Question Number 141163 by Niiicooooo last updated on 16/May/21 Answered by mindispower last updated on 16/May/21 $$\zeta\left({n}\right)=\mathrm{1}+\underset{{k}\geqslant\mathrm{2}} {\sum}\frac{\mathrm{1}}{{k}^{{n}} } \\ $$$${k}^{{n}} \geqslant{nk}^{\mathrm{2}} \:{easy}\:{to}\:{see}\:{that}\:,{k}\geqslant\mathrm{2} \\ $$$${for}\:{n}>>\mathrm{2}…
Question Number 75627 by liki last updated on 14/Dec/19 Answered by $@ty@m123 last updated on 14/Dec/19 $$\boldsymbol{{Initially}}, \\ $$$${Length}={l} \\ $$$${Breadth}={b} \\ $$$$\therefore\:{Diagonal}\:\:\:\boldsymbol{{d}}=\sqrt{{l}^{\mathrm{2}} +{b}^{\mathrm{2}} }\:\:……\left(\mathrm{1}\right)…
Question Number 141152 by ZiYangLee last updated on 16/May/21 $$\begin{pmatrix}{\mathrm{3}}&{\mathrm{5}}\\{\mathrm{2}}&{\mathrm{4}}\end{pmatrix}\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}=\begin{pmatrix}{{x}'}\\{{y}'}\end{pmatrix}\:\mathrm{and}\:\begin{pmatrix}{{a}}&{{b}}\\{{c}}&{{d}}\end{pmatrix}\begin{pmatrix}{{x}'}\\{{y}'}\end{pmatrix}=\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix} \\ $$$$\mathrm{are}\:\mathrm{true}\:\mathrm{for}\:\mathrm{any}\:\mathrm{values}\:\mathrm{of}\:{x},{y},{x}',{y}', \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{values}\:\mathrm{of}\:{a},{b},{c},{d}. \\ $$ Answered by iloveisrael last updated on 16/May/21 $$\Rightarrow\begin{pmatrix}{\mathrm{3}\:\:\:\:\mathrm{5}}\\{\mathrm{2}\:\:\:\:\mathrm{4}}\end{pmatrix}\:\begin{pmatrix}{{a}\:\:\:\:{b}}\\{{c}\:\:\:\:{d}}\end{pmatrix}\:\begin{pmatrix}{{x}'}\\{{y}'}\end{pmatrix}\:=\:\begin{pmatrix}{{x}'}\\{{y}'}\end{pmatrix} \\…
Question Number 75606 by liki last updated on 13/Dec/19 Commented by mathmax by abdo last updated on 15/Dec/19 $$\begin{cases}{{x}−{y}=\mathrm{3}}\\{\mathrm{2}{x}−\mathrm{2}{y}\:={k}\:\:\:\:\:{we}\:{have}\:\Delta_{{s}} =\begin{vmatrix}{\mathrm{1}\:\:\:\:\:\:−\mathrm{1}}\\{\mathrm{2}\:\:\:\:\:\:\:\:\:−\mathrm{2}}\end{vmatrix}}\end{cases}=−\mathrm{2}+\mathrm{2}=\mathrm{0} \\ $$$$\Delta_{{x}} =\begin{vmatrix}{\mathrm{3}\:\:\:\:\:\:\:\:\:−\mathrm{1}}\\{{k}\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{2}}\end{vmatrix}=−\mathrm{6}+{k} \\ $$$$\Delta_{{y}}…
Question Number 10062 by ridwan balatif last updated on 22/Jan/17 Commented by ridwan balatif last updated on 22/Jan/17 $$\mathrm{find}\:\mathrm{the}\:\mathrm{remainder}\:\mathrm{of}\:\mathrm{the}\:\mathrm{division}\:\mathrm{20}^{\mathrm{2017}} +\mathrm{1}^{\mathrm{2017}} +\mathrm{17}^{\mathrm{2017}} +\mathrm{72}^{\mathrm{2017}} \:\mathrm{by}\:\mathrm{2017} \\ $$…
Question Number 10050 by ridwan balatif last updated on 22/Jan/17 $$\mathrm{h}\:\mathrm{ow}\:\mathrm{do}\:\mathrm{we}\:\mathrm{know}\:\mathrm{2017}\:\mathrm{and}\:\mathrm{another}\:\mathrm{number}\:\:\mathrm{is}\:\mathrm{a}\:\mathrm{prime}\:\mathrm{number}? \\ $$ Commented by prakash jain last updated on 22/Jan/17 $$\mathrm{2017}\:\mathrm{is}\:\mathrm{a}\:\mathrm{prime}\:\mathrm{number}.\: \\ $$$$\mathrm{Since}\:\mathrm{it}\:\mathrm{has}\:\mathrm{only}\:\mathrm{two}\:\mathrm{factors} \\…
Question Number 75570 by naka3546 last updated on 12/Dec/19 Commented by naka3546 last updated on 12/Dec/19 $$\frac{\left[\:{blue}\:\:{area}\:\right]}{\left[\:{square}\:\:{area}\:\right]}\:\:=\:\:? \\ $$ Answered by mr W last updated…
Question Number 75569 by naka3546 last updated on 12/Dec/19 Commented by MJS last updated on 13/Dec/19 $$\mathrm{was}\:\mathrm{my}\:\mathrm{answer}\:\mathrm{to}\:\mathrm{the}\:\mathrm{other}\:\mathrm{one}\:\mathrm{ok}? \\ $$$$\mathrm{it}'\mathrm{s}\:\mathrm{hard}\:\mathrm{to}\:\mathrm{understand}… \\ $$ Commented by naka3546 last…
Question Number 10033 by ridwan balatif last updated on 21/Jan/17 Answered by mrW1 last updated on 21/Jan/17 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}\left(\mathrm{cos}\:\mathrm{4}{x}^{\mathrm{2}} −\mathrm{1}\right)}{\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \:\mathrm{2}{x}}\right)} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}\left(\mathrm{cos}\:\mathrm{4}{x}^{\mathrm{2}} −\mathrm{1}\right)\mathrm{cos}^{\mathrm{2}}…
Question Number 10031 by ridwan balatif last updated on 21/Jan/17 Commented by ridwan balatif last updated on 21/Jan/17 Terms of Service Privacy Policy Contact: info@tinkutara.com