Question Number 164549 by wwww last updated on 18/Jan/22 Answered by alephzero last updated on 18/Jan/22 $$\left({a}\right)\:{a}−\frac{\mathrm{1}}{{a}}=\mathrm{4} \\ $$$$\frac{{a}^{\mathrm{2}} −\mathrm{1}}{{a}}\:=\:\mathrm{4} \\ $$$$\frac{{a}^{\mathrm{2}} −\mathrm{1}−\mathrm{4}{a}}{{a}}\:=\:\mathrm{0} \\ $$$$\Rightarrow\:{a}^{\mathrm{2}}…
Question Number 33479 by mondodotto@gmail.com last updated on 17/Apr/18 $$\boldsymbol{\mathrm{A}}\:\boldsymbol{\mathrm{man}}\:\boldsymbol{\mathrm{takes}}\:\mathrm{15}\boldsymbol{\mathrm{days}}\:\boldsymbol{\mathrm{to}} \\ $$$$\boldsymbol{\mathrm{dig}}\:\mathrm{6}\:\boldsymbol{\mathrm{hectres}}.\boldsymbol{\mathrm{how}}\:\boldsymbol{\mathrm{long}}\:\boldsymbol{\mathrm{would}} \\ $$$$\mathrm{10}\:\boldsymbol{\mathrm{boys}}\:\boldsymbol{\mathrm{take}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{dig}}\:\mathrm{81}\:\boldsymbol{\mathrm{hectres}} \\ $$$$\boldsymbol{\mathrm{if}}\:\mathrm{2}\:\boldsymbol{\mathrm{boys}}\:\boldsymbol{\mathrm{do}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{same}}\:\boldsymbol{\mathrm{amount}}\:\boldsymbol{\mathrm{of}} \\ $$$$\boldsymbol{\mathrm{work}}\:\boldsymbol{\mathrm{as}}\:\boldsymbol{\mathrm{one}}\:\boldsymbol{\mathrm{man}}? \\ $$ Commented by MJS last updated…
Question Number 164546 by mathls last updated on 18/Jan/22 Commented by mathls last updated on 18/Jan/22 $${Dom}\left({g}\right)=? \\ $$ Commented by mkam last updated on…
Question Number 33466 by mondodotto@gmail.com last updated on 17/Apr/18 $$\:\boldsymbol{\mathrm{if}}\:\boldsymbol{\mathrm{tan}\beta}=\frac{\boldsymbol{\mathrm{r}}}{\:\sqrt{\boldsymbol{\mathrm{s}}}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{sin}\beta}=\frac{\sqrt{\boldsymbol{\mathrm{s}}}}{\boldsymbol{\mathrm{r}}} \\ $$$$\:\boldsymbol{\mathrm{show}}\:\boldsymbol{\mathrm{that}}\:\boldsymbol{\mathrm{cos}\beta}=\sqrt{\boldsymbol{\mathrm{r}}^{\mathrm{2}} +\boldsymbol{\mathrm{s}}} \\ $$ Commented by MJS last updated on 17/Apr/18 $$\mathrm{cos}\:\beta=\sqrt{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:\beta}=\frac{\sqrt{{r}^{\mathrm{2}} −{s}}}{\mid{r}\mid}…
Question Number 164520 by ZiYangLee last updated on 18/Jan/22 $$\mathrm{A}\:\mathrm{circle}\:\mathrm{is}\:\mathrm{drawn}\:\mathrm{with}\:\mathrm{center}\:\left(\mathrm{0},\mathrm{1}\right)\:\mathrm{and}\:\mathrm{radius}\:\mathrm{1}. \\ $$$$\mathrm{A}\:\mathrm{line}\:\mathrm{OAB}\:\mathrm{is}\:\mathrm{drawn},\:\mathrm{making}\:\mathrm{an}\:\mathrm{angle}\:\theta\:\mathrm{with}\:\mathrm{the} \\ $$$${x}-\mathrm{axis}\:\mathrm{to}\:\mathrm{cut}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{at}\:{A}\:\mathrm{and}\:\mathrm{the}\:\mathrm{tangent}\:\mathrm{to}\:\mathrm{the} \\ $$$$\mathrm{circle}\:\mathrm{at}\:\left(\mathrm{0},\mathrm{2}\right)\:\mathrm{at}\:{B}.\:\mathrm{Lines}\:\mathrm{are}\:\mathrm{now}\:\mathrm{drawn}\:\mathrm{through} \\ $$$${A}\:\mathrm{and}\:{B}\:\:\mathrm{parallel}\:\mathrm{to}\:\mathrm{the}\:{x}-\:\mathrm{and}\:{y}-\mathrm{axes}\:\mathrm{respectively} \\ $$$$\mathrm{to}\:\mathrm{intersect}\:\mathrm{at}\:{P}.\:\:\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\left(\mathrm{i}\right)\:\mathrm{OA}=\mathrm{2}\:\mathrm{sin}\:\theta\:\:\:\:\:\mathrm{and}\: \\ $$$$\left(\mathrm{ii}\right)\mathrm{the}\:\mathrm{coordinates}\:\mathrm{of}\:{P}\:\mathrm{are}\:\left(\mathrm{2}\:\mathrm{cot}\:\theta,\:\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \theta\right)…
Question Number 98984 by otchereabdullai@gmail.com last updated on 17/Jun/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 98968 by Brokris last updated on 17/Jun/20 $$\mathrm{if}\:\mathrm{x}^{\mathrm{x}^{\mathrm{x}^{\mathrm{x}^{\mathrm{2020}} } } } =\:\mathrm{2020}.\:\:\mathrm{Solve}\:\mathrm{for}\:\mathrm{x} \\ $$$$ \\ $$ Commented by mr W last updated on…
Question Number 164507 by mathlove last updated on 18/Jan/22 Commented by mathlove last updated on 18/Jan/22 $$\mathrm{tan}\theta=\frac{\mathrm{1}}{\mathrm{3}}\:\:{and}\:\:\:\mathrm{tan}\:\beta=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$${faind}\:\:{a}\:\:{and}\:\:\:{b} \\ $$ Commented by mr W…
Question Number 98967 by M±th+et+s last updated on 17/Jun/20 $${solve}: \\ $$$$\left(\frac{\int_{\mathrm{2}} ^{\mathrm{6}} {x}\sqrt{\mathrm{1}+\mathrm{9}\lfloor{x}\rfloor^{\mathrm{2}} }{dx}}{\int_{\mathrm{0}} ^{\mathrm{1}} {x}\left\{\frac{\mathrm{1}}{{x}}\right\}\lceil\frac{\mathrm{1}}{{x}}\rceil{dx}}\right)\left(\underset{{n}\geqslant\mathrm{1}} {\sum}\left(−\mathrm{1}\right)^{{n}} \frac{\prod_{{j}=\mathrm{1}} ^{{n}} \left(\frac{\mathrm{3}}{\mathrm{2}}−{j}\right)}{\left(\mathrm{2}{n}+\mathrm{1}\right){n}!}\right) \\ $$ Answered by…
Question Number 33410 by NECx last updated on 15/Apr/18 $${please}\:{is}\:{there}\:{any}\:{general}\:{way}\:{for} \\ $$$${calculating}\:{the}\:{error}\:{or}\:{uncertainty} \\ $$$${in}\:{g}\:{when} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:{m}=\frac{\mathrm{4}\pi^{\mathrm{2}} }{{g}}\:{where}\:{m}={slope}\:{and} \\ $$$${g}={acceleration}\:{due}\:{to}\:{gravity} \\ $$$$ \\ $$$$…