Question Number 101582 by Rohit@Thakur last updated on 03/Jul/20 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\:{r}\:{e}^{{r}/{n}} \:= \\ $$ Answered by mathmax by abdo last updated on…
Question Number 35877 by 0123456789 last updated on 25/May/18 $$\mathrm{If}\:\:\mathrm{cos}\:{A}=\frac{\mathrm{3}}{\mathrm{4}}\:,\:\mathrm{then}\:\mathrm{32}\:\mathrm{sin}\:\left(\frac{{A}}{\mathrm{2}}\right)\:\mathrm{sin}\:\left(\frac{\mathrm{5}{A}}{\mathrm{2}}\right)= \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 25/May/18 $$\mathrm{32}{sin}\left(\frac{{A}}{\mathrm{2}}\right){sin}\left(\frac{\mathrm{5}{A}}{\mathrm{2}}\right) \\ $$$$=\mathrm{16}×\left({cos}\mathrm{2}{A}−{cos}\mathrm{3}{A}\right) \\ $$$$=\mathrm{16}×\left\{\left(\mathrm{2}{cos}^{\mathrm{2}} {A}−\mathrm{1}\right)−\left(\mathrm{4}{cos}^{\mathrm{3}}…
Question Number 35791 by spherecity@gmail.com last updated on 23/May/18 $$\mathrm{A}\:\mathrm{5100}\:\mathrm{cm}^{\mathrm{2}} \:\mathrm{trapezium}\:\mathrm{has}\:\mathrm{the}\:\mathrm{perpendicular} \\ $$$$\mathrm{distance}\:\mathrm{between}\:\mathrm{the}\:\mathrm{two}\:\mathrm{parallel}\:\mathrm{sides} \\ $$$$\mathrm{60}\:\mathrm{m}.\:\mathrm{If}\:\mathrm{one}\:\mathrm{of}\:\mathrm{the}\:\mathrm{parallel}\:\mathrm{sides}\:\mathrm{be}\:\mathrm{40}\:\mathrm{m}\: \\ $$$$\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{length}\:\mathrm{of}\:\mathrm{the}\:\mathrm{other}\:\mathrm{parallel}\:\mathrm{side}. \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on…
Question Number 101082 by 67549972 last updated on 30/Jun/20 $$\mathrm{The}\:\mathrm{value}\:\mathrm{of} \\ $$$${e}^{\mathrm{log}_{\mathrm{10}} \:\mathrm{tan}\:\mathrm{1}°+\mathrm{log}_{\mathrm{10}} \:\mathrm{tan}\:\mathrm{2}°+\mathrm{log}_{\mathrm{10}} \:\mathrm{tan}\:\mathrm{3}°+…+\mathrm{log}_{\mathrm{10}} \:\mathrm{tan}\:\mathrm{89}°} \\ $$$$\mathrm{is} \\ $$ Commented by Dwaipayan Shikari last…
Question Number 35534 by jonesme last updated on 20/May/18 $$\mathrm{If}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{9}{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{7}=\mathrm{0}\:\mathrm{are}\:\mathrm{2}\:\mathrm{more} \\ $$$$\mathrm{than}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:{ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0},\:\mathrm{then} \\ $$$$\mathrm{4}{a}−\mathrm{2}{b}+{c}\:\:\:\mathrm{can}\:\mathrm{be} \\ $$ Answered by ajfour last updated on 20/May/18…
Question Number 35533 by jonesme last updated on 20/May/18 $$\mathrm{If}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation}\:{x}^{\mathrm{2}} −{px}+{q}=\mathrm{0} \\ $$$$\mathrm{differ}\:\mathrm{by}\:\mathrm{unity},\:\mathrm{then} \\ $$ Answered by ajfour last updated on 20/May/18 $$\left(\alpha−\beta\right)^{\mathrm{2}} ={p}^{\mathrm{2}} −\mathrm{4}{q}=\mathrm{1}…
Question Number 35467 by Issifu1 last updated on 19/May/18 $$\mathrm{If}\:\:{x}\:=\:\mathrm{32}\:−\:\mathrm{16}\:\boldsymbol{\div}\:\mathrm{2}\:×\:\mathrm{4},\:\mathrm{then}\:{x}= \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 19/May/18 $${x}=\mathrm{0} \\ $$ Terms of Service…
Question Number 34446 by dilanho last updated on 06/May/18 $$\:\underset{−\pi/\mathrm{2}} {\overset{\pi/\mathrm{2}} {\int}}\:\mathrm{sin}\:\left\{\mathrm{log}\:\left({x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\:\right)\right\}\:{dx}\:= \\ $$ Commented by math khazana by abdo last updated on 06/May/18…
Question Number 34425 by Vijay kumar prasad last updated on 06/May/18 $${a},\:{b},\:{c}\:\in\:{R},\:{a}\neq\mathrm{0}\:\mathrm{and}\:\mathrm{the}\:\mathrm{quadratic} \\ $$$$\mathrm{equation}\:{ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0}\:\mathrm{has}\:\mathrm{no}\:\mathrm{real} \\ $$$$\mathrm{roots},\:\mathrm{then} \\ $$ Answered by MJS last updated on…
Question Number 99625 by Kalam last updated on 22/Jun/20 $$\mathrm{If}\:\:{f}\left({x}\right)=\mathrm{cos}^{\mathrm{2}} {x}\:+\:\mathrm{sec}^{\mathrm{2}} {x},\:\mathrm{its}\:\mathrm{value} \\ $$$$\mathrm{always}\:\mathrm{is} \\ $$ Answered by maths mind last updated on 22/Jun/20 $${cos}^{\mathrm{2}}…