Question Number 2545 by Rasheed Soomro last updated on 22/Nov/15 $$\mathcal{P}{rove}\:{that}\:\mathrm{3}^{\mathrm{3}{n}} −\mathrm{26}{n}−\mathrm{1}\:{is}\:{divisible}\:{by}\:\mathrm{676}, \\ $$$${where}\:{n}\in\mathbb{Z}^{+} \\ $$ Commented by Yozzi last updated on 22/Nov/15 $${Use}\:{induction}. \\…
Question Number 2544 by Rasheed Soomro last updated on 22/Nov/15 $$\mathcal{C}{an}\:{you}\:\mathcal{G}{eneralize}\:{the}\:{following}? \\ $$$$\mathrm{1}+\mathrm{2}+\mathrm{3}+…+{n}=\frac{\mathrm{1}}{\mathrm{2}}\left({n}\right)\left({n}+\mathrm{1}\right) \\ $$$$\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +…+{n}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{6}}\left({n}\right)\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right. \\ $$$$\mathrm{1}^{\mathrm{3}} +\mathrm{2}^{\mathrm{3}} +\mathrm{3}^{\mathrm{3}} +…+{n}^{\mathrm{3}} =\left[\frac{\mathrm{1}}{\mathrm{2}}\left({n}\right)\left({n}+\mathrm{1}\right)\right]^{\mathrm{2}}…
Question Number 68078 by anaplak last updated on 04/Sep/19 Answered by Smail last updated on 04/Sep/19 $${C}+{iS}=\mathrm{1}+{z}\left({cos}\theta+{isin}\theta\right)+\frac{{z}^{\mathrm{2}} }{\mathrm{2}!}\left({cos}\mathrm{2}\theta+{isin}\mathrm{2}\theta\right)+… \\ $$$$=\mathrm{1}+{ze}^{{i}\theta} +\frac{\left({ze}^{{i}\theta} \right)^{\mathrm{2}} }{\mathrm{2}!}+…={e}^{{ze}^{{i}\theta} } ={e}^{{zcos}\theta}…
Question Number 133612 by benjo_mathlover last updated on 23/Feb/21 $$\mathrm{If}\:\begin{cases}{\mathrm{x}+\mathrm{y}+\mathrm{2z}=\mathrm{k}}\\{\mathrm{x}+\mathrm{2y}+\mathrm{z}=\mathrm{k}}\\{\mathrm{2x}+\mathrm{y}+\mathrm{z}=\mathrm{k}}\end{cases}\:;\:\mathrm{k}\neq\:\mathrm{0} \\ $$$$\mathrm{then}\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{z}^{\mathrm{2}} \:=? \\ $$$$ \\ $$ Commented by mr W last updated…
Question Number 68079 by azizullah last updated on 04/Sep/19 Answered by Rasheed.Sindhi last updated on 07/Sep/19 $$\begin{pmatrix}{}&{\mathrm{son}'\mathrm{s}\:\mathrm{age}}&{\mathrm{father}'\mathrm{s}\:\mathrm{age}}\\{\mathrm{Now}}&{\:\:\:\:\:\:\:\:\:\:\mathrm{x}}&{\:\:\:\:\:\:\mathrm{x}+\mathrm{36}}\\{\mathrm{10}\:\mathrm{years}\:\mathrm{ago}}&{\:\:\:\:\:\:\mathrm{x}−\mathrm{10}}&{\:\:\:\:\:\:\mathrm{x}+\mathrm{26}}\end{pmatrix}\: \\ $$$$ \\ $$$$\:\:\:\:\mathrm{x}+\mathrm{26}=\mathrm{3}\left(\mathrm{x}−\mathrm{10}\right) \\ $$$$\left.\:\:\:\:\mathrm{x}+\mathrm{26}=\mathrm{3x}−\mathrm{30}\right) \\ $$$$\:\:\:\mathrm{2x}=\mathrm{26}+\mathrm{30}=\mathrm{56}…
Question Number 133615 by benjo_mathlover last updated on 23/Feb/21 $$\mathrm{If}\:\mathrm{log}\:_{\mathrm{4}} \left(\mathrm{log}\:_{\mathrm{2}} \:\mathrm{x}\right)+\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{log}\:_{\mathrm{4}} \:\mathrm{x}\right)=\mathrm{2} \\ $$$$\mathrm{then}\:\mathrm{log}\:_{\mathrm{5}} \:\sqrt{\mathrm{x}+\sqrt{\mathrm{x}}\:+\mathrm{5}}\:=\:? \\ $$ Answered by TheSupreme last updated on…
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Question Number 133611 by benjo_mathlover last updated on 23/Feb/21 $$\mathrm{Given}\:\mathrm{system}\:\mathrm{of}\:\mathrm{equation}\: \\ $$$$\begin{cases}{\mathrm{2x}−\mathrm{3y}\:=\:\mathrm{13}}\\{\mathrm{3x}+\mathrm{2y}\:=\:\mathrm{b}}\end{cases}\:,\:\mathrm{where}\:\mathrm{l}\:\leqslant\:\mathrm{b}\leqslant\:\mathrm{100}\:\mathrm{and} \\ $$$$\mathrm{b}\:\mathrm{is}\:\mathrm{integer}.\:\mathrm{Suppose}\:\mathrm{n}^{\mathrm{2}} \:=\:\mathrm{x}+\mathrm{y}\:\mathrm{where} \\ $$$$\mathrm{x},\mathrm{y}\:\mathrm{is}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{given}\:\mathrm{system}\: \\ $$$$\mathrm{of}\:\mathrm{equation}\:,\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{n} \\ $$$$\mathrm{for}\:\mathrm{n}\:\mathrm{is}\:\mathrm{integer}\: \\ $$ Answered by…
Question Number 133610 by benjo_mathlover last updated on 23/Feb/21 $$\mathrm{Given}\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\mid\mathrm{tan}\:\mathrm{x}\mid\:,\:\mathrm{find}\: \\ $$$$\:\frac{\mathrm{df}\left(\mathrm{x}\right)}{\mathrm{dx}}\mid_{\mathrm{x}=\mathrm{k}} \:\mathrm{where}\:\frac{\pi}{\mathrm{2}}<\mathrm{k}<\pi \\ $$ Answered by guyyy last updated on 23/Feb/21 Answered by liberty…
Question Number 68073 by necxxx last updated on 04/Sep/19 $$\:{A}\:{straight}\:{rod}\:{AB}\:{which}\:{is}\:\mathrm{60}{cm}\:{long},{is} \\ $$$${in}\:{equilibrum}\:{when}\:{horizontal}\:{and} \\ $$$${supported}\:{at}\:{a}\:{point}\:{C},\mathrm{10}{cm}\:{from}\:{A}, \\ $$$${with}\:{masses}\:\mathrm{6}{kg}\:{and}\:\mathrm{1}{kg}\:{attached}\:{to}\:{the} \\ $$$${rod}\:{at}\:{A}\:{and}\:{B}\:{respectively}.{It}\:{is}\:{also}\:{in} \\ $$$${equilibrum}\:{and}\:{horizontal}\:{when}\: \\ $$$${supported}\:{at}\:{another}\:{pivott}\:{at}\:{its}\:{mid}- \\ $$$${point},{with}\:{masses}\:{of}\:\mathrm{2}{kg}\:{and}\:\mathrm{5}{kg}\: \\…