Question Number 222356 by wewji12 last updated on 23/Jun/25 $$\int_{\mathrm{0}} ^{\:\infty} \:{f}\left({r}\right)\mathrm{d}{r}=\mathrm{1}\:,\:\int_{\mathrm{0}} ^{\:\infty} \:\mathrm{g}\left({r}\right)\mathrm{d}{r}=\mathrm{1} \\ $$$$\int_{−\infty\boldsymbol{{i}}+\boldsymbol{\gamma}} ^{\:\:\infty\boldsymbol{{i}}+\boldsymbol{\gamma}} \:{F}\left({t}\right){G}\left({t}\right)\mathrm{d}{t}=?? \\ $$$${F}\left({t}\right)=\int_{\mathrm{0}} ^{\:\infty} \:{f}\left({r}\right){e}^{−{rt}} \mathrm{d}{r}\:,\:{G}\left({t}\right)=\int_{\mathrm{0}} ^{\:\infty} \:\mathrm{g}\left({r}\right){e}^{−{rt}}…
Question Number 222359 by MathematicalUser2357 last updated on 23/Jun/25 $$\left(\mathrm{1}\right)\:\left[{ax}^{\mathrm{3}} +{bx}^{\mathrm{2}} +{cx}+{d}\right]_{{x}} ' \\ $$$$\left(\mathrm{2}\right)\:\left[{x}\left({x}−{a}\right)^{\mathrm{2}} \right]_{{x}} ' \\ $$$$\left(\mathrm{3}\right)\:\left[\left({x}^{\mathrm{2}} −{x}\right)\left({x}^{\mathrm{2}} −\mathrm{4}\right)\right]_{{x}} ' \\ $$$$\left(\mathrm{4}\right)\:\left[\left({x}+\mathrm{2}\right)\left({x}−\mathrm{5}\right)\left({x}−\mathrm{1}\right)\right]_{{x}} '…
Question Number 222352 by cryptograph last updated on 23/Jun/25 $${Prove}\:{that}\::\:\left({a}−{b}\right)\left({a}−{c}\right)\left({a}−{d}\right)\left({b}−{c}\right)\left({b}−{d}\right)\left({c}−{d}\right)\:{divisible}\:{by}\:\mathrm{12},\:{with}\:{a},{b},{c},{d}\:\in\mathbb{Z} \\ $$ Answered by vnm last updated on 23/Jun/25 $$\mathrm{Among}\:\mathrm{four}\:\mathrm{integers}\:\mathrm{there}\:\mathrm{will}\:\mathrm{always}\:\mathrm{be}\:\mathrm{two}\: \\ $$$$\mathrm{that}\:\mathrm{are}\:\mathrm{comparable}\:\mathrm{modulo}\:\mathrm{3}\: \\ $$$$\mathrm{and}\:\mathrm{two}\:\mathrm{pairs}\:\mathrm{or}\:\mathrm{three}\:\mathrm{that}\:\mathrm{are} \\…
Question Number 222353 by Nicholas666 last updated on 23/Jun/25 $$ \\ $$$$\:\:\:\:\:\mathrm{solve};\:\:\frac{\mathrm{11}}{\mathrm{29}}\:=\:?\:,\:\mathrm{no}\:\mathrm{fraction}\:\mathrm{and}\:\mathrm{no}\:\mathrm{decimal}\:\:\:\:\: \\ $$$$ \\ $$ Commented by mr W last updated on 23/Jun/25 $${what}\:{is}\:{to}\:{solve}?\:{can}\:{you}\:{make}\:{clear}…
Question Number 222300 by fantastic last updated on 22/Jun/25 $${How}\:{do}\:{you}\:{put}\:{a}\:{box}\:{around}\:{something}??\:{please}\:{tell}\:{me} \\ $$ Commented by mr W last updated on 22/Jun/25 $$\begin{array}{|c|}{{something}\:{like}\:{this}?}\\\hline\end{array} \\ $$ Commented by…
Question Number 222334 by SonGoku last updated on 22/Jun/25 Commented by SonGoku last updated on 22/Jun/25 $$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{Rectangle}? \\ $$ Commented by mr W last updated…
Question Number 222296 by Shrodinger last updated on 22/Jun/25 $$\left(\mathrm{1}+{x}^{\mathrm{4}} \right){y}'−{x}^{\mathrm{3}} {y}\:=\:{x}^{\mathrm{5}} −{x}^{\mathrm{3}} +\mathrm{2}{x}+\mathrm{1} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 222329 by klipto last updated on 22/Jun/25 $$\boldsymbol{\mathrm{lim}}_{\boldsymbol{\mathrm{x}}\rightarrow\infty} \:\mathrm{4}\boldsymbol{\mathrm{x}}+\sqrt{\mathrm{16}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\mathrm{3}\boldsymbol{\mathrm{x}}} \\ $$$$ \\ $$ Commented by mr W last updated on 22/Jun/25 $$=\infty…
Question Number 222331 by Nicholas666 last updated on 22/Jun/25 $$ \\ $$$$\:\:\:\:\:\:\:\mathrm{Find}\:\mathrm{closed}\:\mathrm{form}; \\ $$$$\:\:\:\:\:\:\:\:\:\int_{\:\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\mathrm{Li}_{\mathrm{2}} \left({z}^{\mathrm{2}} \right)\mathrm{Li}_{\mathrm{2}} \left(−{z}^{\mathrm{2}} \right)}{\mathrm{1}\:+\:{z}^{\mathrm{2}} }\:\mathrm{d}{z}\:=\:? \\ $$ Terms of…
Question Number 222299 by fantastic last updated on 22/Jun/25 $${For}\:{what}\:{value}\:{of}\:\:{k}\:\:{the}\:{roots}\:{of}\:{the}\:{equation} \\ $$$$\frac{{x}^{\mathrm{2}} −\mathrm{2}{x}}{\mathrm{4}{x}−\mathrm{1}}=\frac{{k}−\mathrm{1}}{{k}+\mathrm{1}} \\ $$$${will}\:{have}\:{same}\:{value}\:{but}\:\:{with}\:{opposite}\:{symbol}\left({like}\:{x}={a}\:{and}\:−{a}\right) \\ $$$${i}\:{mean}\:{the}\:{two}\:{valuea}\:{of}\:{x}\:{will}\:{be}\:{this}\:{type} \\ $$$${x}=\mathrm{2}\:{and}\:−\mathrm{2}\left({both}\:\mathrm{2}\:{but}\:{opposite}\:{symbols}\right) \\ $$ Answered by mr W…