Question Number 8958 by Sopheak last updated on 07/Nov/16 $${Prove}\:{that}\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+….+\frac{\mathrm{1}}{\mathrm{2009}}=\mathrm{2009}−\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{2}}{\mathrm{3}}+\frac{\mathrm{3}}{\mathrm{4}}+…+\frac{\mathrm{2008}}{\mathrm{2009}}\right) \\ $$ Answered by sou1618 last updated on 07/Nov/16 $$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}…+\frac{\mathrm{1}}{\mathrm{2009}} \\ $$$$=\left(\mathrm{1}−\frac{\mathrm{0}}{\mathrm{1}}\right)+\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\right)+\left(\mathrm{1}−\frac{\mathrm{2}}{\mathrm{3}}\right)+\left(\mathrm{1}−\frac{\mathrm{3}}{\mathrm{4}}\right)…+\left(\mathrm{1}−\frac{\mathrm{2008}}{\mathrm{2009}}\right) \\ $$$$=\mathrm{2009}−\left(\frac{\mathrm{0}}{\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{2}}{\mathrm{3}}+\frac{\mathrm{3}}{\mathrm{4}}…+\frac{\mathrm{2008}}{\mathrm{2009}}\right) \\…
Question Number 8957 by Sopheak last updated on 07/Nov/16 $$\: \\ $$$${Let}\:{n}\:{be}\:{a}\:{positive}\:{integer}\:{such}\:{that}\:{one}\:{of} \\ $$$${the}\:{roofs}\:{of}\:{the}\:{quadratic}\:{equation}\: \\ $$$$\mathrm{4}{x}^{\mathrm{2}} −\left(\mathrm{4}\sqrt{\mathrm{3}}+\mathrm{4}\right){x}+\sqrt{\mathrm{3}}{n}−\mathrm{24}=\mathrm{0}\:{is}\:{an}\:{integer}\: \\ $$$${Find}\:{the}\:{value}\:{of}\:{n}\: \\ $$$$\: \\ $$ Commented by…
Question Number 74483 by liki last updated on 24/Nov/19 Commented by liki last updated on 24/Nov/19 $$…\:{sory}\:{mr}\:{w},{i}\:{tried}\:{to}\:{this}\:{qns}\:{according} \\ $$$$\:{to}\:{your}\:{idea}\:{but}\:{i}\:{didn}'{t}\:{get}\:{the}\:{answer}\:{so}\: \\ $$$$\:{plz}\:{assist}\:{me}! \\ $$ Commented by…
Question Number 8939 by arinto27 last updated on 07/Nov/16 $$\left.\mathrm{1}\right)\:\mathrm{diket}\:\mathrm{lingkaran}\:\mathrm{c}\:\mathrm{dg}\:\mathrm{pers}\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} −\mathrm{6x}+\mathrm{2y}+\mathrm{2}=\mathrm{0} \\ $$$$\mathrm{diket}\:\mathrm{pula}\:\mathrm{titik}\:\mathrm{d}\:\left(\mathrm{a},−\mathrm{3}\right).\:\mathrm{agar}\:\mathrm{d}\:\mathrm{brd}\:\mathrm{di}\:\mathrm{dlm}\:\mathrm{lingkaran} \\ $$$$\mathrm{nilai}\:\mathrm{a}\:\mathrm{yg}\:\mathrm{memenuhi}\:\mathrm{adalah}…. \\ $$ Answered by sandy_suhendra last updated on 09/Nov/16…
Question Number 8946 by ridwan balatif last updated on 07/Nov/16 $$\mathrm{x}−\mathrm{y}=\mathrm{1}\:\mathrm{and}\:\mathrm{x}^{\mathrm{y}} =\mathrm{64},\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x}+\mathrm{y}=…? \\ $$ Answered by Rasheed Soomro last updated on 07/Nov/16 $$\mathrm{x}−\mathrm{y}=\mathrm{1}\:\mathrm{and}\:\mathrm{x}^{\mathrm{y}} =\mathrm{64},\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x}+\mathrm{y}=…? \\…
Question Number 140007 by SOMEDAVONG last updated on 03/May/21 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{3}}{\mathrm{x}}\right)^{\mathrm{x}} =? \\ $$ Answered by Ankushkumarparcha last updated on 03/May/21 $${Solution}:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{3}}{{x}}\right)^{{x}} =\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{e}^{\mathrm{log}_{{e}}…
Question Number 140002 by mohammad17 last updated on 03/May/21 Answered by MJS_new last updated on 03/May/21 $$\int\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\mathrm{ln}\:\frac{\mathrm{1}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{{x}}\:{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\frac{\mathrm{1}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{{x}}\:\rightarrow\:{dx}=−\frac{{x}^{\mathrm{2}} \sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{\mathrm{1}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dt}\right]…
Question Number 74456 by liki last updated on 24/Nov/19 Commented by liki last updated on 24/Nov/19 $$…{plz}\:{anyone}\:{knowing}\:{simple}\:{logic}\:{of} \\ $$$${find}\:{the}\:{possible}\:{number},{because}\: \\ $$$$\:{this}\:{always}\:{i}\:{did}\:{not}\:{understand}\:{so}\:{pl}\: \\ $$$$\:{i}\:{need}\:{simple}\:{way}\:{of}\:{find}\:{the}\:{others}\:{no}. \\ $$…
Question Number 8916 by kuldeep singh raj last updated on 05/Nov/16 Commented by Rasheed Soomro last updated on 05/Nov/16 $$\mathrm{Is}\:\mathrm{O}\:\mathrm{the}\:\mathrm{centre}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}? \\ $$ Commented by kuldeep…
Question Number 74455 by Mr. K last updated on 24/Nov/19 $${if}\:{K}=\left({x}\in\mathbb{R}\:\mathrm{2}{x}−\mathrm{1}+\mid\mathrm{2}{x}−\mathrm{1}\mid=\mathrm{0}\:\right){and} \\ $$$${J}=\left({x}\in\mathbb{R}\:−{x}\left(\mathrm{2}{x}+\mathrm{1}\right)\leqslant−\mathrm{1}\right)\:{find}\:{J}−{K}. \\ $$ Answered by MJS last updated on 24/Nov/19 $${K}=\left\{{x}\in\mathbb{R}\mid{x}\leqslant\frac{\mathrm{1}}{\mathrm{2}}\right\} \\ $$$${J}=\left\{{x}\in\mathbb{R}\mid{x}\leqslant−\mathrm{1}\vee{x}\geqslant\frac{\mathrm{1}}{\mathrm{2}}\right\}…