Question Number 58529 by Tawa1 last updated on 24/Apr/19 $$\mathrm{Solve}\:\mathrm{for}\:\mathrm{x}\:\mathrm{and}\:\mathrm{y} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{x}}\:+\:\frac{\mathrm{1}}{\mathrm{y}}\:=\:\mathrm{4}\:\:\:\:\:\:\:…….\:\left(\mathrm{i}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{y}}\:+\:\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{x}}\:\:=\:\:\mathrm{9}\:\:\:\:\:\:…….\:\left(\mathrm{ii}\right) \\ $$ Commented by MJS last updated on 24/Apr/19…
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Question Number 124056 by Snail last updated on 30/Nov/20 $${If}\:\alpha,\beta,\gamma\:{are}\:{the}\:{real}\:{roots}\:{of}\:{the}\:{equation}\: \\ $$$${a}^{\mathrm{3}} −\mathrm{6}{a}^{\mathrm{2}} +\mathrm{3}{a}+\mathrm{1}=\mathrm{0}\:\:\:\:\:{then}\:{find}\:{all}\:{possible}\:{values} \\ $$$${of}\:{the}?{expression}\:\alpha^{\mathrm{2}} \beta+\beta^{\mathrm{2}} \gamma+\gamma^{\mathrm{2}} \alpha \\ $$ Commented by MJS_new last…
Question Number 189576 by normans last updated on 18/Mar/23 $$ \\ $$$$\:\:\:\:\:\mathrm{2}^{\boldsymbol{{x}}} \:+\:\mathrm{2}^{\boldsymbol{{x}}} −\mathrm{4}\:+\:\boldsymbol{{x}}\:=\:−\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\boldsymbol{{x}}\:=\:??? \\ $$$$ \\ $$ Answered by Gbenga last updated…
Question Number 189564 by Rupesh123 last updated on 18/Mar/23 Answered by HeferH last updated on 18/Mar/23 $$\alpha\:+\:\mathrm{67}°\:=\:\mathrm{180}° \\ $$$$\alpha\:=\:\mathrm{113}° \\ $$ Commented by HeferH last…
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Question Number 189533 by mustafazaheen last updated on 18/Mar/23 $${when}\:\:{a}+{b}=\mathrm{60}^{°} \:\:\:\:\:{find}\:\:\frac{{cos}^{\mathrm{2}} {a}−{sin}^{\mathrm{2}} {b}}{{cos}\left({a}−{b}\right)}=? \\ $$ Answered by som(math1967) last updated on 18/Mar/23 $$\frac{{cos}^{\mathrm{2}} {a}−{sin}^{\mathrm{2}} {b}}{{cos}\left({a}−{b}\right)}…