Question Number 216207 by MATHEMATICSAM last updated on 30/Jan/25 $$\mathrm{Prove}\:\mathrm{that}\:\mathrm{any}\:\mathrm{kind}\:\mathrm{of}\:\mathrm{equation}\:\mathrm{should} \\ $$$$\mathrm{have}\:\mathrm{atleast}\:\mathrm{one}\:\mathrm{root}.\:\left(\mathrm{Algebric}\:\right. \\ $$$$\left.\mathrm{fundamental}\:\mathrm{theorem}\right) \\ $$ Commented by Ghisom last updated on 30/Jan/25 $$\mathrm{not}\:{any}\:{kind}\:{of}\:{equation}.\:\mathrm{for}\:\mathrm{example} \\…
Question Number 216183 by MATHEMATICSAM last updated on 29/Jan/25 Answered by A5T last updated on 29/Jan/25 $$=\mathrm{a}^{\mathrm{3}} −\mathrm{a}^{\mathrm{2}} \left(\mathrm{b}+\mathrm{c}\right)+\mathrm{b}^{\mathrm{3}} +\mathrm{c}^{\mathrm{3}} +\mathrm{2abc}−\mathrm{b}^{\mathrm{2}} \left(\mathrm{a}+\mathrm{c}\right)−\mathrm{c}^{\mathrm{2}} \left(\mathrm{a}+\mathrm{b}\right) \\ $$$$=\mathrm{a}^{\mathrm{2}}…
Question Number 216178 by hardmath last updated on 29/Jan/25 $$\mathrm{B}\:=\:\frac{\mathrm{3}^{\mathrm{4}} \:+\:\mathrm{3}^{\mathrm{2}} \:+\:\mathrm{1}}{\mathrm{3}^{\mathrm{7}} \:-\:\mathrm{3}}\:+\:\frac{\mathrm{4}^{\mathrm{4}} \:+\:\mathrm{4}^{\mathrm{2}} \:+\:\mathrm{1}}{\mathrm{4}^{\mathrm{7}} \:-\:\mathrm{4}}\:+\:…\:+\:\frac{\mathrm{10}^{\mathrm{4}} \:+\:\mathrm{10}^{\mathrm{2}} \:+\:\mathrm{1}}{\mathrm{10}^{\mathrm{7}} \:-\:\mathrm{1}} \\ $$$$\mathrm{Find}:\:\:\:\mathrm{B}\:+\:\frac{\mathrm{1}}{\mathrm{220}}\:=\:? \\ $$ Answered by…
Question Number 216076 by efronzo1 last updated on 27/Jan/25 $$\:\:\:\frac{\lfloor\frac{\mathrm{x}}{\mathrm{3}}\:\rfloor}{\lfloor\:\frac{\mathrm{x}}{\mathrm{4}}\:\rfloor}\:=\:\frac{\mathrm{21}}{\mathrm{16}}\:;\:\mathrm{x}=? \\ $$ Answered by mr W last updated on 28/Jan/25 $$\lfloor\frac{{x}}{\mathrm{3}}\rfloor=\mathrm{21}{k} \\ $$$$\lfloor\frac{{x}}{\mathrm{4}}\rfloor=\mathrm{16}{k}\:\neq\mathrm{0}\:\Rightarrow{k}\neq\mathrm{0} \\ $$$$\frac{{x}}{\mathrm{3}}=\mathrm{21}{k}+{a}\:{with}\:\mathrm{0}\leqslant{a}<\mathrm{1}…
Question Number 216056 by hardmath last updated on 26/Jan/25 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 216046 by MATHEMATICSAM last updated on 26/Jan/25 $${x}\:=\:{bz}\:+\:{cy},\:{y}\:=\:{cx}\:+\:{az}\:\mathrm{and}\:{z}\:=\:{bx}\:+\:{ay} \\ $$$$\mathrm{then}\:\mathrm{prove}\:\mathrm{that}\:{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} \:+\:\mathrm{2}{abc}\:=\:\mathrm{1}. \\ $$ Answered by A5T last updated on 26/Jan/25 $$\mathrm{x}=\mathrm{bz}+\mathrm{cy}…\left(\mathrm{i}\right);\:\mathrm{y}=\mathrm{cx}+\mathrm{az}…\left(\mathrm{ii}\right);\:\mathrm{z}=\mathrm{bx}+\mathrm{ay}…\left(\mathrm{iii}\right)…
Question Number 216050 by hardmath last updated on 26/Jan/25 $$\mathrm{x},\mathrm{y},\mathrm{z}\:\in\:\mathbb{N} \\ $$$$\mathrm{lcd}\left(\mathrm{x};\mathrm{y}\right)=\mathrm{72} \\ $$$$\mathrm{lcd}\left(\mathrm{x};\mathrm{z}\right)=\mathrm{600} \\ $$$$\mathrm{lcd}\left(\mathrm{y};\mathrm{z}\right)=\mathrm{900} \\ $$$$\mathrm{Find}:\:\:\:\mathrm{1}.\left(\mathrm{x};\mathrm{y};\mathrm{z}\right)=? \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}.\left(\mathrm{x};\mathrm{y};\mathrm{z}\right)=? \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{3}.\left(\mathrm{x};\mathrm{y};\mathrm{z}\right)=? \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:… \\…
Question Number 216010 by MATHEMATICSAM last updated on 25/Jan/25 $$\mathrm{Solve}\:\mathrm{for}\:{x}\:\mathrm{and}\:{y} \\ $$$${ax}^{\mathrm{2}} \:+\:{bxy}\:+\:{cy}^{\mathrm{2}} \:=\:{bx}^{\mathrm{2}} \:+\:{cxy}\:+\:{ay}^{\mathrm{2}} \:=\:{d}. \\ $$ Answered by mr W last updated on…
Question Number 215994 by Frix last updated on 25/Jan/25 $$\mathrm{Solve}\:\mathrm{for}\:{z}\in\mathbb{C}:\:\:\:\:\:\mid{z}^{{z}} \mid=\mathrm{1} \\ $$ Answered by mr W last updated on 25/Jan/25 $${say}\:{z}={re}^{{i}\theta} ={r}\left(\mathrm{cos}\:\theta+{i}\:\mathrm{sin}\:\theta\right) \\ $$$${with}\:\mathrm{0}\leqslant\theta<\mathrm{2}\pi,\:{r}>\mathrm{0}…
Question Number 215992 by MATHEMATICSAM last updated on 24/Jan/25 $$\left(\sqrt{\mathrm{3}}\:+\:\sqrt{\mathrm{2}}\right)^{{x}} \:+\:\left(\sqrt{\mathrm{3}}\:−\:\sqrt{\mathrm{2}}\right)^{{x}} \:=\:\mathrm{10}.\:\mathrm{Solve} \\ $$$$\mathrm{for}\:{x}. \\ $$ Answered by Rasheed.Sindhi last updated on 24/Jan/25 $$\left(\sqrt{\mathrm{3}}\:+\:\sqrt{\mathrm{2}}\right)^{{x}} \:+\:\left(\sqrt{\mathrm{3}}\:−\:\sqrt{\mathrm{2}}\right)^{{x}}…