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Category: Algebra

Solve-a-2-c-2-196-i-b-2-c-a-2-169-ii-c-2-b-c-2-225-iii-

Question Number 103736 by I want to learn more last updated on 17/Jul/20 $$\mathrm{Solve}:\:\:\:\:\:\:\mathrm{a}^{\mathrm{2}} \:\:+\:\:\mathrm{c}^{\mathrm{2}} \:\:=\:\:\:\mathrm{196}\:\:\:\:\:\:…\:\left(\mathrm{i}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{b}^{\mathrm{2}} \:\:+\:\:\left(\mathrm{c}\:\:−\:\:\mathrm{a}\right)^{\mathrm{2}} \:\:=\:\:\mathrm{169}\:\:\:\:\:…\:\left(\mathrm{ii}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{c}^{\mathrm{2}} \:\:+\:\:\left(\mathrm{b}\:\:−\:\:\mathrm{c}\right)^{\mathrm{2}} \:\:=\:\:\mathrm{225}\:\:\:\:\:…\:\left(\mathrm{iii}\right) \\…

Question-169271

Question Number 169271 by Shrinava last updated on 27/Apr/22 Answered by mr W last updated on 27/Apr/22 $${a}={r}\:\mathrm{cot}\:\frac{{B}}{\mathrm{2}}+{r}\:\mathrm{cot}\:\frac{{C}}{\mathrm{2}} \\ $$$$\frac{{a}}{{r}}=\mathrm{cot}\:\frac{{B}}{\mathrm{2}}+\mathrm{cot}\:\frac{{C}}{\mathrm{2}}\geqslant\mathrm{2}\left(\mathrm{cot}\:\frac{{B}}{\mathrm{2}}\:\mathrm{cot}\:\frac{{C}}{\mathrm{2}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\left(\frac{{a}}{{r}}\right)^{\mathrm{6}} \geqslant\mathrm{2}^{\mathrm{6}} ×\left(\mathrm{cot}\:\frac{{B}}{\mathrm{2}}\:\mathrm{cot}\:\frac{{C}}{\mathrm{2}}\right)^{\mathrm{3}} \\…

Question-169260

Question Number 169260 by mathlove last updated on 27/Apr/22 Answered by nurtani last updated on 27/Apr/22 $$\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}=\frac{\mathrm{1}}{\mathrm{3}}…\left({i}\right) \\ $$$$\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}=\frac{\mathrm{1}}{\mathrm{6}}…\left({ii}\right) \\ $$$$\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{z}}=\frac{\mathrm{1}}{\mathrm{9}}…\left({iii}\right) \\ $$$$\left({i}\right)+\left({ii}\right)+\left({iii}\right): \\ $$$$\frac{\mathrm{2}}{{x}}+\frac{\mathrm{2}}{{y}}+\frac{\mathrm{2}}{{z}}=\frac{\mathrm{11}}{\mathrm{18}}…

Question-169235

Question Number 169235 by Shrinava last updated on 26/Apr/22 Commented by mr W last updated on 26/Apr/22 $${i}\:{guess}\:\left\{{P}\right\}={KL}\cap{AD}\:{means} \\ $$$${P}\:{is}\:{intersection}\:{of}\:{KL}\:{and}\:{AD}. \\ $$$${but}\:{what}\:{do}\:{you}\:{say}\:{with}\:{m}\left(\measuredangle{DPC}\right)? \\ $$ Commented…

Question-103685

Question Number 103685 by bemath last updated on 16/Jul/20 Answered by bobhans last updated on 16/Jul/20 $${x}−\sqrt{\left(\mathrm{2}−{x}\right)\left(\mathrm{3}−{x}\right)}\:=\:\sqrt{\mathrm{4}−{x}}\left\{\sqrt{\mathrm{3}−{x}}+\sqrt{\mathrm{2}−{x}}\right\} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{6}−\mathrm{2}{x}\sqrt{\left(\mathrm{2}−{x}\right)\left(\mathrm{3}−{x}\right)}\:= \\ $$$$\left(\mathrm{4}−{x}\right)\left(\mathrm{5}−\mathrm{2}{x}\right)+\mathrm{2}\sqrt{\left(\mathrm{3}−{x}\right)\left(\mathrm{2}−{x}\right)} \\ $$$$\mathrm{4}{x}−\mathrm{7}\:= \\…