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This-app-is-now-free-as-several-users-do-not-have-cards-and-are-not-able-to-upgrade-Manual-entry-in-backend-system-had-put-lot-of-overhead-Hence-we-have-decide-to-make-this-app-free-We-issued-sever

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}…

The-acute-angle-of-the-rectangle-trapezius-is-equal-to-90-arcsin0-1-The-bases-measure-10-and-30-Calculate-the-area-of-the-trapezius-

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

The-area-of-the-equilateral-triangle-is-equal-to-16-8-3-pi-Calculate-the-area-of-the-circle-inscribed-in-the-triangle-

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-138477

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)…

If-x-26-521x-13-1-x-10-1-x-10-Any-help-

Question Number 138469 by KwesiDerek last updated on 13/Apr/21 $$\boldsymbol{\mathrm{If}}\:\:\:\boldsymbol{\mathrm{x}}^{\mathrm{26}} −\mathrm{521}\boldsymbol{\mathrm{x}}^{\mathrm{13}} =\mathrm{1} \\ $$$$\:\:\:\:\:\:\boldsymbol{\mathrm{x}}^{\mathrm{10}} +\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}^{\mathrm{10}} }=? \\ $$$$\boldsymbol{\mathrm{Any}}\:\boldsymbol{\mathrm{help}}\: \\ $$ Answered by MJS_new last updated…

Given-x-4-16-y-2-2y-2-0-Show-that-is-a-reunion-of-and-Ellipsis-and-an-hyperbole-then-give-their-equations-

Question Number 138470 by mathocean1 last updated on 13/Apr/21 $${Given}\:\left(\Gamma\right):\:{x}^{\mathrm{4}} −\mathrm{16}\left({y}^{\mathrm{2}} −\mathrm{2}{y}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$${Show}\:{that}\:\left(\Gamma\right)\:{is}\:{a}\:{reunion}\:{of}\:\:{and}\: \\ $$$${Ellipsis}\:{and}\:{an}\:{hyperbole}\:{then}\:{give} \\ $$$${their}\:{equations}. \\ $$ Answered by MJS_new last…