Question Number 203055 by mr W last updated on 08/Jan/24 Commented by mr W last updated on 08/Jan/24 $${is}\:{it}\:{possible}\:{to}\:{determine}\:{the}\: \\ $$$${unknown}\:{area}\:{in}\:{terms}\:{of}\:{other} \\ $$$${partial}\:{areas}\:{A},{B},{C},{D},{E}? \\ $$…
Question Number 203051 by ajfour last updated on 08/Jan/24 $${A}\:{two}\:{digit}\:{number}\:\:\left({AB}\right)_{\mathrm{10}} \\ $$$$\left({AB}\right)_{\mathrm{10}} −{A}^{{B}} =\left({BA}\right)_{\mathrm{10}} \\ $$$${Find}\:{the}\:{number}.\:{There}\:{is}\:{a}\:{poem}. \\ $$$${Having}\:{arrived}\:{at}\:{the}\:{age}\:{of}\:\left({BA}\right)_{\mathrm{10}} . \\ $$ Answered by Rasheed.Sindhi last…
Question Number 203045 by MathematicalUser2357 last updated on 08/Jan/24 $${very}\:{old}\:{q}\:{Q}.\mathrm{2} \\ $$ Commented by lazyboy last updated on 11/Jan/24 $${how}\:{can}\:{find}\:{Q}.\mathrm{2} \\ $$ Terms of Service…
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Question Number 203000 by sonukgindia last updated on 07/Jan/24 Answered by SEKRET last updated on 07/Jan/24 $$\mathrm{3};\mathrm{6} \\ $$ Answered by Rasheed.Sindhi last updated on…
Question Number 203002 by mr W last updated on 07/Jan/24 $${Solve}\:{for}\:{x}\in{R} \\ $$$${x}^{\mathrm{4}} +\mathrm{4}{x}−\mathrm{3}=\mathrm{0} \\ $$ Answered by ajfour last updated on 07/Jan/24 $${p}^{\mathrm{2}} \left({p}^{\mathrm{4}}…
Question Number 203035 by Frix last updated on 07/Jan/24 $$\mathrm{e}=\mathrm{2} \\ $$$$\mathrm{Proof}: \\ $$$$\mathrm{Let}\:{x}=\frac{\mathrm{e}+\mathrm{2}}{\mathrm{2}} \\ $$$$\mathrm{2}{x}=\mathrm{e}+\mathrm{2} \\ $$$$\mathrm{2}{x}\left(\mathrm{e}−\mathrm{2}\right)=\left(\mathrm{e}+\mathrm{2}\right)\left(\mathrm{e}−\mathrm{2}\right) \\ $$$$\mathrm{2e}{x}−\mathrm{4}{x}=\mathrm{e}^{\mathrm{2}} −\mathrm{4} \\ $$$$−\mathrm{4}{x}+\mathrm{4}=−\mathrm{2e}{x}+\mathrm{e}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}}…
Question Number 203031 by sonukgindia last updated on 07/Jan/24 Answered by MathematicalUser2357 last updated on 08/Jan/24 $${find}\:\mathrm{8}\:{roots}\:{using}\:{newton}'{s}\:{method} \\ $$ Commented by Calculusboy last updated on…
Question Number 203017 by mnjuly1970 last updated on 07/Jan/24 Answered by som(math1967) last updated on 07/Jan/24 $${let}\:{ar}\:{ofEDCF}={a}\:,\bigtriangleup{AEB}={b} \\ $$$$\frac{\bigtriangleup{ABD}}{\bigtriangleup{BCD}}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\frac{{Green}+{b}}{{Magenta}+{a}}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\frac{\bigtriangleup{ABF}}{\bigtriangleup{ACF}}=\frac{{x}}{\mathrm{6}} \\ $$$$\Rightarrow\frac{{Majenta}+{b}}{{Green}+{a}}=\frac{{x}}{\mathrm{6}}…
Question Number 203018 by mr W last updated on 07/Jan/24 Commented by mr W last updated on 07/Jan/24 $$\left[{an}\:{unsolved}\:{old}\:{question}\:{Q}\mathrm{202864}\right] \\ $$ Commented by esmaeil last…