Question Number 202306 by hardmath last updated on 24/Dec/23 $$ \\ $$Find the 2023rd term in the sequence 2,3,5,6,7,8,10,11,12,13,14,15,17,18,… obtained by subtracting integer squares from…
Question Number 202307 by MATHEMATICSAM last updated on 24/Dec/23 $$\mathrm{Show}\:\mathrm{that}\:\frac{{a}^{\mathrm{3}} \:+\:{b}^{\mathrm{3}} }{{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} }\:>\:\frac{{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} }{{a}\:+\:{b}} \\ $$ Answered by aleks041103 last updated on 24/Dec/23…
Question Number 202224 by sonukgindia last updated on 23/Dec/23 Answered by witcher3 last updated on 23/Dec/23 $$\mathrm{x}\rightarrow\frac{\mathrm{1}}{\mathrm{a}+\mathrm{bcos}^{\mathrm{2}} \left(\mathrm{x}\right)}=\mathrm{f}\left(\mathrm{x}\right),\mathrm{is}\:\mathrm{p}\:\mathrm{peridic} \\ $$$$\Rightarrow\int_{\mathrm{k}\pi} ^{\left(\mathrm{k}+\mathrm{1}\right)\pi} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\int_{\mathrm{0}} ^{\pi} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx};\mathrm{by}\:\mathrm{k}\pi+\mathrm{y}=\mathrm{x} \\…
Question Number 202257 by cortano12 last updated on 23/Dec/23 Commented by Frix last updated on 23/Dec/23 Each wife is specific �� Commented by mr W last updated on 22/Apr/24…
Question Number 202258 by MATHEMATICSAM last updated on 23/Dec/23 $$\mathrm{If}\:\frac{{x}}{{a}}\:=\:\frac{{y}}{{b}}\:\mathrm{then}\:\mathrm{show}\:\mathrm{that}\: \\ $$$$\frac{{x}^{\mathrm{3}} \:+\:\mathrm{3}{xy}^{\mathrm{2}} }{{a}^{\mathrm{3}} \:+\:\mathrm{3}{ab}^{\mathrm{2}} }\:=\:\frac{\:{y}^{\mathrm{3}} \:+\:\mathrm{3}{x}^{\mathrm{2}} {y}}{{b}^{\mathrm{3}} \:+\:\mathrm{3}{a}^{\mathrm{2}} {b}}\:. \\ $$ Answered by AST…
Question Number 202255 by SANOGO last updated on 23/Dec/23 $${calcul}\:{f}_{{n}} '\left({x}\right) \\ $$$${f}_{{n}} \left({x}\right)={n}^{\alpha} {x}\left(\mathrm{1}−{x}\right)^{{n}\:} \: \\ $$ Answered by SANOGO last updated on 23/Dec/23…
Question Number 202249 by cortano12 last updated on 23/Dec/23 $$\:\:\:\:\mathrm{C}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{x}\:\sqrt[{\mathrm{3}}]{\mathrm{cos}\:\mathrm{x}}\:−\:\mathrm{sin}\:\mathrm{x}}{\mathrm{x}^{\mathrm{5}} }\:=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 202250 by MATHEMATICSAM last updated on 23/Dec/23 $$\left({a}^{\mathrm{2}} \:−\:{bc}\right){x}^{\mathrm{2}} \:+\:\mathrm{2}\left({b}^{\mathrm{2}} \:−\:{ca}\right){x}\:+\:\left({c}^{\mathrm{2}} \:−\:{ab}\right)\:=\:\mathrm{0} \\ $$$$\mathrm{has}\:\mathrm{two}\:\mathrm{equal}\:\mathrm{roots}.\:\mathrm{Show}\:\mathrm{that}\:\mathrm{either}\: \\ $$$${b}\:=\:\mathrm{0}\:\mathrm{or}\:\frac{{a}^{\mathrm{2}} }{{bc}}\:+\:\frac{{b}^{\mathrm{2}} }{{ca}}\:+\:\frac{{c}^{\mathrm{2}} }{{ab}}\:=\:\mathrm{3}. \\ $$ Answered by…
Question Number 202251 by BaliramKumar last updated on 23/Dec/23 $$\frac{\mathrm{1}}{\mathrm{1}×\mathrm{2}×\mathrm{3}}\:+\:\frac{\mathrm{1}}{\mathrm{2}×\mathrm{3}×\mathrm{4}}\:+\:\frac{\mathrm{1}}{\mathrm{3}×\mathrm{4}×\mathrm{5}}\:+\:…………..\:+\:\frac{\mathrm{1}}{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{n}+\mathrm{2}\right)}\:=\:?\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$ Commented by BaliramKumar last updated on 23/Dec/23 $$\frac{\mathrm{1}}{\mathrm{a}\left(\mathrm{a}+\mathrm{d}\right)\left(\mathrm{a}+\mathrm{2d}\right)}\:+\:\frac{\mathrm{1}}{\left(\mathrm{a}+\mathrm{d}\right)\left(\mathrm{a}+\mathrm{2d}\right)\left(\mathrm{a}+\mathrm{3d}\right)}\:+\:\:…….+\:\frac{\mathrm{1}}{\left\{\mathrm{a}+\left(\mathrm{n}−\mathrm{1}\right)\mathrm{d}\right\}\left\{\mathrm{a}+\mathrm{nd}\right\}\left\{\mathrm{a}+\left(\mathrm{n}+\mathrm{1}\right)\mathrm{d}\right\}}\:=\:? \\ $$ Commented by MATHEMATICSAM…
Question Number 202276 by professorleiciano last updated on 23/Dec/23 Answered by professorleiciano last updated on 23/Dec/23 $${Area}\left({retangulo}\right)=\mathrm{4}×\mathrm{6}=\mathrm{24}{m}^{\mathrm{2}} \\ $$$${Area}\left({triangulo}\:{I}\right)=\mathrm{3}×\mathrm{6}=\mathrm{18}/\mathrm{2}=\mathrm{9}{m}^{\mathrm{2}} \\ $$$${Area}\left({triangulo}\:{II}\right)=\mathrm{3}×\mathrm{4}=\mathrm{12}/\mathrm{2}=\mathrm{6}{m}^{\mathrm{2}} \\ $$$${Area}\left({triangulo}\:{III}\right)=\mathrm{3}×\mathrm{4}=\mathrm{12}/\mathrm{2}=\mathrm{6}{m}^{\mathrm{2}} \\ $$$${Area}\left({total}\right)=\mathrm{24}{m}^{\mathrm{2}}…