Answered by A5T last updated on 28/Jun/24 $${a}.\:\frac{\left[{ADM}\right]}{\left[{ABCD}\right]}=\frac{\frac{\mathrm{1}}{\mathrm{2}}×{AD}×{DM}}{{AD}×\left({DM}×\mathrm{2}\right)}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${b}.\:{Let}\:{AM}\:{and}\:{DN}\:{intersect}\:{at}\:{E} \\ $$$$\frac{\left[{DEM}\right]}{\left[{ABCD}\right]}=\frac{\frac{\mathrm{1}}{\mathrm{2}}×\frac{{AD}}{\mathrm{2}}×{DM}}{{AD}×{DM}×\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{8}} \\ $$$${c}.\:\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{8}}=\frac{\mathrm{5}}{\mathrm{8}} \\ $$…
Question Number 208866 by Adeyemi889 last updated on 26/Jun/24 Commented by Adeyemi889 last updated on 26/Jun/24 $${partial}\:{F}\boldsymbol{{Raction}} \\ $$$$ \\ $$ Answered by Sutrisno last…
Question Number 208836 by Adeyemi889 last updated on 24/Jun/24 Answered by Rasheed.Sindhi last updated on 25/Jun/24 $${Ax}^{\mathrm{3}} −{A}+{Bx}^{\mathrm{2}} +{Bx}+{B}+\left({Cx}+{D}\right)\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}\right) \\ $$$$={Ax}^{\mathrm{3}} −{A}+{Bx}^{\mathrm{2}} +{Bx}+{B}+{Cx}^{\mathrm{3}} +{Dx}^{\mathrm{2}}…
Question Number 208789 by Tawa11 last updated on 23/Jun/24 $$\mathrm{Why}\:\mathrm{is}\:\mathrm{surface}\:\mathrm{tension}\:\mathrm{formula}\:\mathrm{divided}\:\mathrm{by}\:\:\mathrm{2L} \\ $$$$\mathrm{that}\:\mathrm{is},\:\:\:\:\mathrm{surface}\:\mathrm{tension}\:\:=\:\:\frac{\mathrm{F}}{\mathrm{2L}} \\ $$$$\mathrm{why}\:\mathrm{divided}\:\mathrm{by}\:\:\mathrm{2L}. \\ $$$$\mathrm{Where}\:\mathrm{did}\:\mathrm{the}\:\mathrm{2}\:\mathrm{come}\:\mathrm{from}? \\ $$$$ \\ $$$$\mathrm{Example}. \\ $$$$\mathrm{Calculate}\:\mathrm{the}\:\mathrm{force}\:\mathrm{required}\:\mathrm{to}\:\mathrm{lift}\:\mathrm{a}\:\mathrm{needle} \\ $$$$\mathrm{4cm}\:\mathrm{long}\:\mathrm{off}\:\mathrm{the}\:\mathrm{surface}\:\mathrm{of}\:\mathrm{water},\:\mathrm{if}\:\mathrm{surface} \\…
Question Number 208759 by Tawa11 last updated on 22/Jun/24 Commented by Tawa11 last updated on 22/Jun/24 $$\mathrm{Will}\:\mathrm{this}\:\mathrm{question}\:\mathrm{have}\:\mathrm{workings}? \\ $$$$\mathrm{Based}\:\mathrm{on}\:\mathrm{the}\:\mathrm{condition}\:\mathrm{in}\:\mathrm{the}\:\mathrm{question}. \\ $$$$ \\ $$$$\mathrm{Or}\:\mathrm{answer}\:\mathrm{is}\:\:\mathrm{E} \\ $$…
Question Number 208741 by Tawa11 last updated on 22/Jun/24 Answered by mr W last updated on 22/Jun/24 Commented by mr W last updated on 22/Jun/24…