Question Number 178512 by Spillover last updated on 17/Oct/22 $$\mathrm{Write}\:\mathrm{the}\:\mathrm{converse},\:\mathrm{contrapostive}\:\mathrm{and} \\ $$$$\mathrm{inverse}\:\mathrm{of}\:\mathrm{the}\:\mathrm{statement} \\ $$$$\left(\mathrm{a}\right)“\mathrm{If}\:\mathrm{two}\:\mathrm{angles}\:\mathrm{are}\:\mathrm{congruent},\mathrm{then} \\ $$$$\mathrm{they}\:\mathrm{have}\:\mathrm{the}\:\mathrm{same}\:\mathrm{measure}'' \\ $$$$\left(\mathrm{b}\right)“\mathrm{If}\:\:\mathrm{a}\:\mathrm{person}\:\mathrm{is}\:\mathrm{18}\:\mathrm{years}\:\mathrm{old},\mathrm{then}\:\mathrm{he}\:\mathrm{is} \\ $$$$\mathrm{a}\:\mathrm{legal}\:\mathrm{adult}'' \\ $$ Answered by Spillover…
Question Number 178513 by Spillover last updated on 17/Oct/22 $$\mathrm{Check}\:\mathrm{if}\:\mathrm{the}\:\mathrm{following}\:\mathrm{argument}\:\mathrm{is}\:\mathrm{valid} \\ $$$$“\mathrm{If}\:\mathrm{am}\:\mathrm{clever}\:\mathrm{then}\:\:\mathrm{I}\:\mathrm{understand}\:\mathrm{physics}. \\ $$$$\mathrm{I}\:\mathrm{dont}\:\mathrm{understand}\:\mathrm{physics}\:.\mathrm{therefore} \\ $$$$\mathrm{I}\:\mathrm{am}\:\mathrm{not}\:\mathrm{clever}'' \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 178511 by Spillover last updated on 17/Oct/22 $$\mathrm{Simplify}\: \\ $$$$\left(\mathrm{p}\downarrow\mathrm{q}\right)\wedge\left(\sim\mathrm{q}\downarrow\mathrm{p}\right) \\ $$ Answered by Spillover last updated on 18/Oct/22 Terms of Service Privacy…
Question Number 178500 by infinityaction last updated on 17/Oct/22 Commented by abdullahoudou last updated on 17/Oct/22 $${it}\:{would}\:{be}\:{nice}\:{to}\:{give}\:{the}\:{rest}\:{of}\:{questions}\:{instead}\:{of}\:{only}\:{number}\:\mathrm{15}\:… \\ $$ Answered by mindispower last updated on…
Question Number 178486 by Acem last updated on 17/Oct/22 $$\:{Reduce}\:\frac{\left(\mathrm{2}{n}\right)!}{\mathrm{1}×\mathrm{3}×\mathrm{5}×…×\left(\mathrm{2}{n}−\mathrm{1}\right)}\: \\ $$ Answered by Rasheed.Sindhi last updated on 17/Oct/22 $$\:{Reduce}\:\frac{\left(\mathrm{2}{n}\right)!}{\mathrm{1}×\mathrm{3}×\mathrm{5}×…×\left(\mathrm{2}{n}−\mathrm{1}\right)}\: \\ $$$$=\frac{\cancel{\mathrm{1}}.\mathrm{2}.\cancel{\mathrm{3}}.\mathrm{4}.\cancel{\mathrm{5}}…\left(\cancel{\mathrm{2}{n}−\mathrm{1}}\right).\mathrm{2}{n}}{\cancel{\mathrm{1}}.\cancel{\mathrm{3}}.\cancel{\mathrm{5}}…..\left(\cancel{\mathrm{2}{n}−\mathrm{1}}\right)} \\ $$$$=\mathrm{2}.\mathrm{4}.\mathrm{6}…..\mathrm{2}{n} \\…
Question Number 178476 by Shrinava last updated on 17/Oct/22 $$\mathrm{Find}\:\:\mathrm{a}\in\mathbb{R} \\ $$$$\mathrm{Such}\:\mathrm{that}\:\:\mathrm{x}_{\mathrm{1}} ^{\mathrm{16}} \:+\:\mathrm{x}_{\mathrm{2}} ^{\mathrm{16}} \:+\:\mathrm{x}_{\mathrm{3}} ^{\mathrm{16}} \:=\:\mathrm{30} \\ $$$$\mathrm{Where}\:\:\mathrm{x}_{\mathrm{1}} ,\mathrm{x}_{\mathrm{2}} ,\mathrm{x}_{\mathrm{3}} −\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{equation}:\:\:\mathrm{x}^{\mathrm{3}}…
Question Number 47394 by gunawan last updated on 09/Nov/18 $${f}\left({z}\right)=\frac{\mathrm{3}{z}+\mathrm{1}}{\mathrm{2}−\mathrm{4}{z}} \\ $$$${f}\left({f}\left({z}\right)\right)=… \\ $$ Commented by maxmathsup by imad last updated on 10/Nov/18 $${fof}\left({z}\right)\:=\frac{\mathrm{3}{f}\left({z}\right)+\mathrm{1}}{\mathrm{2}−\mathrm{4}{f}\left({z}\right)}=\:\frac{\mathrm{3}\frac{\mathrm{3}{z}+\mathrm{1}}{\mathrm{2}−\mathrm{4}{z}}\:+\mathrm{1}}{\mathrm{2}−\mathrm{4}\:\frac{\mathrm{3}{z}+\mathrm{1}}{\mathrm{2}−\mathrm{4}{z}}}\:=\frac{\mathrm{9}{z}+\mathrm{3}+\mathrm{2}−\mathrm{4}{z}}{\mathrm{4}−\mathrm{8}{z}−\mathrm{12}{z}−\mathrm{4}}\:=\frac{\mathrm{5}{z}+\mathrm{5}}{−\mathrm{20}{z}} \\…
Question Number 47391 by gunawan last updated on 09/Nov/18 $${z}_{\mathrm{1}} =\mathrm{3}+{i} \\ $$$${z}_{\mathrm{2}} =\mathrm{1}−\mathrm{2}{i} \\ $$$$\begin{vmatrix}{\frac{\mathrm{2}{z}_{\mathrm{2}} +{z}_{\mathrm{1}} −\mathrm{5}−{i}}{\mathrm{2}{z}_{\mathrm{1}} −{z}_{\mathrm{2}} +\mathrm{3}−{i}}}\end{vmatrix}^{\mathrm{2}} =.. \\ $$ Commented by…
Question Number 47389 by gunawan last updated on 09/Nov/18 $$\left(\sqrt{\mathrm{3}}−{i}\right)^{\mathrm{1}+\mathrm{2}{i}} =… \\ $$ Commented by maxmathsup by imad last updated on 09/Nov/18 $${we}\:{have}\:\mid\sqrt{\mathrm{3}}−{i}\mid=\mathrm{2}\:\Rightarrow\sqrt{\mathrm{3}}−{i}\:=\mathrm{2}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{2}\left({cos}\left(−\frac{\pi}{\mathrm{6}}\right)+{isin}\left(−\frac{\pi}{\mathrm{6}}\right)\right) \\ $$$$=\mathrm{2}{e}^{−\frac{{i}\pi}{\mathrm{6}}}…
Question Number 178452 by cortano1 last updated on 16/Oct/22 $$\:\:\:\:\:\:\:\:\:\mathrm{show}\:\mathrm{that}\: \\ $$$$\:\:\:\left(\left(\mathrm{p}\wedge\mathrm{q}\right)\Rightarrow\mathrm{r}\right)\Rightarrow\left(\left(\mathrm{p}\wedge\sim\mathrm{r}\right)\Rightarrow\sim\mathrm{q}\right) \\ $$$$\:\:\:\mathrm{is}\:\mathrm{tautology}\: \\ $$ Answered by Spillover last updated on 17/Oct/22 Commented by…