Question Number 199729 by Mingma last updated on 08/Nov/23 Answered by mathfreak01 last updated on 08/Nov/23 $${N}\:=\:\underset{{n}\:=\:\mathrm{1}} {\overset{\mathrm{450}} {\sum}}\left(\mathrm{10}^{{n}} \:−\:\mathrm{1}\right) \\ $$$${N}\:=\:\underset{{n}=\mathrm{1}} {\overset{\mathrm{450}} {\sum}}\mathrm{10}^{{n}} \:−\:\mathrm{450}…
Question Number 199659 by cherokeesay last updated on 06/Nov/23 Answered by Frix last updated on 07/Nov/23 $$\mathrm{If}\:\mathrm{we}\:\mathrm{stay}\:\mathrm{in}\:\mathbb{R}\:\Rightarrow\:\sqrt[{\mathrm{3}}]{−{r}}=−\sqrt[{\mathrm{3}}]{{r}} \\ $$$$\mathrm{Obvious}\:\mathrm{solutions} \\ $$$${x}=−\mathrm{1}\vee{x}=\mathrm{0} \\ $$$$\mathrm{Not}\:\mathrm{so}\:\mathrm{obvious}\:\mathrm{solutions} \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}…
Question Number 199621 by Mingma last updated on 06/Nov/23 Answered by AST last updated on 06/Nov/23 $${x}+{y}\overset{\mathrm{3}} {\equiv}\mathrm{2}…\left({i}\right) \\ $$$$\left(\mathrm{6}+\mathrm{6}+{y}\right)−\left(\mathrm{7}+{x}\right)=\mathrm{5}+{y}−{x}\overset{\mathrm{11}} {\equiv}\mathrm{0}\Rightarrow{x}−{y}\overset{\mathrm{11}} {\equiv}\mathrm{5}…\left({ii}\right) \\ $$$$\left({i}\right)\wedge\left({ii}\right)\Rightarrow\left({x},{y}\right)=\left(\mathrm{8},\mathrm{3}\right);\left(\mathrm{5},\mathrm{0}\right);\left(\mathrm{1},\mathrm{7}\right) \\…
Question Number 199405 by mnjuly1970 last updated on 03/Nov/23 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{calculate}\:… \\ $$$$\:\:\mathrm{Q}:\:\:\:\:\:\:\mathrm{I}{f}\:\:,\:\:\:{f}\left({x}\right)\:=\mathrm{2}\:{e}^{{x}} \:−\mathrm{1}\:+\:\lfloor{e}^{{x}} +\:\frac{\mathrm{3}}{\mathrm{2}}\:+\lfloor{e}^{{x}} \rfloor\:\rfloor \\ $$$$\:\:\:\:\:\:\:\:\:\Rightarrow\:\:\:\:{f}\:^{−\mathrm{1}} \:\left(\:\frac{\pi}{\mathrm{4}}\:\right)\:=? \\ $$$$\:\:\:\:\: \\ $$ Answered…
Question Number 199399 by Calculusboy last updated on 03/Nov/23 $$\boldsymbol{{Solve}}:\:\boldsymbol{{log}}_{\mathrm{2}} \boldsymbol{{r}}+\boldsymbol{{log}}_{\mathrm{3}} \boldsymbol{{p}}=\mathrm{3} \\ $$$$\boldsymbol{{p}}+\boldsymbol{{r}}=\mathrm{11}\:\:\:\boldsymbol{{fund}}\:\boldsymbol{{p}}\:\boldsymbol{{and}}\:\boldsymbol{{r}}. \\ $$ Answered by mr W last updated on 03/Nov/23 $${one}\:{solution}\:\left({p}=\mathrm{9},\:{r}=\mathrm{2}\right)\:{can}\:{be}\:“{seen}'',…
Question Number 198949 by ArifinTanjung last updated on 26/Oct/23 $$\:\:\frac{\mathrm{5}!×\mathrm{4}!}{\mathrm{3}!×\mathrm{2}!}\:=\:….\:\: \\ $$ Answered by Rasheed.Sindhi last updated on 26/Oct/23 $$\:\:\frac{\mathrm{5}!×\mathrm{4}!}{\mathrm{3}!×\mathrm{2}!}\:=\:\frac{\mathrm{5}!×\cancel{\overset{\mathrm{2}} {\mathrm{4}}}×\cancel{\mathrm{3}!}}{\cancel{\mathrm{3}!}×\cancel{\underset{\mathrm{1}} {\mathrm{2}}}×\mathrm{1}}=\mathrm{2}×\mathrm{5}! \\ $$$$ \\…
Question Number 198950 by ArifinTanjung last updated on 26/Oct/23 $$\mathrm{3x}+\mathrm{2y}=\mathrm{6} \\ $$$$\mathrm{2x}+\mathrm{3y}=\mathrm{6}\:\rightarrow\:\mathrm{x}\:\mathrm{and}\:\mathrm{y}\:=…? \\ $$ Answered by Rasheed.Sindhi last updated on 26/Oct/23 $$\begin{cases}{\mathrm{3x}+\mathrm{2y}=\mathrm{6}}\\{\mathrm{2x}+\mathrm{3y}=\mathrm{6}}\end{cases}\:\:\:\:\:\:\:\:\:\mathrm{x},\mathrm{y}=? \\ $$$$\mathrm{Adding}:\:\:\mathrm{5}\left(\mathrm{x}+\mathrm{y}\right)=\mathrm{12}\Rightarrow\mathrm{x}+\mathrm{y}=\frac{\mathrm{12}}{\mathrm{5}} \\…
Question Number 198753 by hidaoui1960 last updated on 24/Oct/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 198563 by mr W last updated on 22/Oct/23 $${find}\:{all}\:{numbers}\:\left({with}\:{any}\:{number}\right. \\ $$$$\left.{of}\:{digits}\right)\:{satisfying} \\ $$$$\left(\underline{{abcd}…{xyz}}\right)×\mathrm{2}=\left(\underline{{zyx}…{dcba}}\right) \\ $$ Answered by AST last updated on 25/Oct/23 $$\mathrm{2}{z}\equiv{a}\left({mod}\:\mathrm{10}\right);{a}\leqslant\mathrm{4}\Rightarrow{a}\in\left\{\mathrm{0},\mathrm{2},\mathrm{4}\right\};{a}\geqslant{z}…
Question Number 198511 by Hridiana last updated on 21/Oct/23 $$\mathrm{3}+{xy}\sqrt{\frac{{z}}{{xy}}} \\ $$$${xz}+{zx} \\ $$$${how}\:{to}\:{solve} \\ $$$$\mathrm{3}+\pi\left\{\frac{{x}}{\mathrm{2}}>\mathrm{0}\right\}=\left\{\frac{{zy}}{{x}+{z}}>{x}\right\} \\ $$$${operations} \\ $$$$\left\{{x}^{\mathrm{2}+\frac{{x}}{\mathrm{2}}} >\mathrm{0}\right\}\:{is}\:{the}\:{comparator} \\ $$ Commented by…