Question Number 226635 by Kassista last updated on 08/Dec/25 Answered by mingski last updated on 08/Dec/25 $${C}:{z}^{\mathrm{2}} +\mathrm{6}{z}+\mathrm{10}=\mathrm{0} \\ $$$$\left({z}+\mathrm{3}\right)^{\mathrm{2}} =−\mathrm{1} \\ $$$${z}+\mathrm{3}=\pm\mathrm{i},{z}=−\mathrm{3}\pm\mathrm{i}. \\ $$$${for}\:{z}_{\mathrm{1}}…
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Question Number 226641 by fantastic2 last updated on 08/Dec/25 $${if}\:\mathrm{log}\:_{\mathrm{8}} {a}+\mathrm{log}\:_{\mathrm{4}} {b}^{\mathrm{2}} =\mathrm{5} \\ $$$$\& \\ $$$$\mathrm{log}\:_{\mathrm{8}} ^{{b}} +\mathrm{log}\:_{\mathrm{4}} {a}^{\mathrm{2}} =\mathrm{7} \\ $$$${ab}=? \\ $$…
Question Number 226622 by fantastic2 last updated on 07/Dec/25 Commented by mr W last updated on 07/Dec/25 $$\lambda=\frac{{m}}{{n}}=\frac{{x}−{x}_{\mathrm{1}} }{{x}_{\mathrm{2}} −{x}}=\frac{{y}−{y}_{\mathrm{1}} }{{y}_{\mathrm{2}} −{y}} \\ $$$${if}\:\lambda<\mathrm{0},\:{there}\:{are}\:{only}\:{two}\:{cases}: \\…
Question Number 226619 by fantastic2 last updated on 07/Dec/25 $${factorise} \\ $$$${x}^{\mathrm{2}} −{qx}−{p}^{\mathrm{2}} +\mathrm{5}{pq}−\mathrm{6}{q}^{\mathrm{2}} \\ $$ Commented by Frix last updated on 07/Dec/25 $${x}^{\mathrm{2}} −{qx}−{p}^{\mathrm{2}}…
Question Number 226612 by Spillover last updated on 07/Dec/25 Commented by SonGoku last updated on 07/Dec/25 $$\boldsymbol{\mathrm{Solve}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{system}}\:\begin{cases}{\boldsymbol{\mathrm{m}}\:+\:\boldsymbol{\mathrm{m}}\:=\:\mathrm{20}}\\{\boldsymbol{\mathrm{c}}\:−\:\boldsymbol{\mathrm{l}}\:=\:\mathrm{3}}\\{\boldsymbol{\mathrm{m}}\:−\:\boldsymbol{\mathrm{l}}×\boldsymbol{\mathrm{c}}\:=\:?}\end{cases}. \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 226608 by Spillover last updated on 07/Dec/25 Answered by Spillover last updated on 09/Dec/25 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 226609 by Spillover last updated on 07/Dec/25 Answered by Ghisom_ last updated on 08/Dec/25 $$\underset{−\mathrm{2}} {\overset{−\mathrm{1}} {\int}}\frac{\sqrt{\mathrm{2}+{x}}}{{x}\sqrt{\mathrm{2}−{x}}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\frac{\sqrt{\mathrm{2}+{x}}}{\:\sqrt{\mathrm{2}−{x}}}\:\rightarrow\:{dx}=\frac{\sqrt{\left(\mathrm{2}−{x}\right)^{\mathrm{3}} \left(\mathrm{2}+{x}\right)}}{\mathrm{2}}{dt}\right] \\ $$$$=\mathrm{4}\underset{\mathrm{0}} {\overset{\mathrm{1}/\sqrt{\mathrm{3}}}…
Question Number 226610 by Spillover last updated on 07/Dec/25 Answered by mr W last updated on 07/Dec/25 $$\mathrm{2}\left(\mathrm{12}\:\mathrm{cos}\:\frac{\pi{x}}{\mathrm{3}}+\mathrm{5}\:\mathrm{sin}\:\frac{\pi{x}}{\mathrm{3}}\right)=\mathrm{13} \\ $$$$\frac{\mathrm{12}}{\mathrm{13}}\:\mathrm{cos}\:\frac{\pi{x}}{\mathrm{3}}+\frac{\mathrm{5}}{\mathrm{13}}\:\mathrm{sin}\:\frac{\pi{x}}{\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${with}\:\alpha=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{5}}{\mathrm{12}}=\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{5}}{\mathrm{13}}=\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{12}}{\mathrm{13}}…
Question Number 226601 by mr W last updated on 07/Dec/25 Answered by mr W last updated on 07/Dec/25 $${mass}\:{of}\:{particle}={m} \\ $$$${total}\:{mass}\:{of}\:{earth}={M}_{{E}} \\ $$$${radius}\:{of}\:{earth}={R} \\ $$$${F}=\frac{{GmM}}{{x}^{\mathrm{2}}…