Question Number 207231 by efronzo1 last updated on 10/May/24 Answered by Sutrisno last updated on 11/May/24 $$\bullet{f}\left({x}\right)+{f}\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right)=\mathrm{1}{x}+\mathrm{1}\:…\left(\mathrm{1}\right) \\ $$$$\bullet{x}=\mathrm{1}−\frac{\mathrm{1}}{{x}} \\ $$$$\:\:{f}\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right)+{f}\left(\frac{−\mathrm{1}}{{x}−\mathrm{1}}\right)=\mathrm{2}−\frac{\mathrm{1}}{{x}}\:…\left(\mathrm{2}\right) \\ $$$$\bullet{x}=\frac{−\mathrm{1}}{{x}−\mathrm{1}} \\ $$$${f}\left(\frac{−\mathrm{1}}{{x}−\mathrm{1}}\right)+{f}\left({x}\right)=\frac{−\mathrm{1}}{{x}−\mathrm{1}}+\mathrm{1}\:…\left(\mathrm{3}\right)…
Question Number 207243 by efronzo1 last updated on 10/May/24 Answered by mr W last updated on 10/May/24 $${s}={side}\:{length}\:{of}\:{square} \\ $$$$\left(\frac{{s}}{\mathrm{2}}\right)^{\mathrm{2}} ={s}×\mathrm{1} \\ $$$$\Rightarrow{s}=\mathrm{4} \\ $$$${square}'{s}\:{area}={s}^{\mathrm{2}}…
Question Number 206922 by mr W last updated on 01/May/24 $${solve}\:{for}\:{x},\:{y},\:{z}\:\in{R}^{+} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{xy}\:\mathrm{cos}\:\gamma={c}^{\mathrm{2}} \\ $$$${y}^{\mathrm{2}} +{z}^{\mathrm{2}} −\mathrm{2}{yz}\:\mathrm{cos}\:\alpha={a}^{\mathrm{2}} \\ $$$${z}^{\mathrm{2}} +{x}^{\mathrm{2}} −\mathrm{2}{zx}\:\mathrm{cos}\:\beta={b}^{\mathrm{2}} \\ $$$${with}\:\alpha+\beta+\gamma=\mathrm{360}°…
Question Number 206643 by cortano21 last updated on 21/Apr/24 Answered by mr W last updated on 21/Apr/24 Commented by mr W last updated on 21/Apr/24…
Question Number 205631 by Lindemann last updated on 26/Mar/24 Answered by A5T last updated on 26/Mar/24 $${If}\:{angle}\:{between}\:{x}\:{and}\:{x}+\mathrm{17}\:{is}\:\mathrm{90}° \\ $$$$\left({x}+\mathrm{18}\right)^{\mathrm{2}} ={x}^{\mathrm{2}} +\left({x}+\mathrm{17}\right)^{\mathrm{2}} \Rightarrow{x}=\mathrm{7} \\ $$$$\frac{\mathrm{7}×\mathrm{24}}{\mathrm{2}}={r}\left(\frac{\mathrm{7}+\mathrm{24}+\mathrm{25}}{\mathrm{2}}\right)\Rightarrow{r}=\mathrm{3} \\…
Question Number 205654 by Lindemann last updated on 26/Mar/24 $${We}\:{define}\:{a}\:{domino}\:{as}\:{being}\:{and}\:{ordered}\:{pair}\:{of}\:{distinct} \\ $$$${integers}.\:{A}\:{suitable}\:{sequence}\:{of}\:{dominos}\:{is}\:{a}\:{list}\:{of}\:{distinct} \\ $$$${dominoes}\:{in}\:{which}\:{the}\:{first}\:{coordonate}\:{of}\:{each}\:{pair}\:{after} \\ $$$${the}\:{first}\:{is}\:{equal}\:{to}\:{the}\:{second}\:{coordonate}\:{of}\:{the}\:{immediately} \\ $$$${preceding}\:{pair},\:{and}\:{in}\:{which}\:{the}\:{pairs}\:\left({i};{j}\right)\:{and}\:\left({j};{i}\right) \\ $$$${do}\:{not}\:{both}\:{appear}\:{for}\:{all}\:{i}\:{and}\:{j}.\:{Let}\:{D}_{\mathrm{40}} \:{the} \\ $$$${set}\:{of}\:{all}\:{dominoes}\:{whose}\:{coordonate}\:{are}\:{not}\:{greater} \\ $$$${than}\:\mathrm{40}.\:{Find}\:{the}\:{length}\:{of}\:{the}\:{longest}\:{suitable}\:{sequence}…
Question Number 205517 by BaliramKumar last updated on 23/Mar/24 Answered by mr W last updated on 23/Mar/24 $${AB}=\sqrt{\left(\mathrm{5}−\mathrm{4}\right)^{\mathrm{2}} +\left(\mathrm{7}−\mathrm{3}\right)^{\mathrm{2}} }=\sqrt{\mathrm{17}} \\ $$$${say}\:{D}\:{divides}\:{AB}\:{externally}\:{and} \\ $$$${C}\:{divides}\:{AB}\:{internally}. \\…
Question Number 205479 by cortano12 last updated on 22/Mar/24 Answered by mr W last updated on 22/Mar/24 Commented by mr W last updated on 22/Mar/24…
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Question Number 204145 by cortano12 last updated on 07/Feb/24 Answered by AST last updated on 07/Feb/24 Commented by cortano12 last updated on 08/Feb/24 $$\:\mathrm{i}\:\mathrm{got}\:\mathrm{x}=\mathrm{40}° \\…