Question Number 203414 by professorleiciano last updated on 18/Jan/24 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 203396 by Harkoriola last updated on 19/Jan/24 $$\mathrm{State}\:\mathrm{the}\:\mathrm{completness}\:\mathrm{axioms}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{real}\:\mathrm{number} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 203392 by professorleiciano last updated on 18/Jan/24 Commented by a.lgnaoui last updated on 19/Jan/24 Commented by a.lgnaoui last updated on 20/Jan/24 $$\bigtriangleup\boldsymbol{\mathrm{OAB}}\:\:\:\boldsymbol{\mathrm{a}}^{\mathrm{2}} =\mathrm{41}−\mathrm{40cos}\:\boldsymbol{\mathrm{x}}…
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Question Number 203394 by professorleiciano last updated on 18/Jan/24 Answered by som(math1967) last updated on 18/Jan/24 $$\frac{\bigtriangleup{BPQ}}{\bigtriangleup{ABC}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\frac{{BQ}}{{BC}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$$\Rightarrow\frac{{BQ}}{\mathrm{2}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\Rightarrow{BQ}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{2}}}=\sqrt{\mathrm{2}}\:{cm} \\ $$ Terms of…
Question Number 203354 by SEKRET last updated on 19/Jan/24 $$. \\ $$ Answered by Rasheed.Sindhi last updated on 17/Jan/24 $${Can}\:{be}\:{observed}\:{after}\:{some}\: \\ $$$${experiments}\:{that}: \\ $$$$\mathrm{8}^{{r}+\mathrm{20}{q}} \equiv\mathrm{8}^{{r}}…
Question Number 203374 by otchereabdullai@gmail.com last updated on 17/Jan/24 Answered by Calculusboy last updated on 18/Jan/24 $$\boldsymbol{{Solution}}:\:\boldsymbol{{li}}\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{m}}}\frac{\boldsymbol{{tan}}\mathrm{5}\boldsymbol{{x}}}{\mathrm{3}\boldsymbol{{x}}}\:\:\:\left(\boldsymbol{{by}}\:\boldsymbol{{using}}\:\boldsymbol{{algebraic}}\:\boldsymbol{{methods}}\right) \\ $$$$\boldsymbol{{li}}\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{m}}}\frac{\boldsymbol{{tan}}\mathrm{5}\boldsymbol{{x}}}{\mathrm{3}\boldsymbol{{x}}}×\frac{\mathrm{5}\boldsymbol{{x}}}{\mathrm{5}\boldsymbol{{x}}}=\frac{\mathrm{5}\boldsymbol{{x}}}{\mathrm{3}\boldsymbol{{x}}}×\boldsymbol{{li}}\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{m}}}\frac{\boldsymbol{{tan}}\mathrm{5}\boldsymbol{{x}}}{\mathrm{5}\boldsymbol{{x}}} \\ $$$$\boldsymbol{{NB}}:\:\boldsymbol{{li}}\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{m}}}\frac{\boldsymbol{{tanax}}}{\boldsymbol{{x}}}=\mathrm{1}\:\:\boldsymbol{{then}}\:\boldsymbol{{li}}\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}}…
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Question Number 203313 by CrispyXYZ last updated on 16/Jan/24 $${a}_{{n}+\mathrm{1}} ={a}_{{n}} ^{\mathrm{2}} +\mathrm{2}{a}_{{n}} −\mathrm{2},\:{a}_{\mathrm{1}} =\mathrm{3}. \\ $$$$\mathrm{Prove}\:\mathrm{that}\Sigma\:\frac{\mathrm{1}}{{a}_{{n}} +\mathrm{2}}\:\leqslant\:\frac{\mathrm{3}}{\mathrm{10}}. \\ $$ Answered by witcher3 last updated…
Question Number 203301 by MathematicalUser2357 last updated on 15/Jan/24 $$\mathrm{is} \\ $$$$\int\sqrt{{x}}{e}^{−{x}} {dx} \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\frac{\mathrm{2}}{\mathrm{2}{k}+\mathrm{1}}\right){x}^{{n}} \sqrt{{x}}{e}^{−{x}} \right)+{C}? \\ $$ Answered by…