Question Number 138579 by Jamshidbek last updated on 15/Apr/21 Answered by MJS_new last updated on 16/Apr/21 $${x}\approx\mathrm{5}.\mathrm{40343170} \\ $$ Commented by MJS_new last updated on…
Question Number 138575 by KwesiDerek last updated on 15/Apr/21 $$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{x}} \\ $$$$\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\mathrm{2}^{\boldsymbol{\mathrm{x}}} =\mathrm{1} \\ $$$$\boldsymbol{\mathrm{Any}}\:\boldsymbol{\mathrm{help}} \\ $$ Commented by soudo last updated on 15/Apr/21…
Question Number 138539 by mohammad17 last updated on 14/Apr/21 $${find}\:{the}\:{intigral}\:{of}\:{complex}\:{number} \\ $$$$ \\ $$$${I}_{{j}} =\int_{\gamma{j}} {xdz}\:\:\:{if}\:{j}=\mathrm{1},\mathrm{2}\:{and}\:{Y}_{\mathrm{1}} {he}\:{is}\:{a}\:{circle} \\ $$$$ \\ $$$$\mid{z}\mid={R} \\ $$ Terms of…
Question Number 138536 by Tinku Tara last updated on 14/Apr/21 $$\mathrm{This}\:\mathrm{app}\:\mathrm{is}\:\mathrm{now}\:\mathrm{free}\:\mathrm{as}\:\mathrm{several}\:\mathrm{users} \\ $$$$\mathrm{do}\:\mathrm{not}\:\mathrm{have}\:\mathrm{cards}\:\mathrm{and}\:\mathrm{are}\:\mathrm{not}\:\mathrm{able} \\ $$$$\mathrm{to}\:\mathrm{upgrade}.\:\mathrm{Manual}\:\mathrm{entry}\:\mathrm{in}\:\mathrm{backend} \\ $$$$\mathrm{system}\:\mathrm{had}\:\mathrm{put}\:\mathrm{lot}\:\mathrm{of}\:\mathrm{overhead}. \\ $$$$\mathrm{Hence}\:\mathrm{we}\:\mathrm{have}\:\mathrm{decide}\:\mathrm{to}\:\mathrm{make}\:\mathrm{this} \\ $$$$\mathrm{app}\:\mathrm{free}. \\ $$$$\mathrm{We}\:\mathrm{issued}\:\mathrm{several}\:\mathrm{refunds}\:\mathrm{for}\:\mathrm{users} \\ $$$$\mathrm{who}\:\mathrm{had}\:\mathrm{bought}.\:\mathrm{However}\:\mathrm{Google}…
Question Number 72998 by yannickmendes_33 last updated on 05/Nov/19 $${The}\:{acute}\:{angle}\:{of}\:{the}\:{rectangle}\:{trapezius}\:{is}\:{equal}\:{to}\:\alpha=\mathrm{90}°{arcsin}\mathrm{0}.\mathrm{1} \\ $$$${The}\:{bases}\:{measure}\:\mathrm{10}\:{and}\:\mathrm{30}.\:{Calculate}\:{the}\:{area}\:{of}\:{the}\:{trapezius}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 72997 by yannickmendes_33 last updated on 05/Nov/19 $${The}\:{area}\:{of}\:{the}\:{equilateral}\:{triangle}\:{is}\:{equal}\:{to}\:\frac{\sqrt{\mathrm{16}}\sqrt{\mathrm{8}}}{\mathrm{3}\sqrt{\pi}} \\ $$$${Calculate}\:{the}\:{area}\:{of}\:{the}\:{circle}\:{inscribed}\:{in}\:{the}\:{triangle}. \\ $$$$\: \\ $$ Answered by Kunal12588 last updated on 05/Nov/19 $${area}\:{of}\:{equilateral}\:\bigtriangleup\:=\:\frac{\sqrt{\mathrm{3}}\:{a}^{\mathrm{2}} }{\mathrm{4}}=\frac{\sqrt{\mathrm{16}}\sqrt{\mathrm{8}}}{\mathrm{3}\sqrt{\pi}}=\frac{\mathrm{8}\sqrt{\mathrm{2}}}{\mathrm{3}\sqrt{\pi}}…
Question Number 138499 by yutytfjh67ihd last updated on 14/Apr/21 Commented by yutytfjh67ihd last updated on 14/Apr/21 $$\mathrm{Some}\:\mathrm{missing}\:\mathrm{unsolved}\:\mathrm{problem}. \\ $$ Answered by yutytfjh67ihd last updated on…
Question Number 138495 by yutytfjh67ihd last updated on 14/Apr/21 Commented by yutytfjh67ihd last updated on 14/Apr/21 $$\mathrm{Some}\:\mathrm{missing}\:\mathrm{unsolved}\:\mathrm{probelm}. \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 138477 by mathlove last updated on 14/Apr/21 Answered by Ñï= last updated on 14/Apr/21 $${a}^{\mathrm{2}} +\mathrm{2}{ab}+{b}^{\mathrm{2}} −{d}^{\mathrm{2}} −\mathrm{2}{cd}−{c}^{\mathrm{2}} \\ $$$$=\left({a}+{b}\right)^{\mathrm{2}} −\left({c}+{d}\right)^{\mathrm{2}} \\ $$$$=\left({a}+{b}+{c}+{d}\right)\left({a}+{b}−{c}−{d}\right)…
Question Number 72934 by Maclaurin Stickker last updated on 05/Nov/19 Answered by mr W last updated on 05/Nov/19 Commented by mr W last updated on…