Question Number 73295 by naka3546 last updated on 10/Nov/19 $${How}\:\:{many}\:\:{solution}\:\:{so}\:\:{that} \\ $$$$\mathrm{3}{n}−\mathrm{4},\:\:\mathrm{4}{n}−\mathrm{5},\:\:\mathrm{5}{n}−\mathrm{13} \\ $$$${are}\:\:{prime}\:\:{numbers}\:? \\ $$ Answered by mind is power last updated on 10/Nov/19…
Question Number 138831 by mey3nipaba last updated on 18/Apr/21 $${Why}\:{is}\:{the}\:{ammeter}\:{always}\:{connected}\:{in}\:{series} \\ $$$${and}\:{the}\:{voltmeter}\:{in}\:{parallel}? \\ $$ Answered by TheSupreme last updated on 19/Apr/21 $${series}\:{connection}\:{have}\:{same}\:{current}\:\left({ammeter}\:{messure}\:{current}\right) \\ $$$${parallel}\:{connection}\:{saves}\:{ddp}\:\left({voltmeter}\:{measure}\:{ddp}\right) \\…
Question Number 73293 by ~blr237~ last updated on 10/Nov/19 $${Explicit}\:\:{f}\left({x}\right)=\:\int_{\mathrm{1}} ^{\infty} \:\frac{{lnt}}{\left({x}^{\mathrm{2}} +{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{dt}\: \\ $$ Commented by mathmax by abdo last updated on…
Question Number 138827 by mathdanisur last updated on 18/Apr/21 $${Solve}\:{for}\:{real}\:{numbers}: \\ $$$$\begin{cases}{−{a}^{\mathrm{1}} −{b}^{\mathrm{2}} −{c}^{\mathrm{3}} =\mathrm{2024}^{\mathrm{12}} }\\{−{a}^{−\mathrm{1}} −{b}^{−\mathrm{2}} −{c}^{−\mathrm{3}} =\mathrm{2024}^{−\mathrm{12}} }\\{{a}^{\mathrm{1}} {b}^{\mathrm{2}} {c}^{\mathrm{3}} =\mathrm{2024}^{\mathrm{24}} }\end{cases} \\…
Question Number 7752 by Tawakalitu. last updated on 14/Sep/16 $${If}\:{f}\left({x}\right)\:=\:{x}\:+\:{ax}^{\mathrm{2}} \:+\:{bx}^{\mathrm{3}} \:+\:…\:\:\:\:\:\:.{obtain}\:\sqrt{{f}\left({x}^{\mathrm{3}} \right)} \\ $$$${up}\:{to}\:{x}^{\mathrm{3}} \\ $$ Commented by FilupSmith last updated on 14/Sep/16 $${f}\left({x}\right)={x}^{\mathrm{1}}…
Question Number 138821 by mnjuly1970 last updated on 18/Apr/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…….\:{nice}\:..\:..\:..\:{calculus}…… \\ $$$${find}\:{the}\:{value}\:{of}: \\ $$$$\:\:\:\:\:\:\:\:\:\Theta=\underset{{n}=−\infty\:} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{6}{n}^{\mathrm{2}} +\mathrm{5}{n}+\mathrm{1}}=? \\ $$$$ \\ $$ Commented by Kamel last…
Question Number 7750 by Tawakalitu. last updated on 13/Sep/16 $${Let}\:{n}\:=\:\left(\mathrm{2}^{\mathrm{31}} \right)\:×\:\left(\mathrm{3}^{\mathrm{19}} \right)\:{how}\:{many}\:{positive}\:{integer} \\ $$$${divisors}\:{of}\:{n}^{\mathrm{2}} \:{are}\:{less}\:{than}\:{n}\:{but}\:{do}\:{not}\:{divide}\:{n} \\ $$ Commented by Yozzia last updated on 13/Sep/16 $${n}^{\mathrm{2}}…
Question Number 138823 by mnjuly1970 last updated on 18/Apr/21 $$\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:……\mathscr{N}{ice}\:\:…\:\mathscr{C}{alculus}…… \\ $$$$\:\:\:\:\:\:\:\mathscr{E}{valuate}::\:\:\:\:\:\boldsymbol{\phi}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{}=? \\ $$ Answered by qaz last updated on 18/Apr/21…
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Question Number 7748 by Tawakalitu. last updated on 13/Sep/16 $${Given}\:{that}\:{Z}\:{and}\:{H}\:{are}\:{complex}\:{number}.\: \\ $$$${obtain}\:{the}\:{real}\:{and}\:{imaginary}\:{of}\:{Z}^{{H}} \\ $$ Answered by Yozzia last updated on 13/Sep/16 $${Let}\:{Z}={re}^{{i}\theta} ,\:{H}={c}+{di}\:\:\left({r},\theta,{c},{d}\in\mathbb{R},\:{r}>\mathrm{0},\:{i}=\sqrt{−\mathrm{1}}\right). \\ $$$${Z}^{{H}}…