Question Number 72806 by Rio Michael last updated on 03/Nov/19 $$\underset{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}\rightarrow\mathrm{0}} {\:{show}\:{that}\:\:\mathrm{lim}}\:\left[\:{x}\right]\:\:{does}\:{not}\:{exist}. \\ $$$${Hence}\:{define}\:\:\left[{x}\right]\:\:{and}\:{sketch}\:{a}\:{graph}\:{for}\: \\ $$$$\:{y}\:=\:\mathrm{3}{x}^{\mathrm{2}} \:+\:\left[{x}\right] \\ $$ Commented by mathmax by abdo last…
Question Number 138299 by cherokeesay last updated on 12/Apr/21 Answered by bemath last updated on 12/Apr/21 $${BD}\:=\:{r}_{\mathrm{1}} +\mathrm{2}\:\Leftrightarrow\:{r}_{\mathrm{1}} \left(\sqrt{\mathrm{2}}−\mathrm{1}\right)=\mathrm{2} \\ $$$${r}_{\mathrm{1}} =\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{2}}−\mathrm{1}}\:=\:\frac{\mathrm{2}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)}{\mathrm{2}−\mathrm{1}}=\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{2} \\ $$$${shaded}\:{area}=\left(\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{2}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}\pi\left(\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{2}\right)^{\mathrm{2}}…
Question Number 72734 by Mr Jor last updated on 01/Nov/19 $${A}\:{triangle}\:{ABC}\:{is}\:{inscribed}\:{in}\:{a} \\ $$$${circle}.{AC}=\mathrm{10}{cm},{BC}=\mathrm{7}{cm}\:{and}\: \\ $$$${AB}=\mathrm{10}{cm}.{Find}\:{the}\:{radius}\:{of}\:{the} \\ $$$${circle}. \\ $$ Answered by $@ty@m123 last updated on…
Question Number 138254 by mey3nipaba last updated on 11/Apr/21 Commented by Rasheed.Sindhi last updated on 11/Apr/21 $$\mathrm{40} \\ $$ Commented by Dwaipayan Shikari last updated…
Question Number 72718 by Rio Michael last updated on 01/Nov/19 $${given}\:{that}\: \\ $$$$\:{a}\:\equiv\:{b}\left({mod}\:{n}\right)\: \\ $$$${show}\:{that}\:{a}^{{k}} \:\equiv\:{b}^{{k}} \:\left({mod}\:{n}\right) \\ $$ Commented by prof Abdo imad last…
Question Number 72693 by Rio Michael last updated on 31/Oct/19 $${prove}\:{that}\:{the}\:{arithmetic}\:{mean}\:{of}\:{a}\:{sequence} \\ $$$${is}\:{greater}\:{or}\:{equal}\:{to}\:{the}\:{geometric}\:{mean}. \\ $$$${that}\:\:{is}\:\: \\ $$$$\:\:\:\:\frac{{a}\:+\:{b}}{\mathrm{2}}\:\geqslant\:\sqrt{{ab}}\: \\ $$ Answered by MJS last updated on…
Question Number 72688 by Rio Michael last updated on 31/Oct/19 $${Evaluate}\: \\ $$$$\:\underset{{t}\rightarrow\mathrm{9}} {\:{lim}}\frac{\mathrm{9}−{t}}{\mathrm{3}−\sqrt{{t}}\:} \\ $$ Answered by JDamian last updated on 01/Nov/19 $$\underset{{t}\rightarrow\mathrm{9}} {\mathrm{lim}}\left(\frac{\mathrm{9}−{t}}{\mathrm{3}−\sqrt{{t}}\:}×\frac{\mathrm{3}+\sqrt{{t}}}{\mathrm{3}+\sqrt{{t}}}\right)=\underset{{t}\rightarrow\mathrm{9}}…
Question Number 138193 by mey3nipaba last updated on 10/Apr/21 $${Given}\:{x}\neq{y}\:{and}\:{x}^{\mathrm{2}} =\mathrm{25}{x}+{y},\:{y}^{\mathrm{2}} ={x}+\mathrm{25}{y}\: \\ $$$${solve}\:{for}\:{the}\:{value}\:{of}\:\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{1}}\:{without}\: \\ $$$${using}\:{calculators}\:{or}\:{tools}. \\ $$$${Show}\:{your}\:{method}. \\ $$ Answered by liberty…
Question Number 72634 by Rio Michael last updated on 30/Oct/19 $${help}\:{me}\:{with}\:{the}\:{conditions}\:{please}\: \\ $$$${for}\:{a}\:{function}\:{f}\:{to}\:{be}\:{continuous}\:{at}\:{a}\:{point}\:{a} \\ $$ Commented by Prithwish sen last updated on 31/Oct/19 $$\boldsymbol{\mathrm{If}}\:\boldsymbol{\mathrm{you}}\:\boldsymbol{\mathrm{can}}\:\boldsymbol{\mathrm{able}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{draw}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{function}}\:\boldsymbol{\mathrm{f}}\:\:\boldsymbol{\mathrm{through}} \\…
Question Number 72633 by Rio Michael last updated on 30/Oct/19 $${prove}\:{using}\:{th}\:{sandwich}\:{or}\:{Squeez}\:{theorem}\:{that} \\ $$$${for}\:{any}\:\:{a}\:>\:\mathrm{0} \\ $$$$\:\underset{{x}\rightarrow{a}} {\mathrm{lim}}\:\sqrt{{x}}\:=\:\sqrt{{a}}\: \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com