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Category: Algebra

n-15-find-4n-2-4n-120-n-4-2n-3-n-2-

Question Number 213047 by hardmath last updated on 29/Oct/24 $$\boldsymbol{\mathrm{n}}\:=\:\mathrm{15} \\ $$$$\mathrm{find}:\:\:\:\:\:\frac{\mathrm{4n}^{\mathrm{2}} \:\:+\:\:\mathrm{4n}\:\:+\:\:\mathrm{120}}{\:\sqrt{\mathrm{n}^{\mathrm{4}} \:\:+\:\:\mathrm{2n}^{\mathrm{3}} \:\:+\:\:\mathrm{n}^{\mathrm{2}} }}\:\:=\:\:? \\ $$ Answered by mehdee7396 last updated on 29/Oct/24…

Question-212991

Question Number 212991 by efronzo1 last updated on 28/Oct/24 $$\:\:\:\:\:\underbrace{\downharpoonleft\underline{}\:} \\ $$ Answered by Frix last updated on 28/Oct/24 $${H}\left({n}\right)=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\:{k}!\left({k}^{\mathrm{2}} +{k}+\mathrm{1}\right) \\ $$$$\frac{{H}\left({n}\right)+\mathrm{1}}{\left({n}+\mathrm{1}\right)!}={n}+\mathrm{1}\:\:\:\:\:\left[\mathrm{test}\:\mathrm{it}\:\mathrm{for}\:{n}=\mathrm{1},\:\mathrm{2},\:\mathrm{3},…\right]…

52-6-43-3-2-52-6-43-3-2-18-

Question Number 213002 by efronzo1 last updated on 28/Oct/24 $$\:\:\frac{\left(\mathrm{52}+\mathrm{6}\sqrt{\mathrm{43}}\:\right)^{\mathrm{3}/\mathrm{2}} −\left(\mathrm{52}−\mathrm{6}\sqrt{\mathrm{43}}\right)^{\mathrm{3}/\mathrm{2}} }{\mathrm{18}}=? \\ $$ Answered by Rasheed.Sindhi last updated on 28/Oct/24 $$\:\:\frac{\left(\mathrm{52}+\mathrm{6}\sqrt{\mathrm{43}}\:\right)^{\mathrm{3}/\mathrm{2}} −\left(\mathrm{52}−\mathrm{6}\sqrt{\mathrm{43}}\right)^{\mathrm{3}/\mathrm{2}} }{\mathrm{18}}=? \\…

Question-212834

Question Number 212834 by efronzo1 last updated on 25/Oct/24 $$\:\:\:\:\underbrace{\:} \\ $$ Answered by golsendro last updated on 25/Oct/24 $$\:\:\mathrm{let}\:\begin{cases}{\mathrm{a}=\mathrm{sin}\:\mathrm{x}\:,\:\mathrm{b}=\mathrm{cos}\:\mathrm{x}}\\{\mathrm{c}=\mathrm{sin}\:\mathrm{y}\:,\:\mathrm{d}=\mathrm{cos}\:\mathrm{y}}\end{cases} \\ $$$$\:\:\mathrm{ac}\:+\:\mathrm{bd}\:=\:\mathrm{cos}\:\left(\mathrm{x}−\mathrm{y}\right) \\ $$$$\:\:\mid\mathrm{ac}+\mathrm{bd}\mid\:\leqslant\:\mathrm{1}\: \\…

Question-212790

Question Number 212790 by RojaTaniya last updated on 24/Oct/24 Answered by A5T last updated on 24/Oct/24 $${x}−{y}=\left({y}−\mathrm{1}+{x}−\mathrm{1}\right)\left({y}−\mathrm{1}−{x}+\mathrm{1}\right) \\ $$$$\Rightarrow{x}={y}\:{or}\:−\mathrm{1}={x}+{y}−\mathrm{2}\Rightarrow{x}+{y}=\mathrm{1} \\ $$$$\Rightarrow{x}+\mathrm{1}={x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}\:{or}\:{x}+\mathrm{1}={x}^{\mathrm{2}} \Rightarrow{x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={x}+{y}+\mathrm{2}=\mathrm{3}…

Question-212736

Question Number 212736 by RojaTaniya last updated on 22/Oct/24 Answered by efronzo1 last updated on 22/Oct/24 $$\:\:\:\:\frac{\mathrm{x}}{\mathrm{4}}\:=\:\frac{\mathrm{9}}{\mathrm{15}}\:\Rightarrow\mathrm{x}=\frac{\mathrm{12}}{\mathrm{5}} \\ $$$$\:\:\:\:\frac{\mathrm{y}}{\mathrm{5}}=\:\frac{\mathrm{9}}{\mathrm{15}}\Rightarrow\mathrm{y}=\:\mathrm{3} \\ $$$$\:\:\:\:\:\frac{\mathrm{z}}{\mathrm{6}}=\:\frac{\mathrm{9}}{\mathrm{15}}\:\Rightarrow\mathrm{z}=\frac{\mathrm{18}}{\mathrm{5}} \\ $$ Commented by…