Question Number 15352 by Tinkutara last updated on 09/Jun/17 $$\mathrm{Let}\:\overset{\rightarrow} {{a}}\:=\:\overset{\wedge} {{i}}\:+\:\mathrm{4}\overset{\wedge} {{j}}\:+\:\mathrm{2}\overset{\wedge} {{k}},\:\overset{\rightarrow} {{b}}\:=\:\mathrm{3}\overset{\wedge} {{i}}\:−\:\mathrm{2}\overset{\wedge} {{j}}\:+\:\mathrm{7}\overset{\wedge} {{k}} \\ $$$$\mathrm{and}\:\overset{\rightarrow} {{c}}\:=\:\mathrm{2}\overset{\wedge} {{i}}\:−\:\overset{\wedge} {{j}}\:+\:\mathrm{4}\overset{\wedge} {{k}}\:.\:\mathrm{Find}\:\mathrm{a}\:\mathrm{vector}\:\overset{\rightarrow} {{d}}…
Question Number 15348 by tawa tawa last updated on 09/Jun/17 $$\mathrm{Solve}\:\mathrm{the}\:\mathrm{equation}\:\:\mathrm{z}^{\mathrm{2}} \:+\:\mathrm{2}\left(\mathrm{1}\:+\:\mathrm{j}\right)\mathrm{z}\:+\:\mathrm{2}\:=\:\mathrm{0},\:\mathrm{Givin}\:\mathrm{each}\:\mathrm{result}\:\mathrm{in}\:\mathrm{form}\:\:\mathrm{a}\:+\:\mathrm{jb} \\ $$$$\mathrm{with}\:\:\mathrm{a}\:\:\mathrm{and}\:\:\mathrm{b}\:\:\mathrm{correct}\:\mathrm{to}\:\:\mathrm{2dp}. \\ $$ Answered by RasheedSoomro last updated on 10/Jun/17 $$\mathrm{Let}\:\mathrm{z}=\mathrm{x}+\mathrm{jy} \\…
Question Number 146403 by gsk2684 last updated on 13/Jul/21 $${let}\:{f}\left({x},{y}\right)=\frac{{x}^{\mathrm{5}} {y}^{\mathrm{2}} }{\mathrm{10}}\:{then}\:{find}\: \\ $$$${D}_{{u}} {f}\left(−\mathrm{5},−\mathrm{3}\right)\:{in}\:{the}\:{direction}\:{of}\: \\ $$$${the}\:{vector}\:<\mathrm{0},−\mathrm{2}>? \\ $$ Commented by gsk2684 last updated on…
Question Number 15284 by ajfour last updated on 09/Jun/17 Commented by ajfour last updated on 09/Jun/17 $${Q}.\mathrm{14940}\:\:\left({general}\:{proof}\::\:{sides}\:\right. \\ $$$$\left.\:{of}\:\Delta{ABC}\:{divided}\:{in}\:\boldsymbol{{n}}\:{equal}\:{parts}\right) \\ $$ Commented by RasheedSoomro last…
Question Number 146315 by mnjuly1970 last updated on 12/Jul/21 Answered by Olaf_Thorendsen last updated on 13/Jul/21 $$\sqrt{\frac{{x}\left({x}+\mathrm{5}\right)}{{x}+\mathrm{2}}}+\sqrt{{x}+\mathrm{3}}\:=\:\frac{\sqrt{\mathrm{16}{x}+\mathrm{32}}}{{x}+\mathrm{2}} \\ $$$$\sqrt{\frac{{x}\left({x}+\mathrm{5}\right)}{{x}+\mathrm{2}}}+\sqrt{{x}+\mathrm{3}}\:=\:\frac{\mathrm{4}}{\:\sqrt{{x}+\mathrm{2}}} \\ $$$$\sqrt{{x}\left({x}+\mathrm{5}\right)}+\sqrt{\left({x}+\mathrm{3}\right)\left({x}+\mathrm{2}\right)}\:=\:\mathrm{4} \\ $$$$\mathrm{Let}\:{u}\:=\:{x}+\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$\sqrt{\left({u}−\frac{\mathrm{5}}{\mathrm{2}}\right)\left({u}+\frac{\mathrm{5}}{\mathrm{2}}\right)}+\sqrt{\left({u}+\frac{\mathrm{1}}{\mathrm{2}}\right)\left({u}−\frac{\mathrm{1}}{\mathrm{2}}\right)}\:=\:\mathrm{4}…
Question Number 15206 by tawa tawa last updated on 08/Jun/17 $$\mathrm{If}\:\:\:\mathrm{z}\:=\:\mathrm{x}\:+\:\mathrm{jy},\:\:\mathrm{detemine}\:\mathrm{the}\:\mathrm{cartesian}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{locus}\:\mathrm{of}\:\mathrm{the}\:\mathrm{point}\: \\ $$$$\mathrm{z}\:\mathrm{which}\:\mathrm{moves}\:\mathrm{in}\:\mathrm{the}\:\mathrm{Argrand}\:\mathrm{diagram}\:\mathrm{so}\:\mathrm{that} \\ $$$$\mid\mathrm{z}\:+\:\mathrm{j2}\mid^{\mathrm{2}} \:+\:\mid\mathrm{z}\:−\:\mathrm{j2}\mid^{\mathrm{2}\:} =\:\mathrm{40} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 15202 by tawa tawa last updated on 08/Jun/17 $$\mathrm{if}\:\mathrm{z}\:=\:\mathrm{x}\:+\:\mathrm{jy}\:,\:\mathrm{find}\:\mathrm{the}\:\mathrm{equations}\:\mathrm{of}\:\mathrm{the}\:\mathrm{two}\:\mathrm{loci}\:\mathrm{defined}\:\mathrm{by}: \\ $$$$\left(\mathrm{a}\right)\:\mid\mathrm{z}\:−\:\mathrm{4}\mid\:=\:\mathrm{3}\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\left(\mathrm{b}\right)\:\mathrm{arg}\left(\mathrm{z}\:+\:\mathrm{2}\right)\:=\:\frac{\pi}{\mathrm{6}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 15172 by tawa tawa last updated on 07/Jun/17 $$\mathrm{If}\:\:\mathrm{z}\:=\:\mathrm{x}\:+\:\mathrm{jy},\:\mathrm{show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{locus}\:\mathrm{arg}\left\{\frac{\mathrm{z}\:−\:\mathrm{1}}{\mathrm{z}\:−\:\mathrm{j}}\right\}\:=\:\frac{\pi}{\mathrm{6}}\:\mathrm{is}\:\mathrm{a}\:\mathrm{circle}\:.\:\mathrm{Find}\:\mathrm{its}\:\mathrm{centre} \\ $$$$\mathrm{and}\:\mathrm{radius}.\: \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 80642 by naaz last updated on 05/Feb/20 $$ \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 14939 by Tinkutara last updated on 05/Jun/17 $$\mathrm{A}\:\mathrm{point}\:\mathrm{moves}\:\mathrm{in}\:{x}-{y}\:\mathrm{plane}\:\mathrm{according} \\ $$$$\mathrm{to}\:\mathrm{the}\:\mathrm{law}\:{x}\:=\:\mathrm{4}\:\mathrm{sin}\:\mathrm{6}{t}\:\mathrm{and} \\ $$$${y}\:=\:\mathrm{4}\left(\mathrm{1}\:−\:\mathrm{cos}\:\mathrm{6}{t}\right).\:\mathrm{Find}\:\mathrm{distance}\:\mathrm{traversed} \\ $$$$\mathrm{by}\:\mathrm{the}\:\mathrm{particle}\:\mathrm{in}\:\mathrm{5}\:\mathrm{seconds},\:\mathrm{when}\:{x}\:\mathrm{and} \\ $$$${y}\:\mathrm{are}\:\mathrm{in}\:\mathrm{metres}. \\ $$ Answered by ajfour last updated…