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…
Question Number 206600 by naka3546 last updated on 20/Apr/24 Answered by aleks041103 last updated on 22/Apr/24 $$\begin{bmatrix}{−\mathrm{1}}&{\mathrm{3}}&{\mathrm{3}}\\{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{1}}&{−\mathrm{3}}&{−\mathrm{2}}\end{bmatrix}\overset{\mathrm{3}{rd}\:{row}\:+\:\mathrm{1}{st}\:{row}} {\rightarrow}\begin{bmatrix}{−\mathrm{1}}&{\mathrm{3}}&{\mathrm{3}}\\{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\end{bmatrix} \\ $$$$\Rightarrow\begin{bmatrix}{−\mathrm{1}}&{\mathrm{3}}&{\mathrm{3}}\\{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{1}}&{−\mathrm{3}}&{−\mathrm{2}}\end{bmatrix}=\begin{bmatrix}{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{1}}&{\mathrm{0}}\\{−\mathrm{1}}&{\mathrm{0}}&{\mathrm{1}}\end{bmatrix}\begin{bmatrix}{−\mathrm{1}}&{\mathrm{3}}&{\mathrm{3}}\\{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\end{bmatrix} \\ $$$$\begin{bmatrix}{−\mathrm{1}}&{\mathrm{3}}&{\mathrm{3}}\\{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\end{bmatrix}\overset{\mathrm{1}{st}\:{row}\:+\:\mathrm{2}{nd}\:{row}} {\rightarrow}\begin{bmatrix}{\mathrm{0}}&{\mathrm{3}}&{\mathrm{3}}\\{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\end{bmatrix} \\ $$$$\Rightarrow\begin{bmatrix}{−\mathrm{1}}&{\mathrm{3}}&{\mathrm{3}}\\{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\end{bmatrix}=\begin{bmatrix}{\mathrm{1}}&{−\mathrm{1}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{1}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\end{bmatrix}\begin{bmatrix}{\mathrm{0}}&{\mathrm{3}}&{\mathrm{3}}\\{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\end{bmatrix}…
Question Number 206629 by aras18 last updated on 20/Apr/24 Answered by Frix last updated on 20/Apr/24 $${a}+{b}−{c}=\mathrm{170} \\ $$$${a}−{b}+{c}=\mathrm{130} \\ $$$$======= \\ $$$$\mathrm{2}{a}\:\:\:\:\:\:\:\:\:\:\:=\mathrm{300} \\ $$$$\:\:\:{a}\:\:\:\:\:\:\:\:\:\:\:=\mathrm{150}…
Question Number 206591 by necx122 last updated on 19/Apr/24 $${I}\:{have}\:{a}\:{background}\:{in}\:{building}\:{design} \\ $$$${but}\:{I}\:{have}\:{great}\:{interest}\:{in}\:{Mathematics}. \\ $$$${That}'{s}\:{one}\:{major}\:{reason}\:{why}\:{I}\:{joined} \\ $$$${this}\:{platform}\:{cause}\:{I}\:{did}\:{little}\:{or}\:{no} \\ $$$${maths}\:{in}\:{my}\:{undergraduate}\:{days}\:{as}\:{a} \\ $$$${building}\:{design}\:{student}. \\ $$$$ \\ $$$${This}\:{forum}\:{has}\:{contributed}\:{greatly}\:{to} \\…