Question Number 207687 by efronzo1 last updated on 23/May/24 Answered by A5T last updated on 23/May/24 Commented by A5T last updated on 23/May/24 $$\left[{OAB}\right]=\frac{\mathrm{9}\sqrt{\mathrm{3}}}{\mathrm{4}}; \\…
Question Number 207699 by hardmath last updated on 23/May/24 $$\frac{\mathrm{4}}{\mathrm{2x}−\mathrm{1}}\:\:+\:\:\frac{\mathrm{27}}{\mathrm{3x}−\mathrm{1}}\:\:+\:\:\frac{\mathrm{125}}{\mathrm{5x}−\mathrm{1}}\:\:=\:\:\frac{\mathrm{144}}{\mathrm{4x}−\mathrm{1}} \\ $$$$\mathrm{Find}:\:\:\boldsymbol{\mathrm{x}}\:=\:? \\ $$ Answered by Rasheed.Sindhi last updated on 23/May/24 $$\frac{\mathrm{4}}{\mathrm{2x}−\mathrm{1}}\:+\:\frac{\mathrm{27}}{\mathrm{3x}−\mathrm{1}}\:+\:\frac{\mathrm{125}}{\mathrm{5x}−\mathrm{1}}\:=\:\frac{\mathrm{144}}{\mathrm{4x}−\mathrm{1}} \\ $$$$\frac{\mathrm{4}}{\mathrm{2x}−\mathrm{1}}+\:\frac{\mathrm{125}}{\mathrm{5x}−\mathrm{1}}=\:\frac{\mathrm{144}}{\mathrm{4x}−\mathrm{1}}−\frac{\mathrm{27}}{\mathrm{3x}−\mathrm{1}} \\…
Question Number 207661 by efronzo1 last updated on 22/May/24 $$\:\mathrm{P}=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{7}}+…+\frac{\mathrm{1}}{\mathrm{2023}} \\ $$$$\:\mathrm{Q}=\:\frac{\mathrm{1}}{\mathrm{1}×\mathrm{2023}}+\frac{\mathrm{1}}{\mathrm{3}×\mathrm{2021}}+\frac{\mathrm{1}}{\mathrm{5}×\mathrm{2019}}+…+\frac{\mathrm{1}}{\mathrm{2023}×\mathrm{1}} \\ $$$$\:\:\frac{\mathrm{P}}{\mathrm{Q}}=? \\ $$ Answered by Frix last updated on 22/May/24 $${P}\left({m}\right)=\underset{{k}=\mathrm{1}} {\overset{\frac{{m}+\mathrm{1}}{\mathrm{2}}}…
Question Number 207663 by efronzo1 last updated on 24/May/24 $$\:\:\mathrm{Given}\: \\ $$$$\:\:\:\mathrm{x}_{\mathrm{1}} +\mathrm{x}_{\mathrm{3}} +…+\mathrm{x}_{\mathrm{2023}} \:=\:\mathrm{25}−\left(\mathrm{x}_{\mathrm{2}} +\mathrm{x}_{\mathrm{4}} +…+\mathrm{x}_{\mathrm{2022}} \right) \\ $$$$\:\:\mathrm{x}_{\mathrm{1}} ^{\mathrm{2}\:} +\:\mathrm{x}_{\mathrm{3}} ^{\mathrm{2}} +…+\mathrm{x}_{\mathrm{2023}} ^{\mathrm{2}}…
Question Number 207673 by efronzo1 last updated on 22/May/24 $$\:\:\:\:\underline{ \:} \\ $$$$ \\ $$ Answered by A5T last updated on 22/May/24 $${abcde}={a}+{b}+{c}+{d}+{e}\leqslant\mathrm{5}{e}\Rightarrow{abcd}\leqslant\mathrm{5} \\ $$$${abcd}=\mathrm{1}\Rightarrow{a}={b}={c}={d}=\mathrm{1}…
Question Number 207657 by 123564 last updated on 22/May/24 Commented by 123564 last updated on 22/May/24 $${help}\:{me}\:{please} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 207652 by universe last updated on 22/May/24 $$\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{log}\left(\mathrm{1}+{x}^{\mathrm{3}} \right){dx}\:\:=\:?{and}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{log}\:\left(\mathrm{1}+{x}^{\mathrm{4}} \right){dx}\:=\:? \\ $$$$\:\:\mathrm{and}\:\mathrm{if}\:\mathrm{possible}\:\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\:\mathrm{p} \\ $$$$\:\mathrm{p}\:\:\:=\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{log}\left(\mathrm{1}+{x}^{{n}} \right){dx}\:=\:?\:\:\:\:\:\:{n}\in\mathbb{N} \\ $$…
Question Number 207664 by efronzo1 last updated on 22/May/24 $$\:\:\:\:\mathrm{If}\:\mathrm{p}+\mathrm{q}+\mathrm{r}\:=\:\mathrm{0}\:\mathrm{and}\: \\ $$$$\:\:\:\:\mathrm{A}=\frac{\left(\mathrm{p}^{\mathrm{2}} +\mathrm{q}^{\mathrm{2}} +\mathrm{r}^{\mathrm{2}} \right)^{\mathrm{2}} }{\left(\mathrm{pq}\right)^{\mathrm{2}} +\left(\mathrm{qr}\right)^{\mathrm{2}} +\left(\mathrm{rp}\right)^{\mathrm{2}} }\: \\ $$$$\:\:\:\mathrm{B}=\:\frac{\mathrm{q}^{\mathrm{2}} −\mathrm{pr}}{\mathrm{p}^{\mathrm{2}} +\mathrm{q}^{\mathrm{2}} +\mathrm{r}^{\mathrm{2}} }\:.…
Question Number 207648 by SANOGO last updated on 22/May/24 $${f}_{{n}} \left({x}\right)=\frac{{n}}{\mathrm{1}+{x}^{\mathrm{2}} }{sin}\left(\frac{{x}}{{n}}\right)\:\:/{x}\in\left[\mathrm{0},\mathrm{1}\right]\:,\:\:{n}\geqslant\mathrm{1} \\ $$$${calculer}\:{lim}\:{n}\rightarrow\infty\int_{\mathrm{0}} ^{\mathrm{1}} {f}_{{n}} \left({x}\right){dx} \\ $$ Commented by mr W last updated…
Question Number 207665 by efronzo1 last updated on 22/May/24 $$\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}}\begin{pmatrix}{\mathrm{20}}\\{\:\:\mathrm{0}}\end{pmatrix}\:+\frac{\mathrm{1}}{\mathrm{2}}\begin{pmatrix}{\mathrm{20}}\\{\:\:\mathrm{1}}\end{pmatrix}\:+\frac{\mathrm{1}}{\mathrm{3}}\begin{pmatrix}{\mathrm{20}}\\{\:\:\mathrm{2}}\end{pmatrix}\:+…+\frac{\mathrm{1}}{\mathrm{21}}\:\begin{pmatrix}{\mathrm{20}}\\{\mathrm{20}}\end{pmatrix}\:=? \\ $$ Answered by Tinku Tara last updated on 22/May/24 $$\left(\mathrm{1}+{x}\right)^{\mathrm{20}} =\underset{{n}=\mathrm{0}} {\overset{\mathrm{20}} {\sum}}\begin{pmatrix}{\mathrm{20}}\\{{n}}\end{pmatrix}{x}^{{n}} \\…