Question Number 78888 by zainal tanjung last updated on 21/Jan/20 $$\mathrm{The}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\left({x}−{a}\right)\left({x}−{b}\right)={abx}^{\mathrm{2}} \:\mathrm{are}\:\mathrm{always} \\ $$ Commented by mr W last updated on 21/Jan/20 $$…{are}\:{always}\:{what}?…
Question Number 13128 by Madhuri Dixit last updated on 15/May/17 $$\mathrm{If}\:\:\left[{x}\right]\:\mathrm{stands}\:\mathrm{for}\:\mathrm{the}\:\mathrm{greatest}\:\mathrm{integer} \\ $$$$\mathrm{function},\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\: \\ $$$$\underset{\:\mathrm{4}} {\overset{\:\:\:\:\mathrm{10}} {\int}}\:\frac{\left[{x}^{\mathrm{2}} \right]}{\left[{x}^{\mathrm{2}} −\mathrm{28}{x}+\mathrm{196}\right]+\left[{x}^{\mathrm{2}} \right]}\:{dx}\:\mathrm{is} \\ $$ Terms of Service…
Question Number 13103 by kashyappushpendrak1811@gmail.c last updated on 14/May/17 $$\:\underset{\:−\mathrm{1}} {\overset{\mathrm{2}} {\int}}\:\frac{\:\mid\:{x}\:\mid\:}{{x}}\:{dx}\:=\: \\ $$ Answered by mrW1 last updated on 14/May/17 $$\:\underset{\:−\mathrm{1}} {\overset{\mathrm{2}} {\int}}\:\frac{\:\mid\:{x}\:\mid\:}{{x}}\:{dx}\:=\:\underset{\:−\mathrm{1}} {\overset{\mathrm{0}}…
Question Number 12995 by 433 last updated on 09/May/17 $$\mathrm{If}\:{a},\:{b},\:{c}\:\mathrm{are}\:\mathrm{in}\:\mathrm{GP}\:\mathrm{and}\:\:{a}−{b},\:{c}−{a},\:{b}−{c}\:\mathrm{are} \\ $$$$\mathrm{in}\:\mathrm{HP},\:\mathrm{then}\:{a}+\mathrm{4}{b}+{c}\:\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 12926 by kashyappushpendrak1811@gmail.c last updated on 07/May/17 $$\underset{\:\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\:\mathrm{log}\:\mid\mathrm{tan}\:{x}+\mathrm{cot}\:{x}\mid\:{dx}\:= \\ $$ Answered by ajfour last updated on 07/May/17 $$\mathrm{0}\leqslant{x}\leqslant\pi/\mathrm{2}\:\:\Rightarrow\:\:\:\:\:\:\mathrm{0}\leqslant\mathrm{2}{x}\leqslant\pi \\ $$$$\mathrm{tan}\:{x}+\mathrm{cot}\:{x}=\frac{\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}}+\frac{\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}} \\…
Question Number 12919 by kashyappushpendrak1811@gmail.c last updated on 07/May/17 $$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:\mathrm{the}\:\mathrm{integral}\:\underset{\:\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:{e}^{{x}^{\mathrm{2}} } {dx}\:\:\mathrm{lies} \\ $$$$\mathrm{in}\:\mathrm{the}\:\mathrm{interval} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 12841 by CheaV last updated on 04/May/17 $$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{terms}\:\mathrm{in}\:\mathrm{the}\:\mathrm{expansion}\:\mathrm{of} \\ $$$$\left(\mathrm{1}+\mathrm{5}\:\sqrt{\mathrm{2}}\:{x}\right)^{\mathrm{9}} \:+\:\left(\mathrm{1}−\mathrm{5}\:\sqrt{\mathrm{2}}\:{x}\right)^{\mathrm{9}} \:\mathrm{is} \\ $$ Answered by malwaan last updated on 04/May/17 $$\mathrm{5}\:\mathrm{terms} \\…
Question Number 12828 by 786786AM last updated on 03/May/17 $$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:\mathrm{the}\:\mathrm{infinite}\:\mathrm{product} \\ $$$$\sqrt{\mathrm{3}}\:\centerdot\:\sqrt[{\mathrm{4}}]{\mathrm{9}}\:\centerdot\:\sqrt[{\mathrm{8}}]{\mathrm{27}}\:\centerdot\:\sqrt[{\mathrm{16}}]{\mathrm{81}}\:…\mathrm{to}\:\infty\:\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:\_\_\_\_. \\ $$ Answered by prakash jain last updated on 04/May/17 $$\mathrm{3}^{\mathrm{1}/\mathrm{2}} \centerdot\mathrm{3}^{\mathrm{2}/\mathrm{4}} \centerdot\mathrm{3}^{\mathrm{3}/\mathrm{8}}…
Question Number 12827 by 786786AM last updated on 03/May/17 $$\mathrm{Sum}\:\mathrm{of}\:\mathrm{three}\:\mathrm{numbers}\:\mathrm{in}\:\mathrm{GP}\:\mathrm{be}\:\mathrm{14}.\:\mathrm{If}\:\mathrm{one}\:\mathrm{is} \\ $$$$\mathrm{added}\:\mathrm{to}\:\mathrm{first}\:\mathrm{and}\:\mathrm{second}\:\mathrm{and}\:\mathrm{1}\:\mathrm{is}\:\mathrm{subtracted} \\ $$$$\mathrm{from}\:\mathrm{the}\:\mathrm{third},\:\mathrm{the}\:\mathrm{new}\:\mathrm{numbers}\:\mathrm{are}\:\mathrm{in}\:\mathrm{AP}. \\ $$$$\mathrm{The}\:\mathrm{smallest}\:\mathrm{of}\:\mathrm{them}\:\mathrm{is} \\ $$ Answered by mrW1 last updated on 04/May/17…
Question Number 12470 by biah last updated on 23/Apr/17 $$\sqrt[{\mathrm{5}}]{\mathrm{0}.\mathrm{03125}}\:=\: \\ $$ Answered by sma3l2996 last updated on 23/Apr/17 $$=\sqrt[{\mathrm{5}}]{\frac{\mathrm{3125}}{\mathrm{10}^{\mathrm{5}} }}=\frac{\sqrt[{\mathrm{5}}]{\mathrm{5}^{\mathrm{5}} }}{\mathrm{10}}=\mathrm{0}.\mathrm{5} \\ $$ Terms…