Question Number 13260 by Tinkutara last updated on 17/May/17 $$\mathrm{Number}\:\mathrm{of}\:\mathrm{moles}\:\mathrm{of}\:\mathrm{KMnO}_{\mathrm{4}} \:\mathrm{required} \\ $$$$\mathrm{to}\:\mathrm{oxidise}\:\mathrm{one}\:\mathrm{mole}\:\mathrm{of}\:\mathrm{Fe}\left(\mathrm{C}_{\mathrm{2}} \mathrm{O}_{\mathrm{4}} \right)\:\mathrm{in} \\ $$$$\mathrm{acidic}\:\mathrm{medium}\:\mathrm{is} \\ $$$$\left(\mathrm{a}\right)\:\mathrm{0}.\mathrm{6} \\ $$$$\left(\mathrm{b}\right)\:\mathrm{1}.\mathrm{67} \\ $$$$\left(\mathrm{c}\right)\:\mathrm{0}.\mathrm{2} \\ $$$$\left(\mathrm{d}\right)\:\mathrm{0}.\mathrm{4}…
Question Number 13258 by Tinkutara last updated on 17/May/17 Answered by Tinkutara last updated on 15/Jul/17 Commented by Tinkutara last updated on 15/Jul/17 Terms of…
Question Number 13257 by Tinkutara last updated on 17/May/17 Answered by Tinkutara last updated on 09/Jul/17 Commented by Tinkutara last updated on 09/Jul/17 Terms of…
Question Number 13191 by Tinkutara last updated on 16/May/17 $$\mathrm{Can}\:\mathrm{we}\:\mathrm{ask}\:\mathrm{here}\:\mathrm{Chemistry}\:\mathrm{or}\:\mathrm{Physics} \\ $$$$\mathrm{doubts}\:\mathrm{because}\:\mathrm{I}\:\mathrm{am}\:\mathrm{preparing}\:\mathrm{for}\:\mathrm{JEE}? \\ $$ Commented by FilupS last updated on 16/May/17 $$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{know}\:\mathrm{about}\:\mathrm{everyone}\:\mathrm{else}, \\ $$$$\mathrm{but}\:\mathrm{i}\:\mathrm{know}\:\mathrm{only}\:\mathrm{mathematics} \\…
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Question Number 78683 by berket last updated on 19/Jan/20 $${solve}\:{x}^{\mathrm{4}} −\mathrm{18}{x}−\mathrm{35}=\mathrm{0}\:{by}\:{using}\:{substitution}\:{x}=\:{u}+{v} \\ $$ Commented by john santu last updated on 20/Jan/20 $${x}={u}+{v}\:\Rightarrow{x}^{\mathrm{3}} ={u}^{\mathrm{3}} +{v}^{\mathrm{3}} +\mathrm{3}{uv}\left({u}+{v}\right)…
Question Number 13117 by tawa tawa last updated on 14/May/17 Answered by mrW1 last updated on 14/May/17 $${P}×\mathrm{12}=\mathrm{1000}×\mathrm{6} \\ $$$${P}=\mathrm{500}\:{N} \\ $$ Commented by tawa…
Question Number 13107 by tawa tawa last updated on 14/May/17 $$\mathrm{A}\:\mathrm{man}\:\mathrm{moves}\:\mathrm{20m}\:\mathrm{north}\:\mathrm{then}\:\mathrm{12m}\:\mathrm{east}\:\mathrm{and}\:\mathrm{finally}\:\mathrm{15m}\:\mathrm{south}.\:\:\mathrm{His}\:\mathrm{displacement} \\ $$$$\mathrm{from}\:\mathrm{the}\:\mathrm{starting}\:\mathrm{point}\:\mathrm{is}\:? \\ $$ Answered by RasheedSindhi last updated on 14/May/17 $$\mathrm{Net}\:\mathrm{distance}\:\mathrm{in}\:\mathrm{north}-\mathrm{south} \\ $$$$\mathrm{direction}\:\mathrm{20}−\mathrm{15}=\mathrm{5}\:\mathrm{m}…
Question Number 13097 by FilupS last updated on 14/May/17 $${S}=\underset{{x}_{\mathrm{2}} =\mathrm{1}} {\overset{{x}_{\mathrm{1}} } {\sum}}\underset{{x}_{\mathrm{3}} =\mathrm{1}} {\overset{{x}_{\mathrm{2}} } {\sum}}\centerdot\centerdot\centerdot\underset{{x}_{{n}} =\mathrm{1}} {\overset{{x}_{{n}−\mathrm{1}} } {\sum}}\underset{{t}=\mathrm{1}} {\overset{{x}_{{n}} } {\sum}}{t}…
Question Number 13098 by ketto last updated on 14/May/17 Answered by Joel577 last updated on 14/May/17 $${GCD}\:\left({x}^{\mathrm{2}} {y}^{\mathrm{2}} ,\:\mathrm{4}{xy}^{\mathrm{4}} \right)\:=\:{xy}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} {y}^{\mathrm{2}} \:+\:\mathrm{4}{xy}^{\mathrm{4}} \:=\:{xy}^{\mathrm{2}}…