Question Number 146158 by aliibrahim1 last updated on 11/Jul/21 Commented by MJS_new last updated on 12/Jul/21 $${x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={z} \\ $$$${z}=\mathrm{10}\vee{z}=\mathrm{10}\left(−\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right) \\ $$ Answered by…
Question Number 146146 by EDWIN88 last updated on 11/Jul/21 $$ \\ $$to admin tinku tara. why can't i post in hebrew ? Commented by Tinku…
Question Number 146100 by tabata last updated on 10/Jul/21 Commented by tabata last updated on 10/Jul/21 $${in}\:{the}\:{following}\:{figure}\:,{calculate}\:{the}\:{equivalent}\:{resistance}\:{of}\:{the} \\ $$$${external}\:{circuit}\:{and}\:{then}\:{calculate}\:{the}\:{current}\:{through}\:{it}\:? \\ $$ Answered by Olaf_Thorendsen last…
Question Number 146106 by smallEinstein last updated on 10/Jul/21 Answered by Olaf_Thorendsen last updated on 11/Jul/21 $$\Omega\:=\:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\:\sqrt[{\mathrm{4}}]{\left({x}+\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{1}−{x}\right)^{\mathrm{6}} }}\:{dx} \\ $$$$\mathrm{Let}\:{x}\:=\:\mathrm{cos2}{u} \\ $$$$\Omega\:=\:\int_{\frac{\pi}{\mathrm{4}}}…
Question Number 146102 by tabata last updated on 10/Jul/21 $${prove}\:{by}\:{mathmatical}\:{indiction}\: \\ $$$$\mathrm{5}+\mathrm{7}+\mathrm{9}+…..+\left(\mathrm{4}{n}+\mathrm{1}\right)=\mathrm{2}{n}^{\mathrm{2}} +\mathrm{3}{n} \\ $$ Answered by gsk2684 last updated on 11/Jul/21 $${let}\:{S}\left({n}\right):\mathrm{5}+\mathrm{7}+\mathrm{9}+…+\left(\mathrm{4}{n}+\mathrm{1}\right)=\mathrm{2}{n}^{\mathrm{2}} +\mathrm{3}{n} \\…
Question Number 146076 by Mrsof last updated on 10/Jul/21 $${if}\:{f}\left({x}\right)=\int\left(\mathrm{3}{x}^{\mathrm{2}} −\mathrm{2}\right){dx}\:{and}\:{f}\left(\mathrm{2}\right)=\mathrm{9}\:{then} \\ $$$${f}\left(−\mathrm{2}\right)=\:? \\ $$ Answered by gsk2684 last updated on 10/Jul/21 $${f}\left({x}\right)={x}^{\mathrm{3}} −\mathrm{2}{x}+{c} \\…
Question Number 15006 by ajfour last updated on 06/Jun/17 Commented by RasheedSoomro last updated on 07/Jun/17 $$\mathrm{Which}\:\boldsymbol{\mathrm{app}}\:\:\mathrm{do}\:\mathrm{you}\:\mathrm{use}\:\mathrm{to}\:\mathrm{draw}\:\mathrm{so} \\ $$$$\mathrm{nice}\:\mathrm{figures}\:?! \\ $$ Commented by ajfour last…
Question Number 80529 by TawaTawa last updated on 03/Feb/20 Commented by TawaTawa last updated on 03/Feb/20 $$\mathrm{Only}\:\mathrm{the}\:\:\mathrm{c}\:\:\mathrm{part}\:\mathrm{please} \\ $$ Terms of Service Privacy Policy Contact:…
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Question Number 80501 by naka3546 last updated on 03/Feb/20 $$\frac{\mathrm{5}}{\mathrm{7}}\:\:=\:\:\frac{{a}_{\mathrm{2}} }{\mathrm{2}!}\:+\:\frac{{a}_{\mathrm{3}} }{\mathrm{3}!}\:+\:\frac{{a}_{\mathrm{4}} }{\mathrm{4}!}\:+\:\frac{{a}_{\mathrm{5}} }{\mathrm{5}!}\:+\:\frac{{a}_{\mathrm{6}} }{\mathrm{6}!}\:+\:\frac{{a}_{\mathrm{7}} }{\mathrm{7}!} \\ $$$$\mathrm{0}\:\:\leqslant\:{a}_{{i}} \:<\:{i}\:\:,\:\:{a}_{{i}} \:\:\in\:\mathbb{N} \\ $$$${Find}\:\:{possible}\:\:{value}\:\:{of}\:\:\:{a}_{\mathrm{2}} \:+\:{a}_{\mathrm{3}} \:+\:{a}_{\mathrm{4}} \:+\:{a}_{\mathrm{5}}…