Menu Close

Category: Algebra

for-x-y-z-10-with-x-y-z-N-find-10-x-y-z-

Question Number 135472 by mr W last updated on 13/Mar/21 $${for}\:{x}+{y}+{z}=\mathrm{10}\:{with}\:{x},{y},{z}\in\mathbb{N} \\ $$$${find}\:\Sigma\frac{\mathrm{10}!}{{x}!{y}!{z}!} \\ $$ Answered by Ñï= last updated on 13/Mar/21 $$\Sigma\begin{pmatrix}{\:\:\:\mathrm{10}}\\{{x},{y},{z}}\end{pmatrix}=\left(\mathrm{1}+\mathrm{1}+\mathrm{1}\right)^{\mathrm{10}} =\mathrm{3}^{\mathrm{10}} \\…

1-sin-x-dx-

Question Number 4397 by moussapk last updated on 20/Jan/16 $$\int\left(\frac{\mathrm{1}}{{sin}\left({x}\right)}{dx}\right. \\ $$ Answered by Yozzii last updated on 20/Jan/16 $${I}=\int\frac{\mathrm{1}}{{sinx}}{dx}=\int{cosecxdx} \\ $$$${I}=\int\frac{{cosecx}\left({cosecx}+{cotx}\right)}{{cosecx}+{cotx}}{dx} \\ $$$${I}=\int\frac{{cosec}^{\mathrm{2}} {x}+{cotxcosecx}}{{cosecx}+{cotx}}{dx}…

Find-solution-set-the-inequality-2x-5-25-3x-gt-x-

Question Number 135424 by benjo_mathlover last updated on 13/Mar/21 $${Find}\:{solution}\:{set}\:{the}\:{inequality} \\ $$$$\sqrt{\mathrm{2}{x}−\mathrm{5}}\:+\:\sqrt{\mathrm{25}−\mathrm{3}{x}}\:>\:{x} \\ $$ Answered by EDWIN88 last updated on 13/Mar/21 $$\left(\mathrm{1}\right)\:\mathrm{2x}−\mathrm{5}\geqslant\mathrm{0}\rightarrow\mathrm{x}\geqslant\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{3x}−\mathrm{25}\leqslant\mathrm{0}\rightarrow\mathrm{x}\leqslant\frac{\mathrm{25}}{\mathrm{3}} \\…

Given-x-1-x-5-then-x-4-1-x-4-x-2-3x-1-

Question Number 135418 by liberty last updated on 13/Mar/21 $${Given}\:{x}+\frac{\mathrm{1}}{{x}}\:=\:\mathrm{5}\:{then}\:\frac{{x}^{\mathrm{4}} +\frac{\mathrm{1}}{{x}^{\mathrm{4}} }}{{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{1}}\:? \\ $$ Answered by SEKRET last updated on 13/Mar/21 $$\:\left(\boldsymbol{\mathrm{x}}+\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}\right)^{\mathrm{2}} =\mathrm{5}^{\mathrm{2}} \:\:\:\:\:\:\boldsymbol{\mathrm{x}}^{\mathrm{2}}…

Question-135415

Question Number 135415 by JulioCesar last updated on 13/Mar/21 Commented by liberty last updated on 13/Mar/21 $$\left({nx}+\mathrm{1}\right)^{{p}} \:=\:\underset{{i}=\mathrm{0}} {\overset{{p}} {\sum}}\:{C}_{{i}} ^{\:{p}} \:\left({nx}\right)^{{p}−{i}} \: \\ $$$$\:=\:\left({nx}\right)^{{p}}…