Question Number 208861 by mokys last updated on 25/Jun/24 Commented by mokys last updated on 25/Jun/24 $${solve}\:{this} \\ $$ Commented by mokys last updated on…
Question Number 208862 by THORR last updated on 25/Jun/24 $${x}=\:{b}^{\mathrm{2}} \: \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 208852 by Tawa11 last updated on 25/Jun/24 Answered by Frix last updated on 25/Jun/24 $$\mathrm{3}^{−\mathrm{3}\left({x}^{\mathrm{2}} −{x}\right)} ={x}^{\mathrm{2}} −{x}\:\Rightarrow\:{x}^{\mathrm{2}} −{x}>\mathrm{0}\:\Leftrightarrow\:{x}<\mathrm{0}\vee{x}>\mathrm{1} \\ $$$${t}={x}^{\mathrm{2}} −{x}\:\Leftrightarrow\:{x}=\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{4}{t}+\mathrm{1}}}{\mathrm{2}} \\…
Question Number 208855 by efronzo1 last updated on 25/Jun/24 Answered by mr W last updated on 25/Jun/24 $${let}\:{t}=\mathrm{1}+\mid{x}\mid\geqslant\mathrm{1} \\ $$$$\frac{\mathrm{2024}^{−{t}} −\lambda}{\mathrm{2024}^{−{t}} −\lambda^{−\mathrm{1}} }=\lambda\:\mathrm{2024}^{{t}} \\ $$$$\mathrm{2024}^{−{t}}…
Question Number 208866 by Adeyemi889 last updated on 26/Jun/24 Commented by Adeyemi889 last updated on 26/Jun/24 $${partial}\:{F}\boldsymbol{{Raction}} \\ $$$$ \\ $$ Answered by Sutrisno last…
Question Number 208828 by efronzo1 last updated on 24/Jun/24 Answered by mr W last updated on 24/Jun/24 $${BM}={MC}={a},\:{say} \\ $$$${AM}={DM}=\mathrm{2}{a} \\ $$$$\frac{{BM}}{\mathrm{6}}=\frac{{MQ}}{\mathrm{4}}\:\Rightarrow{MQ}=\frac{\mathrm{2}{a}}{\mathrm{3}}\:\Rightarrow{QD}=\frac{\mathrm{4}{a}}{\mathrm{3}} \\ $$$$\left(\mathrm{6}+\mathrm{4}\right)^{\mathrm{2}} ={a}^{\mathrm{2}}…
Question Number 208842 by NasaSara last updated on 24/Jun/24 $${does}\:{the}\:{rule}\:{of}\:{odd}\:{and}\:{even}\:{functions}\: \\ $$$${can}\:{be}\:{applied}\:{with}\:{improper}\:{integration}? \\ $$$${I}=\int_{−\infty} ^{\infty} {xe}^{−{x}^{\mathrm{2}} } {dx}\: \\ $$$${while}\:\:{f}\left({x}\right)=\:{xe}^{−{x}^{\mathrm{2}} } \:{is}\:{odd} \\ $$$${then}\:{I}\:=\mathrm{0} \\…
Question Number 208836 by Adeyemi889 last updated on 24/Jun/24 Answered by Rasheed.Sindhi last updated on 25/Jun/24 $${Ax}^{\mathrm{3}} −{A}+{Bx}^{\mathrm{2}} +{Bx}+{B}+\left({Cx}+{D}\right)\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}\right) \\ $$$$={Ax}^{\mathrm{3}} −{A}+{Bx}^{\mathrm{2}} +{Bx}+{B}+{Cx}^{\mathrm{3}} +{Dx}^{\mathrm{2}}…
Question Number 208849 by SANOGO last updated on 24/Jun/24 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 208789 by Tawa11 last updated on 23/Jun/24 $$\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} \\…