Question Number 36091 by Rio Mike last updated on 28/May/18 $$\mathrm{Given}\:\mathrm{the}\:\mathrm{position}\:\mathrm{vectors} \\ $$$${v}_{\mathrm{1}} =\:\mathrm{2}{i}\:−\:\mathrm{2}{j}\:{and}\:{v}_{\mathrm{2}} =\:\mathrm{2}{j}, \\ $$$$\mathrm{show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{unit}\:\mathrm{vector}\:\mathrm{in}\: \\ $$$$\mathrm{the}\:\mathrm{direction}\:\mathrm{of}\:\mathrm{the}\:\mathrm{vector}\: \\ $$$${v}_{\mathrm{1}} −\:{v}_{\mathrm{2}\:\:\:} \mathrm{is}\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(\mathrm{i}−\mathrm{2j}\right) \\ $$…
Question Number 36080 by Rio Mike last updated on 28/May/18 $$\:\mathrm{i}\:\mathrm{want}\:\mathrm{to}\:\mathrm{know}\:\mathrm{how}\: \\ $$$$\alpha^{\mathrm{2}} +\:\beta^{\mathrm{2}} =\:\left(\alpha+\beta\right)^{\mathrm{2}} −\:\mathrm{2}\alpha\beta\:\mathrm{why}\:\mathrm{not}\: \\ $$$$\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} =\:\left(\alpha+\beta\right)^{\mathrm{2}} +\:\mathrm{2}\alpha\beta? \\ $$ Commented by…
Question Number 101616 by 175 last updated on 03/Jul/20 Commented by Tinku Tara last updated on 03/Jul/20 $$\mathrm{Please}\:\mathrm{post}\:\mathrm{question}\:\mathrm{where}\:\mathrm{you} \\ $$$$\mathrm{yourself}\:\mathrm{doubt}. \\ $$ Answered by mr…
Question Number 36059 by solihin last updated on 28/May/18 Answered by Rasheed.Sindhi last updated on 28/May/18 $$\frac{\mathrm{z}_{\mathrm{1}} +\mathrm{z}_{\mathrm{2}} }{\mathrm{z}_{\mathrm{3}} }=\frac{\left(\mathrm{1}+\mathrm{2i}\right)+\left(\mathrm{4}−\mathrm{3i}\right)}{−\mathrm{5}−\mathrm{2i}} \\ $$$$=\frac{\mathrm{5}−\mathrm{i}}{−\mathrm{5}−\mathrm{2i}}×\frac{−\mathrm{5}+\mathrm{2i}}{−\mathrm{5}+\mathrm{2i}}=\frac{−\mathrm{25}+\mathrm{10i}+\mathrm{5i}+\mathrm{2}}{\left(−\mathrm{5}\right)^{\mathrm{2}} −\left(\mathrm{2i}\right)^{\mathrm{2}} } \\…
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Question Number 36049 by Rio Mike last updated on 27/May/18 $$\mathrm{A}\:\mathrm{triangle}\:\bigtriangleup\mathrm{ABC}\:\mathrm{is}\:\mathrm{constructed}\: \\ $$$$\mathrm{such}\:\mathrm{that}\:\angle\mathrm{B}=\:\mathrm{90}°\:\:\mathrm{and}\:\mathrm{AB}=\:\mathrm{5cm} \\ $$$$\mathrm{Given}\:\mathrm{that}\:\angle\mathrm{C}=\:\mathrm{45}°\:.\:\mathrm{show}\:\mathrm{that} \\ $$$$\:\mathrm{the}\:\mathrm{point}\:\left(\mathrm{2},\mathrm{0}\right)\:\mathrm{lie}\:\mathrm{on}\:\mathrm{the}\:\mathrm{line}\: \\ $$$$\mathrm{BC}\:\mathrm{and}\:\mathrm{is}\:\mathrm{perpendicular}\:\mathrm{to}\:\mathrm{AB}\:\mathrm{but} \\ $$$$\mathrm{meet}\:\mathrm{at}\:\mathrm{45}^{} °\:\mathrm{with}\:\mathrm{the}\:\mathrm{line}\:\mathrm{AC}. \\ $$$$ \\…
Question Number 36034 by solihin last updated on 27/May/18 Commented by abdo mathsup 649 cc last updated on 27/May/18 $${we}\:{have}\:\overset{−} {{z}}\left({z}−{i}\right)\:=\mathrm{10}\:+\mathrm{2}{i}\:\:{let}\:{put}\:\:{z}\:={x}+{iy}\:\Rightarrow \\ $$$$\left({x}−{iy}\right)\left({x}+{iy}−{i}\right)\:=\mathrm{10}\:+\mathrm{2}{i}\:\Leftrightarrow \\ $$$${x}^{\mathrm{2}}…
Question Number 101555 by 175 last updated on 03/Jul/20 Answered by EuclidKaBaap last updated on 03/Jul/20 $${speed}\:{of}\:{body}\:=\:{n}\:{rounds}\:/\:\mathrm{60}\:{sec} \\ $$$${distance}\:{of}\:\mathrm{1}\:{round}\:=\:\mathrm{2}\Pi{r} \\ $$$${where}\:{r}\:{is}\:{radius}\:{of}\:{circular}\:{track}. \\ $$$${so}\:{distance}\:{covered}\:{in}\:\mathrm{1}\:{sec}\:= \\ $$$$\:\:\:\:\:\:\:\frac{{n}\left(\mathrm{2}\Pi{r}\right)}{\mathrm{60}}=\:\frac{{n}\Pi{r}}{\mathrm{30}}…
Question Number 36011 by solihin last updated on 27/May/18 $$\mathrm{simplify}:\:\:'{interval}\:{number}' \\ $$$$\left(\mathrm{1},\mathrm{6}\right)\cup\left(\mathrm{3},\mathrm{7}\right) \\ $$ Commented by Rasheed.Sindhi last updated on 27/May/18 $$\mathrm{For}\:\mathrm{overlapping}\:\mathrm{intervals}\:\left({a},{b}\right)\:\&\:\left({c},{d}\right) \\ $$$$\left({a},{b}\right)\cup\left({c},{d}\right)=\left(\:\:\:\mathrm{min}\left({a},{c}\right)\:,\:\mathrm{max}\left({b},{d}\right)\:\:\:\right) \\…
Question Number 36003 by rishijagtar@gmail.com last updated on 27/May/18 $${x}\begin{bmatrix}{\mathrm{2}}\\{\mathrm{1}}\end{bmatrix}+{y}\begin{bmatrix}{\mathrm{3}}\\{\mathrm{5}}\end{bmatrix}+\begin{bmatrix}{−\mathrm{8}}\\{−\mathrm{11}}\end{bmatrix}=\mathrm{0} \\ $$$${find}\:{x}\:{and}\:{y} \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 27/May/18 $$\mathrm{2}{x}+\mathrm{3}{y}−\mathrm{8}=\mathrm{0} \\ $$$${x}+\mathrm{5}{y}−\mathrm{11}=\mathrm{0} \\…