Question Number 171165 by MathsFan last updated on 09/Jun/22 $${A}\:{triangle}\:{QRS}\:{is}\:{to}\:{be}\:{constructed}\:{from} \\ $$$${a}\:{line}\:{segment}\:{of}\:{lenght}\:\mathrm{15}{cm}.\: \\ $$$${Construct}\:{the}\:{triangle}\:{using}\:{the} \\ $$$${division}\:{of}\:{the}\:{line}\:{segment}\:{into}\:{the} \\ $$$${ratio}\:\mathrm{5}:\mathrm{4}:\mathrm{3}\:{such}\:{that}\:{QS}\:{and}\:{RS}\:{are} \\ $$$${laegest}\:{and}\:{smallest}\:{ratio}\:{respectively}. \\ $$$${circumscribe}\:{the}\:{triangle}\:{by}\:{locating} \\ $$$${the}\:{circumcenter}. \\…
Question Number 171164 by MathsFan last updated on 08/Jun/22 $$\boldsymbol{\mathrm{show}}\:\boldsymbol{\mathrm{that}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{common}}\:\boldsymbol{\mathrm{chord}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{the}} \\ $$$$\boldsymbol{\mathrm{circle}}\:\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{y}}^{\mathrm{2}} =\mathrm{4}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{y}}^{\mathrm{2}} −\mathrm{4}\boldsymbol{\mathrm{x}}−\mathrm{2}\boldsymbol{\mathrm{y}}−\mathrm{4}=\mathrm{0} \\ $$$$\boldsymbol{\mathrm{passes}}\:\boldsymbol{\mathrm{through}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{origin}}. \\ $$ Answered by thfchristopher last updated…
Question Number 171160 by mokys last updated on 08/Jun/22 Answered by thfchristopher last updated on 09/Jun/22 $$\mathrm{A}.\:\frac{{dy}}{{dx}}=\mathrm{3}{x}^{\mathrm{2}} +\mathrm{9} \\ $$$${y}=\int\left(\mathrm{3}{x}^{\mathrm{2}} +\mathrm{9}\right){dx} \\ $$$$={x}^{\mathrm{3}} +\mathrm{9}{x}+{C} \\…
Question Number 171156 by MathsFan last updated on 08/Jun/22 $$\boldsymbol{\mathrm{show}}\:\boldsymbol{\mathrm{that}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{equation}}\: \\ $$$$\:\mathrm{3}\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\mathrm{3}\boldsymbol{\mathrm{y}}^{\mathrm{2}} −\mathrm{24}\boldsymbol{\mathrm{x}}+\mathrm{12}\boldsymbol{\mathrm{y}}+\mathrm{12}=\mathrm{0} \\ $$$$\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{circle}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 105616 by DeepakMahato last updated on 30/Jul/20 Answered by nimnim last updated on 30/Jul/20 $$\mathrm{217}{x}+\mathrm{131}{y}=\mathrm{913}…..\left(\mathrm{1}\right) \\ $$$$\mathrm{131}{x}+\mathrm{217}{y}=\mathrm{827}…..\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{1}\right)+\left(\mathrm{2}\right)\:\rightarrow\mathrm{348}{x}+\mathrm{348}{y}=\mathrm{1740} \\ $$$$\Rightarrow{x}+{y}=\mathrm{5}………..\left(\mathrm{3}\right) \\ $$$$\left(\mathrm{1}\right)−\left(\mathrm{2}\right)\rightarrow\mathrm{86}{x}−\mathrm{86}{y}=\mathrm{86}…
Question Number 105613 by Study last updated on 30/Jul/20 $${f}\left({x}\right)=\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{{x}!} \:\:\:\:\:\:\:\:\:\:\:\:{f}^{'} \left({x}\right)=???? \\ $$ Answered by mathmax by abdo last updated on 30/Jul/20 $$\mathrm{f}\left(\mathrm{x}\right)\:=\mathrm{e}^{\mathrm{x}!\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}}\right)} \:\Rightarrow\mathrm{f}^{'}…
Question Number 171146 by akolade last updated on 08/Jun/22 Commented by cortano1 last updated on 08/Jun/22 $$\:\mathrm{5}{f}\left(\frac{\mathrm{1}}{{x}}\right)={f}\left({x}\right)+\sqrt{{x}}\:…\left(\mathrm{1}\right) \\ $$$$\:\:{f}\left(\frac{\mathrm{1}}{{x}}\right)=\mathrm{5}{f}\left({x}\right)−\frac{\mathrm{1}}{\:\sqrt{{x}}}\:…\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{1}\right)−\mathrm{5}\left(\mathrm{2}\right) \\ $$$$\Rightarrow\mathrm{0}=−\mathrm{24}{f}\left({x}\right)+\sqrt{{x}}+\frac{\mathrm{5}}{\:\sqrt{{x}}} \\ $$$$\Rightarrow{f}\left({x}\right)=\frac{\sqrt{{x}}}{\mathrm{24}}\:+\frac{\mathrm{5}}{\mathrm{24}\sqrt{{x}}}\:…
Question Number 171147 by akolade last updated on 08/Jun/22 Commented by SLVR last updated on 10/Jun/22 $${have}\:{you}\:{got}\:{answer} \\ $$ Commented by akolade last updated on…
Question Number 105609 by mohammad17 last updated on 30/Jul/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 171136 by pablo1234523 last updated on 08/Jun/22 $$\mathrm{Using}\:\mathrm{Taylor}'\mathrm{s}\:\mathrm{theorem},\:\mathrm{prove}\:\mathrm{that} \\ $$$${x}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}<\mathrm{sin}\:{x}<{x}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}+\frac{{x}^{\mathrm{5}} }{\mathrm{120}}\:\:\:\mathrm{for}\:{x}>\mathrm{0} \\ $$ Commented by pablo1234523 last updated on 08/Jun/22 Commented…