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Category: Algebra

f-x-x-2-x-2-1-then-f-1-1-f-2-1-f-100-1-f-1-2-f-2-2-f-100-2-f-1-100-f-2-100-f-100-100-

Question Number 210737 by mathlove last updated on 18/Aug/24 $${f}\left({x}\right)=\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} +\mathrm{1}}\:\:\:\:{then} \\ $$$${f}\left(\frac{\mathrm{1}}{\mathrm{1}}\right)+{f}\left(\frac{\mathrm{2}}{\mathrm{1}}\right)+…..+{f}\left(\frac{\mathrm{100}}{\mathrm{1}}\right)+{f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$+{f}\left(\frac{\mathrm{2}}{\mathrm{2}}\right)+…+{f}\left(\frac{\mathrm{100}}{\mathrm{2}}\right)+{f}\left(\frac{\mathrm{1}}{\mathrm{100}}\right)+{f}\left(\frac{\mathrm{2}}{\mathrm{100}}\right) \\ $$$$+……+{f}\left(\frac{\mathrm{100}}{\mathrm{100}}\right)=? \\ $$ Answered by mr W last…

If-sin1-a-1-cos1-cos2-1-cos2-cos3-1-cos44-cos45-

Question Number 210718 by hardmath last updated on 17/Aug/24 $$\mathrm{If}\:\:\:\mathrm{sin1}°\:=\:\mathrm{a} \\ $$$$\frac{\mathrm{1}}{\mathrm{cos1}°\centerdot\mathrm{cos2}°}\:+\:\frac{\mathrm{1}}{\mathrm{cos2}°\centerdot\mathrm{cos3}°}\:+…+\:\frac{\mathrm{1}}{\mathrm{cos44}°\centerdot\mathrm{cos45}°}\:=\:? \\ $$ Commented by Spillover last updated on 18/Aug/24 $${i}\:{get}\:\mathrm{22}\:? \\ $$ Terms…

Question-210701

Question Number 210701 by noraell last updated on 17/Aug/24 Answered by Berbere last updated on 18/Aug/24 $${f}'\left({x}\right)=\mathrm{3}{x}^{\mathrm{2}} +\mathrm{1}\geqslant\mathrm{1}\:{Increase}\:{steictly}\: \\ $$$${f}\:{is}\:{bijection}\:\mathbb{R}\rightarrow\mathbb{R}\Rightarrow{f}^{−} \:{existe} \\ $$$$\mathrm{2}{f}\left({x}\right)+\mathrm{3}{f}^{−} \left({x}\right)=\mathrm{10} \\…

Question-210679

Question Number 210679 by peter frank last updated on 16/Aug/24 Answered by mr W last updated on 16/Aug/24 $${RH}_{\mathrm{16}°{C}} =\frac{\mathrm{273}.\mathrm{15}+\mathrm{16}}{\mathrm{273}.\mathrm{15}+\mathrm{20}}×\frac{\mathrm{17}.\mathrm{5}}{\mathrm{13}.\mathrm{6}}×\mathrm{36\%}\approx\mathrm{46\%} \\ $$ Commented by peter…

Question-210689

Question Number 210689 by Tawa11 last updated on 16/Aug/24 Answered by mm1342 last updated on 16/Aug/24 $${AM}={MD}={a}\Rightarrow{BC}=\frac{\mathrm{2}}{\mathrm{3}}{a} \\ $$$$\frac{{s}}{{S}}=\frac{\frac{{BC}×{h}}{\mathrm{2}}}{\frac{\left({BC}+{AD}\right){h}}{\mathrm{2}}}=\frac{{BC}}{{BC}+{AD}}=\frac{\frac{\mathrm{2}}{\mathrm{3}}{a}}{\mathrm{2}{a}+\frac{\mathrm{2}}{\mathrm{3}}{a}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\:\checkmark \\ $$$$ \\ $$…