Question Number 206639 by mathlove last updated on 21/Apr/24 Commented by Frix last updated on 21/Apr/24 $$=\underset{\frac{\pi}{\mathrm{12}}} {\overset{\frac{\pi}{\mathrm{11}}} {\int}}\left(\frac{\sqrt{{s}}}{\mathrm{2}\left(\sqrt{{c}}+\sqrt{{s}}\right)}−\frac{\sqrt{{cs}}}{{c}+\sqrt{{s}}\left(\sqrt{{c}}+\mathrm{1}\right)}+\frac{\left(\sqrt{{c}}−\mathrm{1}\right){s}+\sqrt{{cs}}}{\mathrm{2}\left(\left(\sqrt{{s}}+\mathrm{1}\right){c}+\left(\sqrt{{c}}+\mathrm{1}\right){s}\right)}\right){dx} \\ $$$$\mathrm{With}\:{c}=\mathrm{cos}\:{x}\:\wedge{s}=\mathrm{sin}\:{x} \\ $$$$\mathrm{I}\:\mathrm{doubt}\:\mathrm{we}\:\mathrm{can}\:\mathrm{solve}\:\mathrm{this}. \\ $$…
Question Number 206659 by cherokeesay last updated on 21/Apr/24 Answered by mr W last updated on 21/Apr/24 Commented by mr W last updated on 21/Apr/24…
Question Number 206622 by necx122 last updated on 20/Apr/24 $$\sum_{{n}=\mathrm{2}} ^{\mathrm{4}} \left(\frac{\mathrm{2}{n}}{\mathrm{3}}\:−\mathrm{1}\right)\:=\:? \\ $$ Answered by Frix last updated on 20/Apr/24 $$\mathrm{Seriously}?! \\ $$$$…\mathrm{ok}: \\…
Question Number 206616 by universe last updated on 20/Apr/24 Answered by aleks041103 last updated on 21/Apr/24 $${e}^{\mathrm{3}{x}} \:{is}\:{increasing}\:{and}\:{continuous}\:{for}\:{x}>\mathrm{0} \\ $$$${ln}\left({x}\right)\:{is}\:{incr}.\:{and}\:{cont}.\:{for}\:{x}>\mathrm{0} \\ $$$$\Rightarrow{f}\left({x}\right)\:{is}\:{incr}.\:{and}\:{cont}.\:{for}\:{x}>\mathrm{0} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:{f}\left({x}\right)\rightarrow−\infty…
Question Number 206618 by MATHEMATICSAM last updated on 20/Apr/24 $$\mathrm{If}\:\mathrm{A}\:=\:\mathrm{sin}^{\mathrm{4}} \theta\:+\:\mathrm{cos}^{\mathrm{4}} \theta\:\mathrm{then}\:\mathrm{select}\:\mathrm{the}\: \\ $$$$\mathrm{correct}\:\mathrm{option}: \\ $$$$\left.\mathrm{i}\right)\:\mathrm{0}\:<\:\mathrm{A}\:<\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\left.\mathrm{ii}\right)\:\mathrm{1}\:<\:\mathrm{A}\:<\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\left.\mathrm{iii}\right)\:\frac{\mathrm{1}}{\mathrm{2}}\:\leq\:\mathrm{A}\:\leq\:\mathrm{1} \\ $$$$\left.\mathrm{iv}\right)\:\frac{\mathrm{3}}{\mathrm{2}}\:\leq\:\mathrm{A}\:\leq\:\mathrm{2} \\ $$ Answered…
Question Number 206613 by hardmath last updated on 20/Apr/24 $$\mathrm{If}\:\:\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\begin{cases}{\mathrm{x}^{\mathrm{2}} \:\:\:,\:\:\:\mathrm{x}\:\leqslant\:\mathrm{2}}\\{\mathrm{f}\left(\mathrm{x}−\mathrm{3}\right)\:\:\:,\:\:\:\mathrm{x}\:>\:\mathrm{2}}\end{cases} \\ $$$$\mathrm{Find}\:\:\:\int_{−\mathrm{1}} ^{\:\mathrm{53}} \:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{dx}\:=\:? \\ $$ Answered by MM42 last updated on 20/Apr/24 $$\mathrm{57}…
Question Number 206614 by cortano21 last updated on 20/Apr/24 Commented by BaliramKumar last updated on 20/Apr/24 $$\mathrm{Q}.\:\mathrm{188151} \\ $$ Answered by BaliramKumar last updated on…
Question Number 206615 by cortano21 last updated on 20/Apr/24 Answered by mr W last updated on 20/Apr/24 $$\frac{{R}−{r}}{{R}+{r}}=\mathrm{sin}\:\mathrm{22}.\mathrm{5}°=\frac{\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}}{\mathrm{2}} \\ $$$$\left(\mathrm{2}−\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}{R}\right){R}=\left(\mathrm{2}+\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}\right){r} \\ $$$$\frac{{r}}{{R}}=\frac{\mathrm{2}−\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}}{\mathrm{2}+\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}}\approx\mathrm{0}.\mathrm{4465} \\ $$ Terms…
Question Number 206608 by BaliramKumar last updated on 20/Apr/24 Answered by Rasheed.Sindhi last updated on 20/Apr/24 $${Number}\:{of}\:{classes}\:{before}\:{recruit}.:\:\mathrm{15}−\mathrm{4}=\mathrm{11} \\ $$$${Number}\:{of}\:{old}\:{students}:\:\mathrm{11}×\mathrm{30}=\mathrm{330} \\ $$$${Number}\:{of}\:{all}\:{students}\:{now}:\mathrm{15}×\mathrm{25}=\mathrm{375} \\ $$$${New}\:{students}:\:\mathrm{375}−\mathrm{330}=\mathrm{45} \\ $$…
Question Number 206609 by universe last updated on 20/Apr/24 $$\:\:\:\mathrm{let}\:\mathrm{matrix}\:\:\mathrm{A}_{\mathrm{n}×\mathrm{n}} \:\mathrm{and}\:\mathrm{B}_{\mathrm{n}×\mathrm{n}} \:\mathrm{satisfying} \\ $$$$\:\:\:\mathrm{A}^{\mathrm{2}} =\:\mathrm{A}\:\:\&\:\:\mathrm{B}^{\mathrm{2}} =\:\mathrm{B}\:\:\mathrm{then}\:\mathrm{prove} \\ $$$$\:\:\:\rho\left(\mathrm{A}−\mathrm{B}\right)\:=\:\rho\left(\mathrm{A}−\mathrm{AB}\right)+\:\:\rho\left(\mathrm{B}−\mathrm{AB}\right) \\ $$$$\:\:\:\mathrm{here}\:\rho\:=\:\mathrm{rank} \\ $$ Terms of Service…