Question Number 191775 by Mingma last updated on 30/Apr/23 Answered by AST last updated on 30/Apr/23 $$\overset{\_\_\_\_\_\_\_\_\_\_\_\_} {{ABCDEF}}=\mathrm{10000}\overset{\_\_\_\_\_\_\_} {{ABCD}}+\overset{\_\_\_} {{EF}} \\ $$$$\mathrm{17}\left[\mathrm{10000}\overset{\_\_\_} {{AB}}+\overset{\_\_\_\_\_\_\_} {{CDEF}}\right]=\mathrm{12}\left[\mathrm{100}\overset{\_\_\_\_\_\_\_} {{CDEF}}+\overset{\_\_\_}…
Question Number 191753 by Abdullahrussell last updated on 30/Apr/23 Answered by Frix last updated on 30/Apr/23 $${t}^{\mathrm{3}} +{pt}^{\mathrm{2}} +{qt}+{r}=\mathrm{0} \\ $$$${s}_{\mathrm{1}} ={t}_{\mathrm{1}} ^{\mathrm{2}} {t}_{\mathrm{2}} +{t}_{\mathrm{2}}…
Question Number 191735 by Spillover last updated on 29/Apr/23 $${Verify}\:{that} \\ $$$$\:\urcorner\left({p}\rightarrow{q}\right)\rightarrow\left({p}\wedge^{\urcorner} {q}\right)\:{is}\:{tautology}\:{using}\:{laws}\:{of} \\ $$$${algebra} \\ $$ Answered by manxsol last updated on 30/Apr/23 $$\sim\left(\sim{p}…
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Question Number 126180 by benjo_mathlover last updated on 18/Dec/20 $${solve}\:\mid\:\mid{x}−\mathrm{1}\mid\:−\mathrm{2}\mid\:=\:\mid\:{x}−\mathrm{3}\:\mid\: \\ $$ Answered by bobhans last updated on 18/Dec/20 $$\left(\mathrm{1}\right)\:{for}\:{x}\geqslant\mathrm{3}\:\Rightarrow\:\mid\:{x}−\mathrm{3}\mid={x}−\mathrm{3} \\ $$$$\left(\mathrm{2}\right)\:{for}\:{x}<\mathrm{3}\:\Rightarrow\mid\mathrm{1}−{x}−\mathrm{2}\mid=\mathrm{3}−{x} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid−\mathrm{1}−{x}\mid\:=\:\mathrm{3}−{x} \\…
Question Number 191706 by BaliramKumar last updated on 29/Apr/23 Answered by Skabetix last updated on 29/Apr/23 $$\rightarrow\left({b}\right)\:\frac{\mathrm{97}}{\mathrm{36}} \\ $$ Answered by Skabetix last updated on…
Question Number 126165 by TheMexicanTacosG last updated on 17/Dec/20 $${if}:\:\:\:{b}^{\mathrm{2}} \:=\:−\mathrm{3}\:−{c}^{\mathrm{2}} \:+\mathrm{2}\left({a}−{b}−{c}\right)\:−{a}^{\mathrm{2}} \\ $$$$ \\ $$$${calculate}:\:{a}\:+\:{b}\:+\:{c}\:=\:? \\ $$ Commented by PRITHWISH SEN 2 last updated…
Question Number 126161 by TheMexicanTacosG last updated on 17/Dec/20 $${find}\:\:“{a}'' \\ $$$$ \\ $$$$\:^{{a}} \sqrt{{a}^{\mathrm{2}} \:}\:=\:\left(\left(\mathrm{4}^{−\frac{\mathrm{1}}{\mathrm{2}}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} \right)^{−\mathrm{2}} \:\:\:\:\: \\ $$ Answered by Olaf last…
Question Number 191675 by MATHEMATICSAM last updated on 28/Apr/23 $$\mathrm{Solve}\:\mathrm{for}\:{x}\:: \\ $$$$\left({x}\:−\:\frac{\mathrm{1}}{{x}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:+\:\left(\mathrm{1}\:−\:\frac{\mathrm{1}}{{x}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:=\:{x} \\ $$ Commented by mehdee42 last updated on 29/Apr/23 $${the}\:{graph}\:{of}\:{the}\:{function}\:{is}\:{as}\:{follows}.{therefore}\:.{the}\:{given}\:\: \\…