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lim-x-1-2-x-sin-x-1-2cos-x-1-arctan-x-1-ln-x-

Question Number 126195 by bramlexs22 last updated on 18/Dec/20 $$\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{2}\sqrt{{x}}−\mathrm{sin}\:\left({x}−\mathrm{1}\right)−\mathrm{2cos}\:\left({x}−\mathrm{1}\right)}{\mathrm{arctan}\:\left({x}−\mathrm{1}\right)−\mathrm{ln}\:{x}}\:? \\ $$ Answered by liberty last updated on 18/Dec/20 $${let}\:{w}={x}−\mathrm{1}\: \\ $$$${L}=\underset{{w}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2}\sqrt{{w}+\mathrm{1}}−\mathrm{sin}\:{w}−\mathrm{2cos}\:{w}}{\mathrm{arctan}\:{w}−\mathrm{ln}\:\left({w}+\mathrm{1}\right)} \\…

Use-laws-of-algebra-to-prove-the-following-a-B-A-u-A-B-AuB-AnB-b-A-AnB-A-B-

Question Number 191731 by Spillover last updated on 29/Apr/23 $$\mathrm{Use}\:\mathrm{laws}\:\mathrm{of}\:\mathrm{algebra}\:\mathrm{to}\:\mathrm{prove}\:\mathrm{the}\:\mathrm{following} \\ $$$$\left(\mathrm{a}\right)\left[\left(\mathrm{B}−\mathrm{A}\right)\mathrm{u}\left(\mathrm{A}−\mathrm{B}\right)\right]=\left[\left(\mathrm{AuB}\right)−\left(\mathrm{AnB}\right)\right] \\ $$$$\left(\mathrm{b}\right)\mathrm{A}\bigtriangledown\left(\mathrm{AnB}\right)=\mathrm{A}−\mathrm{B} \\ $$ Answered by AST last updated on 29/Apr/23 $$\left({a}\right)\:{B}−{A}={B}\cap{A}^{'} \:\:\:{and}\:\:{A}−{B}={A}\cap{B}^{'}…

Question-126186

Question Number 126186 by benjo_mathlover last updated on 18/Dec/20 Commented by benjo_mathlover last updated on 18/Dec/20 $$\left({A}\right)\:\frac{\mathrm{23}}{\mathrm{257}}\:\:\:\:\:\:\:\left({B}\right)\:\frac{\mathrm{24}}{\mathrm{257}}\:\:\:\:\:\left({C}\right)\:\frac{\mathrm{25}}{\mathrm{257}} \\ $$$$\left({D}\right)\:\frac{\mathrm{26}}{\mathrm{257}}\:\:\:\:\:\:\:\left({E}\right)\:\frac{\mathrm{27}}{\mathrm{257}} \\ $$ Terms of Service Privacy…

0-2pi-3sintcost-dt-

Question Number 191723 by SANOGO last updated on 29/Apr/23 $$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \mathrm{3}{sintcost}\:{dt} \\ $$ Answered by mehdee42 last updated on 29/Apr/23 $$\left.{I}=\frac{\mathrm{3}}{\mathrm{2}}{sin}^{\mathrm{2}} {t}\:\right]_{\mathrm{0}} ^{\mathrm{2}\pi} \:=\mathrm{0}…