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Author: Tinku Tara

find-the-first-root-8i-1-2-

Question Number 139608 by mohammad17 last updated on 29/Apr/21 $${find}\:{the}\:{first}\:{root}\:\left(−\mathrm{8}{i}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$ Answered by floor(10²Eta[1]) last updated on 29/Apr/21 $$\sqrt{−\mathrm{8i}}=\mathrm{a}+\mathrm{bi} \\ $$$$−\mathrm{8i}=\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} +\mathrm{2abi} \\…

Question-74075

Question Number 74075 by ajfour last updated on 18/Nov/19 Commented by ajfour last updated on 18/Nov/19 $${To}\:{Tinkutara}\: \\ $$$${I}\:{was}\:{going}\:{to}\:{delete}\:{it}\:{on}\:{second} \\ $$$${thought}\:{but}\:{i}\:{touched}\:{upvote} \\ $$$${and}\:{cannot}\:{delete}\:{it}\:{now}, \\ $$$$\left({please}\:{help}\:{delete}\:{it}.\right)…

A-2-b-translation-T-1-3-b-followed-by-translation-T-2-a-4-A-b-1-a-3-determine-the-value-of-a-b-

Question Number 8528 by suci last updated on 14/Oct/16 $${A}\left(\mathrm{2},{b}\right)\:{translation}\:{T}_{\mathrm{1}} =\begin{pmatrix}{−\mathrm{3}}\\{\:\:\:{b}}\end{pmatrix} \\ $$$${followed}\:{by}\:{translation}\:{T}_{\mathrm{2}} =\begin{pmatrix}{{a}}\\{\mathrm{4}}\end{pmatrix} \\ $$$${A}'=\left({b}−\mathrm{1},{a}−\mathrm{3}\right) \\ $$$${determine}\:{the}\:{value}\:{of}\:{a}+{b}=…? \\ $$ Answered by sandy_suhendra last updated…

6-32-

Question Number 8527 by suci last updated on 14/Oct/16 $$\sqrt{\mathrm{6}−\sqrt{\mathrm{32}}}\:=…? \\ $$ Answered by prakash jain last updated on 14/Oct/16 $$\sqrt{\mathrm{6}−\sqrt{\mathrm{32}}}=\sqrt{\mathrm{6}−\mathrm{4}\sqrt{\mathrm{2}}} \\ $$$$\sqrt{\mathrm{6}−\mathrm{4}\sqrt{\mathrm{2}}}=\sqrt{\mathrm{a}}−\sqrt{\mathrm{b}} \\ $$$$\mathrm{squaring}…

8-2-5-

Question Number 8526 by suci last updated on 14/Oct/16 $$\sqrt{\mathrm{8}+\mathrm{2}\sqrt{\mathrm{5}}}\:=\:…? \\ $$ Answered by ridwan balatif last updated on 14/Oct/16 $$\sqrt{\mathrm{8}+\mathrm{2}\sqrt{\mathrm{5}}}=\sqrt{\mathrm{k}}+\sqrt{\mathrm{p}} \\ $$$$\left(\sqrt{\mathrm{8}+\mathrm{2}\sqrt{\mathrm{5}}}\right)^{\mathrm{2}} =\left(\sqrt{\mathrm{k}}+\sqrt{\mathrm{p}}\right)^{\mathrm{2}} \\…