Question Number 133761 by liberty last updated on 24/Feb/21 $$\mathrm{Solve}\:\mathrm{the}\:\mathrm{linear}\:\mathrm{congruence}\: \\ $$$$\mathrm{19}{x}\:\equiv\:\mathrm{4}\:\left(\mathrm{mod}\:\mathrm{141}\:\right) \\ $$ Answered by bobhans last updated on 24/Feb/21 $${Using}\:{Euclidean}\:{algorithm}\: \\ $$$$\:\mathrm{141}\:=\:\mathrm{7}×\mathrm{19}+\mathrm{8} \\…
Question Number 68226 by Tinku Tara last updated on 07/Sep/19 $$\mathrm{We}\:\mathrm{are}\:\mathrm{working}\:\mathrm{on}\:\mathrm{problems} \\ $$$$\mathrm{reported}\:\mathrm{on}\:\mathrm{post}\:\mathrm{67927}. \\ $$$$ \\ $$$$\mathrm{We}\:\mathrm{will}\:\mathrm{update}\:\mathrm{on}\:\mathrm{the}\:\mathrm{resolution} \\ $$$$\mathrm{as}\:\mathrm{soon}\:\mathrm{as}\:\mathrm{possible}. \\ $$ Commented by Rasheed.Sindhi last…
Question Number 133757 by liberty last updated on 24/Feb/21 Answered by bobhans last updated on 24/Feb/21 $${Let}\:{x}\:{be}\:{the}\:{least}\:{number}\:{of}\:{marbles} \\ $$$${in}\:{the}\:{box}\:,\:{such}\:{that}\:\begin{cases}{{x}\equiv\mathrm{5}\:\left({mod}\:\mathrm{7}\right)}\\{{x}\equiv\mathrm{6}\:\left({mod}\:\mathrm{11}\right)}\\{{x}\equiv\:\mathrm{8}\:\left({mod}\:\mathrm{13}\right)}\end{cases} \\ $$$${We}\:{can}\:{use}\:{Chinese}\:{remainder}\:{theorem} \\ $$$${a}_{\mathrm{1}} =\mathrm{5}\:;\:{a}_{\mathrm{2}} =\mathrm{6}\:;\:{a}_{\mathrm{3}}…
Question Number 68222 by necxxx last updated on 07/Sep/19 $${Sketch}\:{the}\:{shear}\:{and}\:{moment}\:{diagrams} \\ $$$${of}\:{a}\:{simply}\:{supported}\:{beam}\:{of}\:\mathrm{6}{m}.{The} \\ $$$${load}\:{on}\:{the}\:{beam}\:{consists}\:{of}\:{UDL}\:{of} \\ $$$$\mathrm{15}{KN}/{m}\:{over}\:{the}\:{left}\:{half}\:{of}\:{the}\:{span}. \\ $$$$ \\ $$ Commented by necxxx last updated…
Question Number 68220 by ~ À ® @ 237 ~ last updated on 07/Sep/19 $$\:\:\:{Let}\:{consider}\:\left({a}_{{n}} \right)_{{n}} \:{and}\:\left({u}_{{n}} \right)_{{n}} \:{two}\:{reals}\:\:{sequence}\:\: \\ $$$${defined}\:{such}\:{as}\:\:\:{a}_{\mathrm{0}} =\mathrm{1}\:,\:\forall\:{n}>\mathrm{1}\:\:{a}_{{n}+\mathrm{1}} =\underset{{p}=\mathrm{0}} {\overset{{n}} {\sum}}{a}_{{p}}…
Question Number 133758 by EDWIN88 last updated on 24/Feb/21 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}^{\mathrm{2}} \:\mathrm{ln}\:\left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}/\mathrm{3}} }\right)\left(\mathrm{tan}^{−\mathrm{1}} \sqrt[{\mathrm{3}}]{{x}+\mathrm{1}}\:−\mathrm{tan}^{−\mathrm{1}} \sqrt[{\mathrm{3}}]{{x}−\mathrm{1}}\:\right)=? \\ $$ Answered by liberty last updated on 24/Feb/21 $$\left(\mathrm{i}\right)\:\mathrm{ln}\:\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}/\mathrm{3}}…
Question Number 68219 by ~ À ® @ 237 ~ last updated on 07/Sep/19 $$\:\:\:{Let}\:{consider}\:\left({a}_{{n}} \right)_{{n}} \:{and}\:\left({u}_{{n}} \right)_{{n}} \:{two}\:{reals}\:\:{sequence}\:\: \\ $$$${defined}\:{such}\:{as}\:\:\:{a}_{\mathrm{0}} =\mathrm{1}\:,\:\forall\:{n}>\mathrm{1}\:\:{a}_{{n}+\mathrm{1}} =\underset{{p}=\mathrm{0}} {\overset{{n}} {\sum}}{a}_{{p}}…
Question Number 68212 by peter frank last updated on 07/Sep/19 Answered by $@ty@m123 last updated on 07/Sep/19 $${Let}\:{required}\:{equation}\:{of}\:{line}: \\ $$$${y}={m}_{\mathrm{1}} {x}+{c}\:\:\:…..\left(\mathrm{1}\right) \\ $$$${Given}\:{line}:\:\mathrm{4}{x}+\mathrm{3}{y}=\mathrm{21} \\ $$$${Its}\:{slope}:\:{m}_{\mathrm{2}}…
Question Number 68210 by peter frank last updated on 07/Sep/19 Answered by $@ty@m123 last updated on 08/Sep/19 $${Let}\:\frac{\mathrm{sin}\:{A}}{{a}}=\frac{\mathrm{sin}\:{B}}{{b}}=\frac{\mathrm{sin}\:{C}}{{c}}={R} \\ $$$$\Rightarrow\mathrm{sin}\:{A}={aR},\:\mathrm{sin}\:{B}={bR},\:\mathrm{sin}\:{C}={cR}\:…\left(\mathrm{1}\right) \\ $$$$\left(\mathrm{2}\right)\:\frac{\mathrm{sin}\:\left({A}−{B}\right)\mathrm{sin}\:{C}}{\mathrm{1}+\mathrm{cos}\:\left({A}−{B}\right)\mathrm{cos}\:{C}} \\ $$$$=\:\frac{\mathrm{sin}\:\left({A}−{B}\right)\mathrm{sin}\:\left({A}+{B}\right)}{\mathrm{1}−\mathrm{cos}\:\left({A}−{B}\right)\mathrm{cos}\:\left({A}+{B}\right)} \\…
Question Number 2675 by prakash jain last updated on 24/Nov/15 $$\mathrm{Bases}\:\mathrm{on}\:\mathrm{suggestion}\:\mathrm{from}\:\mathrm{Filup}\:\mathrm{and}\:\mathrm{some} \\ $$$$\mathrm{discussion}\:\mathrm{on}\:\mathrm{that}\:\mathrm{I}\:\mathrm{am}\:\mathrm{suggesting}\:\mathrm{that}\:\mathrm{we} \\ $$$$\mathrm{sequence},\:\mathrm{series}\:\mathrm{and}\:\mathrm{related}\:\mathrm{function}\:\mathrm{as}\:\mathrm{a} \\ $$$$\mathrm{topic}\:\mathrm{for}\:\mathrm{this}\:\mathrm{month}. \\ $$$$\zeta\left({x}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{n}^{−{x}} ,\:{x}\in\mathbb{R},\:{x}>\mathrm{1} \\ $$$$\mathrm{Show}\:\mathrm{that} \\…