Question Number 23897 by ajfour last updated on 09/Nov/17 Commented by ajfour last updated on 09/Nov/17 $$\:{The}\:{string}\:{is}\:{wrapped}\:{around} \\ $$$${the}\:{outer}\:{cicumference}\:{of}\:{spool} \\ $$$${of}\:{radius}\:\mathrm{2}{r}\:,\:{while}\:{it}\:{rolls}\:{on} \\ $$$${the}\:{track}\:{on}\:{its}\:{inner}\:{circle}\:{of} \\ $$$${radius}\:{r}.\:{Relate}\:\boldsymbol{{v}}\:{with}\:\boldsymbol{{u}}\:{in}\:{terms}…
Question Number 154966 by n0y0n last updated on 23/Sep/21 $$\:\mathrm{we}\:\mathrm{know}\:\:\mathrm{y}=\mathrm{asin}\left(\omega\mathrm{t}−\mathrm{kx}\right)\:\mathrm{is}\: \\ $$$$\mathrm{equation}\:\mathrm{of}\:\mathrm{wave} \\ $$$$\:\mathrm{which}\:\mathrm{velocity}\:\mathrm{will}\:\mathrm{be}\:\frac{\omega}{\mathrm{k}} \\ $$$$\:\:\mathrm{But}\:\mathrm{what}\:\:\mathrm{will}\:\mathrm{be}\:\mathrm{the}\:\mathrm{velocity}\:\mathrm{of}\: \\ $$$$\:\:\mathrm{a}\:\mathrm{wave}\:\mathrm{that}\:\mathrm{is}\:\mathrm{created}\:\mathrm{from} \\ $$$$\:\mathrm{superposition}\:\mathrm{of}\:\mathrm{two}\:\mathrm{sine}\:\mathrm{wave}\:\mathrm{of}\: \\ $$$$\mathrm{different}\:\mathrm{velocity}\:\mathrm{like}\:\mathrm{bellow} \\ $$$$\:\:\:\mathrm{y}=\mathrm{5sin}\left(\mathrm{6t}−\mathrm{x}\right)\mathrm{cos}\left(\mathrm{13t}−\mathrm{6x}\right)\:?\: \\…
Question Number 23891 by ketto last updated on 09/Nov/17 Commented by math solver last updated on 09/Nov/17 $$\mathrm{write}\:,\:\mathrm{0}.\mathrm{75}\:\mathrm{as}\:\frac{\mathrm{75}}{\mathrm{100}}\: \\ $$$$\mathrm{so}\:\mathrm{by}\:\mathrm{the}\:\mathrm{properties}\:\mathrm{of}\:\mathrm{log}\:, \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{write}\: \\ $$$$\mathrm{log}\:\mathrm{75}\:−\:\mathrm{log}\:\mathrm{100}\: \\…
Question Number 89425 by M±th+et£s last updated on 17/Apr/20 $$\int_{\mathrm{1}} ^{\mathrm{4}} \sqrt{\mathrm{1}+\left(\frac{{y}^{\mathrm{3}} }{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}{y}^{−\mathrm{1}} \right)^{\mathrm{2}} }\:{dy} \\ $$$$ \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 89422 by naka3546 last updated on 17/Apr/20 Commented by jagoll last updated on 17/Apr/20 $${f}\left({x}\right)\:=\:\left({x}+\mathrm{2}\right)\left({x}+\mathrm{1}\right) \\ $$$${f}\left({A}\right)\:=\:\left({A}+\mathrm{2}{I}\right)\left({A}+{I}\right)\: \\ $$$$ \\ $$ Answered by…
Question Number 154959 by mathdanisur last updated on 23/Sep/21 $$\mathrm{Consider}\:\mathrm{the}\:\mathrm{polynomial} \\ $$$$\mathrm{P}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{10}} -\mathrm{6x}^{\mathrm{9}} -\mathrm{11x}^{\mathrm{8}} +\mathrm{3x}^{\mathrm{2}} -\mathrm{18x}-\mathrm{7} \\ $$$$\mathrm{Calculate}:\:\:\mathrm{P}\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{5}}\right) \\ $$ Commented by prakash jain last…
Question Number 154958 by n0y0n last updated on 23/Sep/21 $$\:\: \\ $$$$\:\:\mathrm{if}\:,\:\mathrm{sinA}+\mathrm{sinB}=\frac{\mathrm{1}}{\mathrm{5}}\: \\ $$$$\:\mathrm{then}\:\:\:\frac{\mathrm{6cosA}+\mathrm{13cosB}}{\mathrm{cosA}+\mathrm{6cosB}}=? \\ $$$$\:\: \\ $$ Commented by prakash jain last updated on…
Question Number 23884 by Tinkutara last updated on 09/Nov/17 $$\mathrm{If}\:{C}_{{r}} \:\mathrm{stands}\:\mathrm{for}\:^{{n}} {C}_{{r}} \:=\:\frac{{n}!}{{r}!\:{n}\:−\:{r}!}\:\mathrm{and} \\ $$$$\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{r}.{C}_{{r}} ^{\mathrm{2}} \:=\:\lambda\:\mathrm{for}\:{n}\:\geqslant\:\mathrm{2},\:\mathrm{then}\:\lambda\:\mathrm{is}\:\mathrm{divisible} \\ $$$$\mathrm{by} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{3}\:\left({n}\:−\:\mathrm{1}\right) \\ $$$$\left(\mathrm{2}\right)\:{n}\:+\:\mathrm{1}…
Question Number 23883 by lizan 123 last updated on 09/Nov/17 $${plz}\:{anyone}\:\:\:{answer}\:{the}\:{question}\:\mathrm{23870}\:.{plz}\:{plz}\:{plz}….. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 154948 by mathdanisur last updated on 23/Sep/21 $$\mathrm{If}\:\:\overset{\rightarrow} {\mathrm{a}}\:=\:\left(-\mathrm{1};\mathrm{0};-\mathrm{3};\mathrm{4}\right) \\ $$$$\:\:\:\:\:\:\overset{\rightarrow} {\mathrm{b}}\:=\:\left(\mathrm{1};-\mathrm{4};\mathrm{0};-\mathrm{3}\right) \\ $$$$\:\:\:\:\:\:\overset{\rightarrow} {\mathrm{c}}\:=\:\frac{\mid\overset{\rightarrow} {\mathrm{a}}\mid}{\mid\overset{\rightarrow} {\mathrm{b}}\mid}\:\centerdot\:\left(\mathrm{2}\overset{\rightarrow} {\mathrm{a}}\:+\:\overset{\rightarrow} {\mathrm{b}}\right) \\ $$$$\:\:\:\:\:\:\:\overset{\rightarrow} {\mathrm{d}}\:=\:\frac{\mid\overset{\rightarrow} {\mathrm{b}}\mid}{\mid\overset{\rightarrow}…