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n-n-1-n-1-n-n-n-n-1-n-1-n-n-n-2-n-1-n-3-

Question Number 217482 by Rasheed.Sindhi last updated on 14/Mar/25 $$\left(\:{n}\sqrt{\left({n}+\mathrm{1}\right)}\:+\left({n}+\mathrm{1}\right)\sqrt{{n}}\:\:\right)^{{n}} +\left(\:{n}\sqrt{\left({n}+\mathrm{1}\right)}\:−\left({n}+\mathrm{1}\right)\sqrt{{n}}\:\:\right)^{{n}} ={n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)\left({n}+\mathrm{3}\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com

Question-217447

Question Number 217447 by Tawa11 last updated on 14/Mar/25 Commented by mr W last updated on 14/Mar/25 $${question}\:{is}\:{wrong}.\:{the}\:{friction}\: \\ $$$${coefficient}\:{between}\:{cylinder}\:{and} \\ $$$${stick}\:{may}\:{not}\:{be}\:{the}\:{same}\:{as}\:{the} \\ $$$${friction}\:{coefficient}\:{between}\:{cylinder} \\…

Question-217431

Question Number 217431 by peter frank last updated on 13/Mar/25 Answered by Frix last updated on 13/Mar/25 $$\mathrm{Simply}\:\mathrm{by}\:\mathrm{parts}: \\ $$$${u}'=\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:\rightarrow\:{u}=−\frac{\mathrm{1}}{{x}+\mathrm{1}} \\ $$$${v}={x}\mathrm{e}^{{x}} \:\rightarrow\:{v}'=\left({x}+\mathrm{1}\right)\mathrm{e}^{{x}} \\…