Question Number 78334 by john santu last updated on 16/Jan/20 $$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\left[\mathrm{cos}\:^{\mathrm{2}} \left(\mathrm{cos}\:{x}\right)+\mathrm{sin}^{\mathrm{2}} \:\left(\mathrm{sin}\:{x}\right)\:\right]\:{dx} \\ $$ Commented by MJS last updated on 16/Jan/20 $$\mathrm{we}\:\mathrm{cannot}\:\mathrm{solve}\:\mathrm{this}\:\mathrm{integral}\:\mathrm{but}\:\mathrm{we}\:\mathrm{can}…
Question Number 143868 by ajfour last updated on 19/Jun/21 Commented by ajfour last updated on 19/Jun/21 $${Find}\:{minimum}\:{h}\:{such}\:{that} \\ $$$${stick}\:{just}\:{loses}\:{contact}\:{with} \\ $$$${table}\:{upon}\:{being}\:{hit}\:{by}\:{a}\:{small} \\ $$$${ball}\:{of}\:{mass}\:{m}\:{released}\:{a} \\ $$$${height}\:{h}\:{above}\:{the}\:{table}\:{top}…
Question Number 78335 by john santu last updated on 16/Jan/20 $${find}\:{the}\:{solution}\:{of} \\ $$$$\frac{{x}}{\left({x}−\mathrm{2}\right)^{\mathrm{3}} +\left({x}−\mathrm{3}\right)^{\mathrm{3}} −\mathrm{1}}\:\geqslant\:\mathrm{0} \\ $$ Answered by MJS last updated on 16/Jan/20 $$\Rightarrow\:{x}>\mathrm{0}\wedge\left(\left({x}−\mathrm{2}\right)^{\mathrm{3}}…
Question Number 78332 by Lontum Hans last updated on 16/Jan/20 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{values}\:\mathrm{of}\:\mathrm{k}\:\mathrm{and}\:\mathrm{n}\:\mathrm{for}\:\mathrm{which}\:\mathrm{x}^{\mathrm{3}} \mathrm{and}\:\mathrm{higher}\:\mathrm{powers}\:\mathrm{of}\:\mathrm{x}\:\mathrm{are}\:\mathrm{negligeble} \\ $$$$\mathrm{given}\:\mathrm{that}\:\left(\mathrm{1}+\mathrm{kx}\right)^{\mathrm{n}} =\mathrm{1}+\mathrm{2x}+\mathrm{6x}^{\mathrm{2}} . \\ $$ Commented by mr W last updated on…
Question Number 12797 by tawa last updated on 01/May/17 $$\mathrm{1}\:+\:\mathrm{x}\:+\:\mathrm{x}^{\mathrm{2}} \:+\:…\:\mathrm{x}^{\mathrm{49}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{x}^{\mathrm{49}} \:−\:\frac{\mathrm{1}}{\mathrm{x}}\right) \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x} \\ $$ Commented by prakash jain last updated on 01/May/17…
Question Number 78333 by Lontum Hans last updated on 16/Jan/20 $$\mathrm{prove}\:\mathrm{by}\:\mathrm{contradiction}\:\mathrm{that}\:\sqrt{\mathrm{2}\:}\:\mathrm{is}\:\mathrm{irrational}. \\ $$ Answered by MJS last updated on 16/Jan/20 $$\sqrt{\mathrm{2}}\in\mathbb{Q}\wedge\sqrt{\mathrm{2}}>\mathrm{0}\:\Rightarrow\:\sqrt{\mathrm{2}}=\frac{{p}}{{q}} \\ $$$${p},\:{q}\:\in\mathbb{N}\wedge\mathrm{gcd}\:\left({p},\:{q}\right)\:=\mathrm{1}\:\Leftrightarrow\:{p}\nmid{q}\wedge{q}\nmid{p} \\ $$$$\left(\sqrt{\mathrm{2}}=\frac{{p}}{{q}}\right)^{\mathrm{2}}…
Question Number 12796 by tawa last updated on 01/May/17 $$\mathrm{The}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{two}\:\mathrm{positive}\:\:\mathrm{numbers}\:\mathrm{is}\:\mathrm{20}.\:\mathrm{find}\:\mathrm{the}\:\mathrm{numbers} \\ $$$$\left(\mathrm{i}\right)\:\:\mathrm{If}\:\mathrm{their}\:\mathrm{product}\:\mathrm{is}\:\mathrm{maximum} \\ $$$$\left(\mathrm{ii}\right)\:\:\mathrm{If}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{their}\:\mathrm{square}\:\mathrm{is}\:\mathrm{maximum} \\ $$$$\left(\mathrm{iii}\right)\:\mathrm{If}\:\mathrm{the}\:\mathrm{product}\:\mathrm{of}\:\mathrm{the}\:\mathrm{square}\:\mathrm{of}\:\mathrm{one}\:\mathrm{and}\:\mathrm{the}\:\mathrm{cube}\:\mathrm{of}\:\mathrm{the}\:\mathrm{other}\:\mathrm{is}\:\mathrm{maximum} \\ $$ Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 01/May/17…
Question Number 78330 by john santu last updated on 16/Jan/20 $${given}\:{the}\:{regular}\:{pyramid} \\ $$$${T}.{ABCD}\:{with}\:{a}\:{square}\:{base} \\ $$$$.\:{length}\:{AB}\:=\:\mathrm{8}\:,\:{TC}\:=\:\mathrm{6}.\:{point} \\ $$$${P}\:{is}\:{mid}\:{BC}.\:{if}\:{x}\:{is}\:{the}\:{angle}\: \\ $$$${between}\:{TP}\:{and}\:{BD}.\:{determine} \\ $$$${the}\:{value}\:{of}\:\mathrm{cos}\:{x}. \\ $$ Terms of…
Question Number 12794 by tawa last updated on 01/May/17 $$\mathrm{Let}\:\mathrm{R}\:\mathrm{be}\:\mathrm{a}\:\mathrm{cummutative}\:\mathrm{ring}\:\mathrm{with}\:\mathrm{1},\:\mathrm{and}\:\:\mathrm{a},\mathrm{b}\in\mathrm{R}.\:\mathrm{suppose}\:\mathrm{a}\:\mathrm{is}\:\mathrm{ivertible}\:\mathrm{and} \\ $$$$\mathrm{b}\:\mathrm{is}\:\mathrm{nilpotent}.\:\mathrm{Show}\:\mathrm{that}\:\:\mathrm{a}\:+\:\mathrm{b}\:\:\mathrm{is}\:\mathrm{ivertible}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 143860 by mathmax by abdo last updated on 19/Jun/21 $$\mathrm{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\mathrm{x}\:\mathrm{e}^{−\mathrm{x}^{\mathrm{2}} } \mathrm{log}\left(\mathrm{1}+\mathrm{e}^{\mathrm{x}} \right)\mathrm{dx} \\ $$ Answered by mathmax by abdo last…