Question Number 167189 by rexford last updated on 09/Mar/22 Commented by MJS_new last updated on 09/Mar/22 $${a}<{b}\wedge{c}<{d}\:\Rightarrow\:\left({a}+{c}\right)<\left({b}+{d}\right) \\ $$$${x}<{y}\wedge−{x}<−{y}\:\Rightarrow\:\left({x}−{x}\right)<\left({y}−{y}\right)\:\Rightarrow\:\mathrm{0}<\mathrm{0}\:\mathrm{false} \\ $$ Terms of Service Privacy…
Question Number 167110 by nimnim last updated on 06/Mar/22 $$\:\:{Find}\:{the}\:{remainder}\:{when}:− \\ $$$$\:\:\left({a}\right)\:\:\mathrm{41}!\:{is}\:{divided}\:{by}\:\mathrm{1681} \\ $$$$\:\:\left({b}\right)\:\mathrm{225}!\:{is}\:{divided}\:{by}\:\mathrm{227} \\ $$$$\:\:\left({c}\right)\:\mathrm{15}!\:{is}\:{divided}\:{by}\:\mathrm{19} \\ $$ Answered by Rasheed.Sindhi last updated on 06/Mar/22…
Question Number 36021 by Rasheed.Sindhi last updated on 27/May/18 $$\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{hypotenuse}\:\mathrm{never}\:\mathrm{be} \\ $$$$\mathrm{even}\:\mathrm{of}\:\mathrm{a}\:\mathrm{right}\:\mathrm{angled}\:\mathrm{triangle}\:\mathrm{whose} \\ $$$$\mathrm{positive}\:\mathrm{integer}\:\mathrm{sides}\:\mathrm{are}\:\mathrm{relatively}\:\mathrm{prime}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 167056 by diskoquestion2 last updated on 05/Mar/22 $$\mathrm{help}\:\mathrm{me}!\: \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}cosec}^{−\mathrm{1}} \left(\mathrm{2}\right)\sqrt{{x}\sqrt{\mathrm{2}\sqrt{{x}}}} \\ $$ Answered by diskoquestion2 last updated on 05/Mar/22 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}cosec}^{−\mathrm{1}}…
Question Number 166994 by null last updated on 04/Mar/22 $$\mathrm{1}+\mathrm{1}\neq?? \\ $$ Answered by null last updated on 04/Mar/22 $$\mathrm{1}+\mathrm{1}\neq\Bbbk \\ $$ Answered by null…
Question Number 101418 by john santu last updated on 02/Jul/20 $$\mathrm{Find}\:\mathrm{2020}\:\mathrm{term}\:\mathrm{from}\:\mathrm{series} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}},\frac{\mathrm{2}}{\mathrm{1}},\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{3}}{\mathrm{1}},\frac{\mathrm{2}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{3}},\frac{\mathrm{4}}{\mathrm{1}},\frac{\mathrm{3}}{\mathrm{2}},\frac{\mathrm{2}}{\mathrm{3}},\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$,\frac{\mathrm{5}}{\mathrm{1}},\frac{\mathrm{4}}{\mathrm{2}},…\:\mathrm{is}\:\_\_\_ \\ $$$$\left(\mathrm{A}\right)\:\frac{\mathrm{2019}}{\mathrm{2020}}\:\:\:\:\:\:\left(\mathrm{B}\right)\:\frac{\mathrm{61}}{\mathrm{4}}\:\:\:\:\:\left(\mathrm{C}\right)\frac{\mathrm{63}}{\mathrm{1}} \\ $$$$\left(\mathrm{D}\right)\:\frac{\mathrm{96}}{\mathrm{4}}\:\:\:\:\:\:\left(\mathrm{E}\right)\:\frac{\mathrm{2020}}{\mathrm{2019}} \\ $$ Commented by PRITHWISH SEN…
Question Number 35872 by Rasheed.Sindhi last updated on 25/May/18 Commented by Tinkutara last updated on 25/May/18 Yes this was also my doubt! Answered by Rasheed.Sindhi last updated on 26/May/18 $$\mathrm{Note}:\:\mathrm{Capital}\:\mathrm{C}\:\mathrm{and}\:\mathrm{small}\:\mathrm{c}\:\mathrm{has}\:\mathrm{been}…
Question Number 166910 by rexford last updated on 02/Mar/22 $${given}\:{that}\:{is}\:{prime},{proof}\:{that}\:\sqrt{{p}}\:{is}\: \\ $$$${irrational} \\ $$ Commented by rexford last updated on 02/Mar/22 $${thank}\:{you} \\ $$ Answered…
Question Number 166911 by rexford last updated on 02/Mar/22 Answered by MathsFan last updated on 02/Mar/22 $$\boldsymbol{{suppose}}\:\sqrt{\mathrm{2}}\:\boldsymbol{{is}}\:\boldsymbol{{rational}} \\ $$$$\sqrt{\mathrm{2}}=\frac{\boldsymbol{{p}}}{\boldsymbol{{q}}}\:\Rightarrow\:\mathrm{2}=\frac{\boldsymbol{{p}}^{\mathrm{2}} }{\boldsymbol{{q}}^{\mathrm{2}} }\:\Rightarrow\:\:\mathrm{2}\boldsymbol{{q}}^{\mathrm{2}} =\boldsymbol{{p}}^{\mathrm{2}} …..\left(\mathrm{1}\right) \\ $$$$\:\mathrm{2}\:\boldsymbol{{divides}}\:\boldsymbol{{p}}…
Question Number 166881 by amin96 last updated on 01/Mar/22 Terms of Service Privacy Policy Contact: info@tinkutara.com