Question Number 26374 by tawa tawa last updated on 24/Dec/17 $$\mathrm{show}\:\mathrm{that}\:\:\mathrm{if}\:\:\:\:\mathrm{arg}\left(\frac{\mathrm{z}_{\mathrm{1}} \:+\:\mathrm{z}_{\mathrm{2}} }{\mathrm{z}_{\mathrm{1}} \:−\:\mathrm{z}_{\mathrm{2}} }\right)\:=\:\frac{\pi}{\mathrm{2}}\:\:\:\:\mathrm{then}\:\:\:\:\mid\mathrm{z}_{\mathrm{1}} \mid\:=\:\mid\mathrm{z}_{\mathrm{2}} \mid \\ $$ Answered by Rasheed.Sindhi last updated on…
Question Number 25783 by lizan 123 last updated on 14/Dec/17 $${If}\:\:\:{r}\:\:{is}\:\:{a}\:\:{unit}\:\:{vector}\:{then}\:\:{show}\:{that}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mid{r}×\frac{{dr}}{{dt}}\mid\:\:=\:\:\mid\frac{{dr}}{{dt}}\mid \\ $$ Answered by ajfour last updated on 15/Dec/17 $$\bar {{r}}=\mathrm{cos}\:\theta\hat {{i}}+\mathrm{sin}\:\theta\hat…
Question Number 25245 by adityapratap2585@gmail.com last updated on 06/Dec/17 $$\mathrm{yadi}\:\mathrm{kisi}\:\mathrm{equilatter}\:\mathrm{triangle}\:\mathrm{ke}\: \\ $$$$\mathrm{parivrit}\:\mathrm{aur}\:\mathrm{antahvrit}\:\mathrm{ke}\:\mathrm{areas}\:\mathrm{ka}\: \\ $$$$\mathrm{difference}\:\mathrm{44cm}^{\mathrm{2}} \:\mathrm{hai}\:\mathrm{to}\:\mathrm{triangle}\:\mathrm{ka}\: \\ $$$$\mathrm{area}\:\mathrm{kya}\:\mathrm{hoga}? \\ $$ Answered by $@ty@m last updated on…
Question Number 156082 by mnjuly1970 last updated on 07/Oct/21 $$ \\ $$$$\:\:\:\:\:\mathrm{0}<\:\alpha\:<\frac{\pi}{\mathrm{2}}\:\:\: \\ $$$$\left.\:\:\sqrt[{\:\mathrm{3}}]{\:{sin}\left(\alpha\right)}\:+\:\sqrt[{\mathrm{3}}]{{cos}\left(\alpha\right.}\right)=\:\sqrt[{\mathrm{3}}]{\:{tan}\left(\alpha\right)} \\ $$$$\: \\ $$$$\:\:\:\:\:\:\:\frac{\:{tan}\:\left(\alpha\:\right)\:+\:{cot}\:\left(\alpha\:\right)}{\mathrm{2}}\:=? \\ $$ Terms of Service Privacy Policy…
Question Number 90468 by ar247 last updated on 23/Apr/20 Commented by ar247 last updated on 23/Apr/20 $${help}\:{me}\:{please}.\:{stap}\:{by}\:{stap}\: \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 24622 by ajfour last updated on 22/Nov/17 Commented by ajfour last updated on 22/Nov/17 $${If}\:\bar {{b}}\:{is}\:{the}\:{result}\:{of}\:{rotating}\:\bar {{a}} \\ $$$${about}\:\bar {{r}}\:{by}\:{an}\:{angle}\:\theta\:{in}\:{the} \\ $$$${manner}\:{shown},\:{then},\:{express} \\…
Question Number 24569 by ajfour last updated on 21/Nov/17 Commented by ajfour last updated on 21/Nov/17 $${In}\:{a}\:{gravity}\:{free}\:{cubical}\:{box},\:{a} \\ $$$${ball}\:{bounces}\:{elastically}.\:{If}\:{it}\:{be} \\ $$$${projected}\:{from}\:{origin}\:{O}\:{towards} \\ $$$${A}\left(\mathrm{4},\:\mathrm{13},\:\mathrm{16}\right),\:{a}\:{point}\:{on}\:{the}\:{roof} \\ $$$${plane}\:{then}\:{find}\:{the}\:{second}\:{and}…
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Question Number 89033 by M±th+et£s last updated on 14/Apr/20 Answered by ajfour last updated on 15/Apr/20 $${let}\:{center}\:{of}\:{square}\:{be}\:{origin}. \\ $$$${eq}.\:{of}\:{left}\:{circle}: \\ $$$$\left({x}+\frac{{r}}{\mathrm{2}}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$${let}\:{side}\:{of}\:{square}=\mathrm{2}{s}…
Question Number 23442 by chernoaguero@gmail.com last updated on 30/Oct/17 $$\mathrm{Find}\:\mathrm{a}\:\mathrm{unit}\:\mathrm{vector}\:\mathrm{which}\:\mathrm{is}\:\mathrm{perpendicula} \\ $$$$\mathrm{to}\:\mathrm{a}\:\mathrm{vector}\:\mathrm{A}\left(\mathrm{3coma5coma1}\right) \\ $$$$ \\ $$$$\mathrm{Sorry}\:\mathrm{for}\:\mathrm{writing}\:\mathrm{coma}\:\mathrm{cuz}\:\mathrm{i}\:\mathrm{dnt}\:\mathrm{see}\:\mathrm{a}\:\mathrm{key}\:\mathrm{for}\:\mathrm{it} \\ $$ Commented by $@ty@m last updated on 31/Oct/17…