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psin-con-2-a-pcos-sin-2-b-p-0-0-pi-2-prove-p-a-2-b-2-3-2-ab-

Question Number 202295 by liuxinnan last updated on 24/Dec/23 $${psin}\theta{con}^{\mathrm{2}} \theta={a} \\ $$$${pcos}\theta{sin}^{\mathrm{2}} \theta={b} \\ $$$${p}\neq\mathrm{0}\:\:\:\:\theta\in\left(\mathrm{0}\:,\frac{\pi}{\mathrm{2}}\right) \\ $$$${prove}\:{p}=\frac{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{{ab}} \\ $$ Answered by…

If-n-2-and-U-n-3-5-n-3-5-n-then-prove-that-U-n-1-6U-n-4U-n-1-

Question Number 202328 by MATHEMATICSAM last updated on 24/Dec/23 $$\mathrm{If}\:{n}\:\geqslant\:\mathrm{2}\:\mathrm{and}\:\mathrm{U}_{{n}} \:=\:\left(\mathrm{3}\:+\:\sqrt{\mathrm{5}}\right)^{{n}} \:+\:\left(\mathrm{3}\:−\:\sqrt{\mathrm{5}}\right)^{{n}} \\ $$$$\mathrm{then}\:\mathrm{prove}\:\mathrm{that}\:\mathrm{U}_{{n}\:+\:\mathrm{1}} \:=\:\mathrm{6U}_{{n}} \:−\:\mathrm{4U}_{{n}\:−\:\mathrm{1}} \:. \\ $$ Commented by aleks041103 last updated on…

Question-202224

Question Number 202224 by sonukgindia last updated on 23/Dec/23 Answered by witcher3 last updated on 23/Dec/23 $$\mathrm{x}\rightarrow\frac{\mathrm{1}}{\mathrm{a}+\mathrm{bcos}^{\mathrm{2}} \left(\mathrm{x}\right)}=\mathrm{f}\left(\mathrm{x}\right),\mathrm{is}\:\mathrm{p}\:\mathrm{peridic} \\ $$$$\Rightarrow\int_{\mathrm{k}\pi} ^{\left(\mathrm{k}+\mathrm{1}\right)\pi} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\int_{\mathrm{0}} ^{\pi} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx};\mathrm{by}\:\mathrm{k}\pi+\mathrm{y}=\mathrm{x} \\…

Question-202276

Question Number 202276 by professorleiciano last updated on 23/Dec/23 Answered by professorleiciano last updated on 23/Dec/23 $${Area}\left({retangulo}\right)=\mathrm{4}×\mathrm{6}=\mathrm{24}{m}^{\mathrm{2}} \\ $$$${Area}\left({triangulo}\:{I}\right)=\mathrm{3}×\mathrm{6}=\mathrm{18}/\mathrm{2}=\mathrm{9}{m}^{\mathrm{2}} \\ $$$${Area}\left({triangulo}\:{II}\right)=\mathrm{3}×\mathrm{4}=\mathrm{12}/\mathrm{2}=\mathrm{6}{m}^{\mathrm{2}} \\ $$$${Area}\left({triangulo}\:{III}\right)=\mathrm{3}×\mathrm{4}=\mathrm{12}/\mathrm{2}=\mathrm{6}{m}^{\mathrm{2}} \\ $$$${Area}\left({total}\right)=\mathrm{24}{m}^{\mathrm{2}}…