Question Number 78316 by mathocean1 last updated on 15/Jan/20 $$\mathrm{the}\:\mathrm{circle}\:\mathrm{represents}\:\mathrm{a}\:\mathrm{farm}\:\mathrm{where} \\ $$$$\left(\mathrm{LK}\right)\:\mathrm{is}\:\mathrm{symetric}\:\mathrm{axe}\:\mathrm{of}\:\mathrm{circle}\:\mathrm{such} \\ $$$$\mathrm{as}\:\forall\:\mathrm{M}\:\mathrm{of}\:\mathrm{this}\:\mathrm{circle}\:\mathrm{verifying} \\ $$$$\mathrm{ML}^{\mathrm{2}} −\mathrm{4MK}^{\mathrm{2}} =\mathrm{0}\:\:\mathrm{with}\:\mathrm{LK}=\mathrm{150m}. \\ $$$$\mathrm{calculate}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{circle}. \\ $$$$\mathrm{please}\:\mathrm{help}\:\mathrm{me}… \\ $$ Commented…
Question Number 12752 by tawa last updated on 30/Apr/17 $$\mathrm{Let}\:\mathrm{V}\:\mathrm{and}\:\mathrm{W}\:\mathrm{be}\:\mathrm{4}\:\mathrm{dimensional}\:\mathrm{subspaces}\:\mathrm{of}\:\mathrm{a}\:\mathrm{7}\:\mathrm{dimensional}\:\mathrm{vector}\:\mathrm{space}\:\mathrm{X}. \\ $$$$\mathrm{Which}\:\mathrm{of}\:\mathrm{the}\:\mathrm{following}\:\mathrm{CANNOT}\:\mathrm{be}\:\mathrm{the}\:\mathrm{dimension}\:\mathrm{of}\:\mathrm{the}\:\mathrm{subspace}\:\mathrm{V}\cap\mathrm{W}. \\ $$$$\left(\mathrm{A}\right)\:\mathrm{0}\:\left(\mathrm{B}\right)\:\mathrm{1}\:\left(\mathrm{C}\right)\:\mathrm{2}\:\left(\mathrm{D}\right)\:\mathrm{3}\:\left(\mathrm{E}\right)\:\mathrm{4} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 78060 by aliesam last updated on 13/Jan/20 Commented by aliesam last updated on 13/Jan/20 $${the}\:{triangle}\:{is}\:{equilateral} \\ $$ Answered by ajfour last updated on…
Question Number 12442 by tawa last updated on 22/Apr/17 $$\mathrm{Solve}:\:\:\:\mathrm{z}^{\mathrm{4}} \:=\:−\:\mathrm{16} \\ $$ Answered by ajfour last updated on 22/Apr/17 $${z}^{\mathrm{4}} =\mathrm{2}^{\mathrm{4}} {e}^{{i}\left(\pi+\mathrm{2}{k}\pi\right)} \\ $$$${z}=\mathrm{2}{e}^{{i}\left(\frac{\pi}{\mathrm{4}}+\frac{\mathrm{2}{k}\pi}{\mathrm{4}}\:\right)}…
Question Number 77745 by mathocean1 last updated on 09/Jan/20 $$\mathrm{ABC}\:\mathrm{is}\:\mathrm{any}\:\mathrm{triangle}. \\ $$$$\mathrm{C}'\:.\:\mathrm{B}'\:\:.\mathrm{A}'\:\:\mathrm{are}\:\mathrm{respectively}\:\mathrm{middles} \\ $$$$\mathrm{of}\:\left[\mathrm{AB}\right]\:.\:\left[\mathrm{AC}\right]\:\:\mathrm{and}\:\:\left[\mathrm{BC}\right]. \\ $$$$\mathrm{we}\:\mathrm{suppose}\:\mathrm{that}\: \\ $$$$\mathrm{AB}=\mathrm{c}\:\:\:\mathrm{AC}=\mathrm{b}\:\:\:\:\mathrm{BC}=\mathrm{a}. \\ $$$$\left.\mathrm{1}\right)\:\overset{\rightarrow\:} {\mathrm{u}}=\mathrm{a}^{\mathrm{2}} \mathrm{B}\overset{\rightarrow} {\mathrm{C}}+\mathrm{b}^{\mathrm{2}\:} \overset{\rightarrow} {\mathrm{C}A}+\mathrm{c}^{\mathrm{2}}…
Question Number 77676 by mathocean1 last updated on 08/Jan/20 $$\mathrm{Please}\:\mathrm{how}\:\mathrm{can}\:\mathrm{we}\:\mathrm{demonstrate} \\ $$$$\mathrm{that}\:\mathrm{a}\:\mathrm{vector}\:\mathrm{is}\:\mathrm{null}… \\ $$$$\mathrm{hello} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 143127 by liberty last updated on 10/Jun/21 Answered by EDWIN88 last updated on 11/Jun/21 $$\:\mathrm{Let}\:\begin{cases}{\overset{\rightarrow} {\mathrm{a}}=\mathrm{QR}=\left(\mathrm{1},\mathrm{3},−\mathrm{1}\right)}\\{\overset{\rightarrow} {\mathrm{b}}=\mathrm{QS}=\left(−\mathrm{3},\mathrm{2},\mathrm{3}\right)}\\{\overset{\rightarrow} {\mathrm{c}}=\mathrm{QP}=\left(\mathrm{0},\mathrm{3},\mathrm{3}\right)}\end{cases} \\ $$$$\mathrm{distance}\:\mathrm{d}\:\mathrm{from}\:\mathrm{P}\:\mathrm{to}\:\mathrm{plane}\:\mathrm{is}\: \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{d}\:=\:\frac{\mid\overset{\rightarrow} {\mathrm{a}}.\left(\overset{\rightarrow}…
Question Number 142970 by mnjuly1970 last updated on 08/Jun/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:……….{CALCULUS}……….. \\ $$$$\:\:\:\:\:\:\:{prove}\:{that}::\:\: \\ $$$$\:\:\:\:\:\:\boldsymbol{\phi}:=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left(\left({n}−\mathrm{1}\right)!\right)^{\mathrm{2}} }{\left(\mathrm{2}{n}\right)!}=\mathrm{2}{log}^{\mathrm{2}} \left(\varphi\right) \\ $$$$\:\:\:\:\varphi={golden}\:{ratio}…. \\ $$$$\:\:\:\:…………. \\ $$…
Question Number 142927 by bramlexs22 last updated on 07/Jun/21 Answered by Olaf_Thorendsen last updated on 07/Jun/21 $$\mathrm{Let}\:\left(\mathrm{C},\:\overset{\rightarrow} {{i}},\overset{\rightarrow} {{j}}\right) \\ $$$$\mathrm{with}\:\mathrm{C}\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix} \\ $$$$\mathrm{G}\begin{pmatrix}{{c}_{\mathrm{2}} }\\{\mathrm{0}}\end{pmatrix}\:=\:{c}_{\mathrm{2}} \overset{\rightarrow}…
Question Number 77356 by mathocean1 last updated on 05/Jan/20 $$\mathrm{The}\:\mathrm{plan}\:\mathrm{is}\:\mathrm{provided}\:\mathrm{with}\:\mathrm{an}\: \\ $$$$\mathrm{orthonormal}\:\mathrm{reference}\:\left(\:\mathrm{O}.\mathrm{I}.\mathrm{J}\right). \\ $$$$\mathrm{the}\:\mathrm{following}\:\mathrm{points}\:\mathrm{are}\:\mathrm{given} \\ $$$$\mathrm{A}\left(\mathrm{1},\mathrm{2}\right)\:\mathrm{B}\left(−\mathrm{2},\mathrm{3}\right)\:\mathrm{C}\left(\mathrm{1},\mathrm{9}\right). \\ $$$$\mathrm{We}\:\mathrm{assume}\:\mathrm{that}\:\mathrm{the}\:\mathrm{point}\:\mathrm{O}\:\mathrm{is}\:\mathrm{the} \\ $$$$\mathrm{barycenter}\:\mathrm{of}\:\mathrm{the}\:\mathrm{point}\:\mathrm{A},\mathrm{B},\mathrm{C}. \\ $$$$\rightarrow\mathrm{O}=\mathrm{bar}\left\{\left(\mathrm{A};\mathrm{3}\right),\left(\mathrm{B};\mathrm{1}\right),\left(\mathrm{C};−\mathrm{1}\right)\right\} \\ $$$$ \\…