Question Number 210689 by Tawa11 last updated on 16/Aug/24 Answered by mm1342 last updated on 16/Aug/24 $${AM}={MD}={a}\Rightarrow{BC}=\frac{\mathrm{2}}{\mathrm{3}}{a} \\ $$$$\frac{{s}}{{S}}=\frac{\frac{{BC}×{h}}{\mathrm{2}}}{\frac{\left({BC}+{AD}\right){h}}{\mathrm{2}}}=\frac{{BC}}{{BC}+{AD}}=\frac{\frac{\mathrm{2}}{\mathrm{3}}{a}}{\mathrm{2}{a}+\frac{\mathrm{2}}{\mathrm{3}}{a}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\:\checkmark \\ $$$$ \\ $$…
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 210666 by universe last updated on 15/Aug/24 $$\:\:\mathrm{prove}\:\mathrm{that}\:\mathrm{p}\left(\mathrm{n}\right)\:\mathrm{is}\:\mathrm{integer}\:\forall\:\mathrm{n}\in\mathbb{Z} \\ $$$$\:\:\:\mathrm{p}\left(\mathrm{n}\right)\:=\:\frac{\mathrm{3n}^{\mathrm{7}} +\mathrm{7n}^{\mathrm{3}} +\mathrm{11n}}{\mathrm{21}} \\ $$ Answered by A5T last updated on 15/Aug/24 $${Let}\:{f}\left({n}\right)=\mathrm{3}{n}^{\mathrm{7}} +\mathrm{7}{n}^{\mathrm{3}}…
Question Number 210667 by mr W last updated on 15/Aug/24 Answered by ajfour last updated on 16/Aug/24 $${let}\:{C}\:{be}\:{origin}\:\:{and}\:{A}\left({p},{q}\right) \\ $$$${q}=\frac{{p}}{\:\sqrt{\mathrm{3}}} \\ $$$${q}={p}−{a} \\ $$$$\mathrm{0}−{q}={m}\left(\mathrm{2}{a}−{p}\right) \\…
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Question Number 210663 by Erico last updated on 15/Aug/24 $$\mathrm{Calcul} \\ $$$$\underset{\:\mathrm{1}} {\int}^{\:+\infty} \left(\frac{\mathrm{lnt}}{\mathrm{t}−\mathrm{1}}\right)^{\mathrm{2}} \mathrm{dt}=??? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 210674 by noraell last updated on 15/Aug/24 Answered by BHOOPENDRA last updated on 16/Aug/24 $${f}\left({x}\right)=\frac{{x}^{\mathrm{2}} }{{k}^{\mathrm{2}} \:\left\{−\mathrm{1}+\sqrt{\left.−\mathrm{4}{x}^{\mathrm{4}} −\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}\right\}}\:\right.} \\ $$$$\left\{{lim}\:{x}\rightarrow\mathrm{0}\:\frac{{sinx}}{{x}}=\mathrm{1},{limx}\rightarrow\mathrm{0}\:\frac{{tanx}}{{x}}=\mathrm{1}\right\} \\ $$$${conjugate}\:{multiplication}…
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