Menu Close

Category: None

how-to-show-f-x-x-4-2x-3-5x-2-16x-20-in-the-form-of-x-2-x-a-2-4-x-b-2-

Question Number 162351 by ZiYangLee last updated on 29/Dec/21 $$\mathrm{how}\:\mathrm{to}\:\mathrm{show}\: \\ $$$${f}\left({x}\right)={x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{3}} +\mathrm{5}{x}^{\mathrm{2}} −\mathrm{16}{x}−\mathrm{20}\: \\ $$$$\mathrm{in}\:\mathrm{the}\:\mathrm{form}\:\mathrm{of}\:\left({x}^{\mathrm{2}} +{x}+{a}\right)^{\mathrm{2}} −\mathrm{4}\left({x}+{b}\right)^{\mathrm{2}} . \\ $$ Answered by Rasheed.Sindhi…

Hello-please-show-it-a-0-pi-4-a-tan-a-2a-

Question Number 162326 by stelor last updated on 28/Dec/21 $${Hello}\:{please}\:{show}\:{it}… \\ $$$$\:{a}\:\in\:\left[\mathrm{0}\:,\:\frac{\pi}{\mathrm{4}}\right]\:\:\:\:\:\:\:\:{a}\:\leqslant{tan}\:{a}\:\leqslant\:\mathrm{2}{a} \\ $$ Answered by mindispower last updated on 28/Dec/21 $$\mathrm{1}\leqslant\mathrm{1}+{tg}^{\mathrm{2}} \left({a}\right)\leqslant\mathrm{2},\forall{a}\in\left[\mathrm{0},\frac{\pi}{\mathrm{4}}\right] \\ $$$$\int_{\mathrm{0}}…

1-cos-4-x-sin-4-y-1-x-y-find-dy-dx-2-solve-2-x-4-x-6-3-x-3-lim-x-4-cos-x-x-sin-x-x-cos-2x-x-4-

Question Number 96782 by  M±th+et+s last updated on 04/Jun/20 $$\left.\mathrm{1}\right)\frac{{cos}^{\mathrm{4}} \left(\theta\right)}{{x}}−\frac{{sin}^{\mathrm{4}} \left(\theta\right)}{{y}}=\frac{\mathrm{1}}{{x}+{y}} \\ $$$${find}\:\frac{{dy}}{{dx}} \\ $$$$ \\ $$$$\left.\mathrm{2}\right){solve}:\mathrm{2}\lfloor{x}−\mathrm{4}+\lfloor{x}\rfloor\rfloor=\mathrm{6}−\mathrm{3}\lfloor{x}\rfloor \\ $$$$ \\ $$$$\left.\mathrm{3}\right)\underset{{x}\rightarrow\mathrm{4}} {{lim}}\frac{\left({cos}\left({x}\right)\right)^{{x}} −\left({sin}\left({x}\right)\right)^{{x}} −{cos}\left(\mathrm{2}{x}\right)}{\left({x}−\mathrm{4}\right)}…

Question-31188

Question Number 31188 by mondodotto@gmail.com last updated on 03/Mar/18 Commented by Tinkutara last updated on 03/Mar/18 $$\mathrm{tan}\:\theta=\frac{\mathrm{4}}{\mathrm{3}} \\ $$$$\mathrm{sin}\:\theta=\pm\frac{\mathrm{4}}{\mathrm{5}};\mathrm{cos}\:\theta=\pm\frac{\mathrm{3}}{\mathrm{5}} \\ $$$$\pm\frac{\mathrm{3}}{\mathrm{5}}=\mathrm{1}−\mathrm{2sin}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}} \\ $$$$\mathrm{sin}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{5}}…

Question-162249

Question Number 162249 by SANOGO last updated on 28/Dec/21 Answered by mr W last updated on 28/Dec/21 $$\left(\mathrm{1}\right) \\ $$$$\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}\:{x}^{{k}} =\left(\mathrm{1}+{x}\right)^{{n}} \\ $$$${set}\:{x}=−\mathrm{1}…

The-tangent-of-a-parabola-y-2-4ax-at-the-point-P-ap-2-2ap-intersects-the-line-x-a-0-at-T-i-If-M-is-the-midpoint-of-PT-find-the-coordinates-of-M-in-terms-of-a-and-p-ii-Prove-that-t

Question Number 162253 by ZiYangLee last updated on 28/Dec/21 $$\mathrm{The}\:\mathrm{tangent}\:\mathrm{of}\:\mathrm{a}\:\mathrm{parabola}\:{y}^{\mathrm{2}} =\mathrm{4}{ax}\:\mathrm{at}\:\mathrm{the}\:\mathrm{point} \\ $$$${P}\:\left({ap}^{\mathrm{2}} ,\:\mathrm{2}{ap}\right)\:\mathrm{intersects}\:\mathrm{the}\:\mathrm{line}\:{x}+{a}=\mathrm{0}\:\mathrm{at}\:{T}\:. \\ $$$$\left(\mathrm{i}\right)\:\mathrm{If}\:{M}\:\mathrm{is}\:\mathrm{the}\:\mathrm{midpoint}\:\mathrm{of}\:{PT}\:,\:\mathrm{find}\:\mathrm{the}\: \\ $$$$\:\:\:\:\:\:\mathrm{coordinates}\:\mathrm{of}\:{M}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:{a}\:\mathrm{and}\:{p}. \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{locus}\:\mathrm{of}\:{M}\:\mathrm{is} \\ $$$$\:\:\:\:\:\:\:\:{y}^{\mathrm{2}} \left(\mathrm{2}{x}+{a}\right)={a}\left(\mathrm{3}{x}+{a}\right)^{\mathrm{2}} \\ $$…

2x-5-x-2-4x-5-dx-

Question Number 162240 by Gbenga last updated on 27/Dec/21 $$\int\frac{\mathrm{2}\boldsymbol{{x}}−\mathrm{5}}{\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{4}\boldsymbol{{x}}+\mathrm{5}}\boldsymbol{{dx}} \\ $$ Answered by Ar Brandon last updated on 27/Dec/21 $$\mathrm{2}{x}−\mathrm{5}=\alpha\left(\mathrm{2}{x}+\mathrm{4}\right)+\beta=\mathrm{2}\alpha{x}+\left(\mathrm{4}\alpha+\beta\right) \\ $$$$\Rightarrow\alpha=\mathrm{1},\:\mathrm{4}+\beta=−\mathrm{5},\:\beta=−\mathrm{9} \\…