Question Number 158081 by Eric002 last updated on 30/Oct/21 $${determine}\:{the}\:{angle}\:{between}\:{two}\:{vectors} \\ $$$${A}=\mathrm{4}{ax}+{ay}−\mathrm{3}{az}\:\:{and}\:\:{B}=\mathrm{2}{ax}+\mathrm{4}{ay}−\mathrm{3}{az} \\ $$ Answered by ajfour last updated on 30/Oct/21 $$\mathrm{cos}\:\theta=\frac{\mathrm{4}\left(\mathrm{2}\right)+\mathrm{1}\left(\mathrm{4}\right)−\mathrm{3}\left(−\mathrm{3}\right)}{\:\sqrt{\mathrm{4}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} +\left(−\mathrm{3}\right)^{\mathrm{2}} }\sqrt{\mathrm{2}^{\mathrm{2}}…
Question Number 158066 by zainaltanjung last updated on 30/Oct/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 92521 by hovero clinton last updated on 07/May/20 Commented by hovero clinton last updated on 07/May/20 $${my}\:{post}\:{here}\:{never}\: \\ $$$${get}\:{answers}\:{y}?? \\ $$ Commented by…
Question Number 92235 by ~blr237~ last updated on 05/May/20 $${let}\:\:\mathrm{0}<{p}<\mathrm{1}\:\:{and}\:\:{x}>\mathrm{0} \\ $$$${prove}\:{that}\:\:\:\:{x}^{\mathrm{2}} \leqslant\:\left(\mathrm{1}−{p}\right)\left(\:\:\:^{\left(\mathrm{1}−{p}\right)} \sqrt{{x}}\:\right)\:+{p}\:\left(\:^{{p}} \sqrt{{x}}\right) \\ $$$$ \\ $$$$ \\ $$ Terms of Service Privacy…
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} \\…