Question Number 92566 by Tinku Tara last updated on 08/May/20 $$\mathrm{Posting}\:\mathrm{Question}\:\mathrm{with}\:\mathrm{Images} \\ $$$$\mathrm{Preferably}\:\mathrm{you}\:\mathrm{should}\:\mathrm{type}\:\mathrm{the}\:\mathrm{question}. \\ $$$$\mathrm{However}\:\mathrm{if}\:\mathrm{you}\:\mathrm{are}\:\mathrm{using}\:\mathrm{pictures}\:\mathrm{then} \\ $$$$\mathrm{please}\:\mathrm{do}\:\mathrm{the}\:\mathrm{following}\:\mathrm{steps} \\ $$$$\mathrm{which}\:\mathrm{posting}\:\mathrm{a}\:\mathrm{photo}\:\mathrm{of}\:\mathrm{printed} \\ $$$$\mathrm{question}. \\ $$$$\mathrm{A}.\:\mathrm{Use}\:\mathrm{camscanner}\:\mathrm{app}\:\mathrm{to}\:\mathrm{take}\: \\ $$$$\mathrm{pictures}\:\left(\mathrm{search}\:\mathrm{for}\:\mathrm{camscanner}\:\mathrm{in}\right.…
Question Number 158097 by zainaltanjung last updated on 31/Oct/21 Answered by Rasheed.Sindhi last updated on 31/Oct/21 $${x}^{\mathrm{2}} +{px}+\mathrm{12}:{Roots}\:\alpha\:,\:\mathrm{4} \\ $$$$\mathrm{product}\:{of}\:{roots}:\mathrm{4}\alpha=\mathrm{12}\Rightarrow\alpha=\mathrm{3}\:\left({other}\:{root}\right) \\ $$$${sum}\:{of}\:{roots}:\mathrm{4}+\mathrm{3}=−\mathrm{p}\Rightarrow\mathrm{p}=−\mathrm{7} \\ $$$${x}^{\mathrm{2}} +{px}+{q}=\mathrm{0}\Rightarrow{x}^{\mathrm{2}}…
Question Number 158098 by zainaltanjung last updated on 31/Oct/21 $$\: \\ $$$$\mathrm{The}\:\mathrm{result}\:\mathrm{of}: \\ $$$$\:\left[\mathrm{5}\:+\:\left\{\left(\mathrm{8}\right)\:×\:\left(−\:\:\mathrm{2}\right)\right\}\right]\:+\mathrm{3}\left\{\mathrm{6}\:+\:\left(\:\:\mathrm{2}\right)\right\}\:\mathrm{is}\:.\:.\: \\ $$ Commented by Rasheed.Sindhi last updated on 31/Oct/21 $${I}\:{think},\:{the}\:{forum}\:{is}\:{not}\:{for}\:{primary} \\…
Question Number 92557 by I want to learn more last updated on 07/May/20 $$\mathrm{If}\:\:\:\mathrm{a}_{\mathrm{1}} \:\:=\:\:\mathrm{5},\:\:\:\:\:\mathrm{a}_{\mathrm{2}} \:\:=\:\:\mathrm{13}\:\:\:\:\:\mathrm{and}\:\:\:\:\mathrm{a}_{\mathrm{n}\:\:+\:\:\mathrm{2}} \:\:\:=\:\:\mathrm{5a}_{\mathrm{n}\:\:+\:\:\mathrm{1}} \:−\:\:\mathrm{6a}_{\mathrm{n}} . \\ $$$$\mathrm{Find}\:\:\:\:\:\mathrm{a}_{\mathrm{n}} \\ $$ Commented by…
Question Number 158085 by HongKing last updated on 31/Oct/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 27000 by abdo imad last updated on 01/Jan/18 $${P}\:{is}\:{a}\:{polynomial}\:{havng}\:{n}\:{roots}\:\left({x}_{{i}} \right)_{\mathrm{1}\leqslant{i}\leqslant{n}} \:\:{with}\:{x}_{{i}} \neq\:{x}_{{j}} \:{for}\:{i}\neq\:{j} \\ $$$${find}\:{the}\:{values}\:{of}\:\sum_{{k}\mathrm{1}} ^{{k}={n}} \frac{\mathrm{1}}{{x}−{x}_{{k}} }\:\:{and}\:\:\sum_{{k}=\mathrm{1}} ^{{n}} \frac{\mathrm{1}}{\left({x}−{x}_{{k}} \right)^{\mathrm{2}} }\:. \\…
Question Number 27002 by math solver last updated on 01/Jan/18 Commented by moxhix last updated on 01/Jan/18 $${x}\neq\mathrm{0}\:\left(\because\mathrm{0}^{\mathrm{2}} +\mathrm{0}+\mathrm{1}\neq\mathrm{0}\right) \\ $$$$\frac{{x}^{\mathrm{2}} +{x}+\mathrm{1}}{{x}}=\frac{\mathrm{0}}{{x}} \\ $$$${x}+\frac{\mathrm{1}}{{x}}=−\mathrm{1} \\…
Question Number 26999 by abdo imad last updated on 01/Jan/18 $${calculate}\:\prod_{{k}=\mathrm{1}} ^{{n}} {cos}\left(\frac{{a}}{\mathrm{2}^{{k}} }\right)\:\:{and}\mathrm{0}<{a}<\pi\:\:{then}\:{find}\:{the}\:{value}\:{of} \\ $$$${lim}_{{n}−>\propto} \:\sum_{{k}=\mathrm{1}} ^{{n}} {ln}\left({cos}\left(\frac{{a}}{\mathrm{2}^{{k}} }\right)\right). \\ $$ Answered by prakash…
Question Number 26997 by abdo imad last updated on 01/Jan/18 $${let}\:{give}\:\xi\:\in\mathbb{C}\:{and}\:\xi^{{n}} =\mathrm{1}\:\left(\xi\:{is}\:{the}\:{n}^{{me}} \:{root}\:{of}\:\mathrm{1}\right) \\ $$$${simplify}\:\:{A}=\:\mathrm{1}+\xi^{{p}} +\xi^{\mathrm{2}{p}} +…\:+\xi^{\left({n}−\mathrm{1}\right){p}} \\ $$$${and}\:{B}=\:\mathrm{1}+\mathrm{2}\xi\:+\mathrm{3}\xi^{\mathrm{2}} +…+{n}\xi^{{n}−\mathrm{1}} . \\ $$ Commented by…
Question Number 26998 by abdo imad last updated on 01/Jan/18 $${smlify}\:{X}=\:\:\prod_{{p}=\mathrm{2}} ^{{n}} \frac{{p}^{\mathrm{3}} −\mathrm{1}}{{p}^{\mathrm{3}} \:+\mathrm{1}}\:{by}\:{using}\:\mathrm{1},{j},{j}^{\mathrm{2}} {and}\:{j}={e}^{\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} . \\ $$ Terms of Service Privacy Policy Contact:…