Question Number 226697 by Linton last updated on 10/Dec/25 Commented by Linton last updated on 10/Dec/25 $${The}\:{letters}\:{in}\:{TWO}\:{UV}\:{PAIRS}\: \\ $$$${have}\:{the}\:{values}\:\mathrm{0}\:\mathrm{1}\:\mathrm{2}…\:\mathrm{9}\:{in}\:{some} \\ $$$${order}\:{with}\:{each}\:{letter}\:{represent} \\ $$$${ing}\:{a}\:{different}\:{digit}. \\ $$…
Question Number 226705 by mr W last updated on 10/Dec/25 Answered by fantastic2 last updated on 11/Dec/25 Commented by fantastic2 last updated on 11/Dec/25 $$\bigtriangleup{OPQ}\cong\bigtriangleup{SRQ}…
Question Number 226675 by hardmath last updated on 10/Dec/25 Commented by hardmath last updated on 10/Dec/25 $$ \\ $$The image is not fully displayed, the…
Question Number 226668 by Spillover last updated on 09/Dec/25 Answered by Frix last updated on 09/Dec/25 $$=\underset{\mathrm{0}} {\overset{\pi} {\int}}\left(\mathrm{2cos}\:\mathrm{16}{x}\:+\mathrm{4cos}\:\mathrm{14}{x}\:+\mathrm{2cos}\:\mathrm{12}{x}\:−\mathrm{4cos}\:\mathrm{10}{x}\:−\mathrm{8cos}\:\mathrm{8}{x}\:−\mathrm{8cos}\:\mathrm{6}{x}\:−\mathrm{2cos}\:\mathrm{4}{x}\:+\mathrm{8cos}\:\mathrm{2}{x}\:+\mathrm{7}\right){dx}= \\ $$$$… \\ $$$$=\mathrm{7}\pi \\ $$…
Question Number 226653 by MrAjder last updated on 09/Dec/25 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{Si}\left(\pi{n}\right)}{{n}^{\mathrm{2}} }=? \\ $$ Commented by Kassista last updated on 10/Dec/25 Terms of Service…
Question Number 226638 by Lara2440 last updated on 08/Dec/25 Commented by Lara2440 last updated on 08/Dec/25 $$\mathrm{Boy}'\mathrm{s}\:\mathrm{Surface}\:\mathrm{can}\:\mathrm{be}\:\mathrm{parametrized}\:\mathrm{in}\:\mathrm{several}\:\mathrm{ways} \\ $$$$\mathrm{Given}\:\mathrm{complex}\:\mathrm{number}\:{w}\:\mathrm{whose}\:\mid\mid{w}\mid\mid<\mathrm{1}, \\ $$$$\mathrm{Let}\:{g}_{\mu} =\begin{cases}{{g}_{\mathrm{1}} =−\frac{\mathrm{3}}{\mathrm{2}}\mathrm{Im}\left[\:\frac{{w}\left(\mathrm{1}−{w}^{\mathrm{4}} \right)}{{w}^{\mathrm{6}} +\sqrt{\mathrm{5}}{w}^{\mathrm{3}}…
Question Number 226651 by Hanuda354 last updated on 08/Dec/25 Commented by Hanuda354 last updated on 08/Dec/25 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{shaded}\:\mathrm{area}\:\left(\mathrm{without}\:\mathrm{using}\:\mathrm{integration}\right). \\ $$ Answered by fantastic2 last updated on…
Question Number 226635 by Kassista last updated on 08/Dec/25 Answered by mingski last updated on 08/Dec/25 $${C}:{z}^{\mathrm{2}} +\mathrm{6}{z}+\mathrm{10}=\mathrm{0} \\ $$$$\left({z}+\mathrm{3}\right)^{\mathrm{2}} =−\mathrm{1} \\ $$$${z}+\mathrm{3}=\pm\mathrm{i},{z}=−\mathrm{3}\pm\mathrm{i}. \\ $$$${for}\:{z}_{\mathrm{1}}…
Question Number 226640 by fantastic2 last updated on 08/Dec/25 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 226641 by fantastic2 last updated on 08/Dec/25 $${if}\:\mathrm{log}\:_{\mathrm{8}} {a}+\mathrm{log}\:_{\mathrm{4}} {b}^{\mathrm{2}} =\mathrm{5} \\ $$$$\& \\ $$$$\mathrm{log}\:_{\mathrm{8}} ^{{b}} +\mathrm{log}\:_{\mathrm{4}} {a}^{\mathrm{2}} =\mathrm{7} \\ $$$${ab}=? \\ $$…