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Question Number 210664 by Adeyemi889 last updated on 15/Aug/24 Commented by Adeyemi889 last updated on 15/Aug/24 $$\boldsymbol{{pls}}\:\boldsymbol{{help}}\:\boldsymbol{{me}}\:\boldsymbol{{with}}\:\boldsymbol{{this}}\:\boldsymbol{{partial}}\:\boldsymbol{{fraction}}\: \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 210542 by SANOGO last updated on 12/Aug/24 Commented by Rasheed.Sindhi last updated on 12/Aug/24 $${x}+{y}+{z}=\mathrm{1}\Rightarrow\left({x}+{y}+{z}\right)^{\mathrm{2}} =\mathrm{1} \\ $$ Commented by Frix last updated…
Question Number 210473 by Ismoiljon_008 last updated on 10/Aug/24 Answered by mrdiane last updated on 11/Aug/24 $${on}\:{a}\:\begin{cases}{\mathrm{7}^{\mathrm{2}} ={x}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} −\mathrm{10}{xcos}\left(\theta\right)}\\{\mathrm{7}^{\mathrm{2}} ={x}^{\mathrm{2}} +\left({x}+\mathrm{5}\right)^{\mathrm{2}} −\mathrm{2}{x}\left({x}+\mathrm{5}\right){cos}\left(\Pi−\theta\right)}\end{cases} \\ $$$$\Leftrightarrow\begin{cases}{\mathrm{7}^{\mathrm{2}}…
Question Number 210416 by Adeyemi889 last updated on 08/Aug/24 Commented by Adeyemi889 last updated on 08/Aug/24 $$\boldsymbol{{pls}}\:\boldsymbol{{someone}}\:\boldsymbol{{should}}\:\boldsymbol{{help}}\:\boldsymbol{{me}}\:\boldsymbol{{with}}\:\boldsymbol{{this}}\:\boldsymbol{{partail}}\:\boldsymbol{{fraction}}\: \\ $$$$\boldsymbol{{its}}\:\boldsymbol{{an}}\:\boldsymbol{{assignment}}\:\boldsymbol{{plss}}\:\boldsymbol{{help}}\:\boldsymbol{{me}}\: \\ $$ Commented by mr W…
Question Number 210354 by klipto last updated on 08/Aug/24 $$\int_{\mathrm{0}} ^{\boldsymbol{\alpha}} \frac{\boldsymbol{\mathrm{x}}}{\left(\mathrm{1}+\boldsymbol{\mathrm{x}}^{\mathrm{2}} \right)\left(\mathrm{1}+\boldsymbol{\alpha\mathrm{x}}\right)}\boldsymbol{\mathrm{dx}} \\ $$ Answered by klipto last updated on 08/Aug/24 $$ \\ $$…
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Question Number 210314 by Ansu last updated on 06/Aug/24 $$\mathrm{2}{x}+\mathrm{4}=\mathrm{1} \\ $$ Answered by Spillover last updated on 06/Aug/24 $$\mathrm{2}{x}=−\mathrm{3} \\ $$$${x}=−\frac{\mathrm{3}}{\mathrm{2}} \\ $$ Terms…
Question Number 210291 by Ismoiljon_008 last updated on 05/Aug/24 Answered by mr W last updated on 06/Aug/24 $${x}=\sqrt{{x}^{\mathrm{2}} }=\sqrt{\mathrm{2}^{\mathrm{2}} +{x}^{\mathrm{2}} −\mathrm{2}^{\mathrm{2}} }=\sqrt{\mathrm{2}^{\mathrm{2}} +\left({x}−\mathrm{2}\right)\left({x}+\mathrm{2}\right)} \\ $$$$=\sqrt{\mathrm{2}^{\mathrm{2}}…