Question Number 207718 by hardmath last updated on 24/May/24 $$\mid\mathrm{x}−\mathrm{3}\mid\:+\:\mid\mathrm{x}\:+\mathrm{2}\mid\:=\:\mathrm{9} \\ $$$$\mathrm{find}:\:\:\boldsymbol{\mathrm{x}}\:=\:? \\ $$ Answered by TonyCWX08 last updated on 24/May/24 $${There}\:{are}\:\mathrm{4}\:{possible}\:{cases}. \\ $$$${x}−\mathrm{3}+{x}+\mathrm{2}=\mathrm{9} \\…
Question Number 207710 by marycakessandara last updated on 23/May/24 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 207699 by hardmath last updated on 23/May/24 $$\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 $$\frac{\mathrm{4}}{\mathrm{2x}−\mathrm{1}}\:+\:\frac{\mathrm{27}}{\mathrm{3x}−\mathrm{1}}\:+\:\frac{\mathrm{125}}{\mathrm{5x}−\mathrm{1}}\:=\:\frac{\mathrm{144}}{\mathrm{4x}−\mathrm{1}} \\ $$$$\frac{\mathrm{4}}{\mathrm{2x}−\mathrm{1}}+\:\frac{\mathrm{125}}{\mathrm{5x}−\mathrm{1}}=\:\frac{\mathrm{144}}{\mathrm{4x}−\mathrm{1}}−\frac{\mathrm{27}}{\mathrm{3x}−\mathrm{1}} \\…
Question Number 207663 by efronzo1 last updated on 24/May/24 $$\:\:\mathrm{Given}\: \\ $$$$\:\:\:\mathrm{x}_{\mathrm{1}} +\mathrm{x}_{\mathrm{3}} +…+\mathrm{x}_{\mathrm{2023}} \:=\:\mathrm{25}−\left(\mathrm{x}_{\mathrm{2}} +\mathrm{x}_{\mathrm{4}} +…+\mathrm{x}_{\mathrm{2022}} \right) \\ $$$$\:\:\mathrm{x}_{\mathrm{1}} ^{\mathrm{2}\:} +\:\mathrm{x}_{\mathrm{3}} ^{\mathrm{2}} +…+\mathrm{x}_{\mathrm{2023}} ^{\mathrm{2}}…
Question Number 207661 by efronzo1 last updated on 22/May/24 $$\:\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 $${P}\left({m}\right)=\underset{{k}=\mathrm{1}} {\overset{\frac{{m}+\mathrm{1}}{\mathrm{2}}}…
Question Number 207673 by efronzo1 last updated on 22/May/24 $$\:\:\:\:\underline{ \:} \\ $$$$ \\ $$ Answered by A5T last updated on 22/May/24 $${abcde}={a}+{b}+{c}+{d}+{e}\leqslant\mathrm{5}{e}\Rightarrow{abcd}\leqslant\mathrm{5} \\ $$$${abcd}=\mathrm{1}\Rightarrow{a}={b}={c}={d}=\mathrm{1}…
Question Number 207664 by efronzo1 last updated on 22/May/24 $$\:\:\:\:\mathrm{If}\:\mathrm{p}+\mathrm{q}+\mathrm{r}\:=\:\mathrm{0}\:\mathrm{and}\: \\ $$$$\:\:\:\:\mathrm{A}=\frac{\left(\mathrm{p}^{\mathrm{2}} +\mathrm{q}^{\mathrm{2}} +\mathrm{r}^{\mathrm{2}} \right)^{\mathrm{2}} }{\left(\mathrm{pq}\right)^{\mathrm{2}} +\left(\mathrm{qr}\right)^{\mathrm{2}} +\left(\mathrm{rp}\right)^{\mathrm{2}} }\: \\ $$$$\:\:\:\mathrm{B}=\:\frac{\mathrm{q}^{\mathrm{2}} −\mathrm{pr}}{\mathrm{p}^{\mathrm{2}} +\mathrm{q}^{\mathrm{2}} +\mathrm{r}^{\mathrm{2}} }\:.…
Question Number 207641 by hardmath last updated on 21/May/24 $$\mathrm{Find}:\:\:\:\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\mathrm{40}\:−\:\frac{\mathrm{1}}{\mathrm{cos}\:\mathrm{20}}\:\:=\:\:? \\ $$ Commented by Frix last updated on 22/May/24 $$\mathrm{4}−\mathrm{2}\sqrt{\mathrm{7}}\mathrm{cos}\:\frac{\pi+\mathrm{2sin}^{−\mathrm{1}} \:\frac{\mathrm{37}\sqrt{\mathrm{7}}}{\mathrm{98}}}{\mathrm{6}} \\ $$ Answered…
Question Number 207639 by hardmath last updated on 21/May/24 Commented by Berbere last updated on 21/May/24 $${false}\:{not}\:{true}\: \\ $$$$\begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}\sim\frac{\mathrm{4}^{{n}} }{\:\sqrt{\pi{n}}} \\ $$$$\frac{\mathrm{2}^{−\mathrm{2}\left({n}−\mathrm{1}\right)} \left({n}+\mathrm{1}\right)\left({n}+\mathrm{3}\right)}{\left({n}+\mathrm{2}\right)\left({n}+\mathrm{4}\right)}\begin{pmatrix}{{n}}\\{\mathrm{2}{n}}\end{pmatrix}^{\mathrm{2}} \sim\frac{\mathrm{4}^{{n}} }{\pi{n}}\:{diverge}…
Question Number 207638 by hardmath last updated on 21/May/24 Commented by TonyCWX08 last updated on 22/May/24 $${arcsin}\left({sin}\mathrm{5}\right)=\mathrm{5} \\ $$$${arccis}\left({cos}\mathrm{6}\right)=\mathrm{6} \\ $$$${arctan}\left({tan}\mathrm{2}\right)=\mathrm{2} \\ $$$$\mathrm{5}+\mathrm{6}+\mathrm{2}=\mathrm{13} \\ $$…