Question and Answers Forum
All Questions Topic List
AllQuestion and Answers: Page 325
Question Number 176942 Answers: 1 Comments: 0
Question Number 176936 Answers: 1 Comments: 0
Question Number 176933 Answers: 1 Comments: 0
Question Number 182216 Answers: 3 Comments: 2
$$\:{Find}\:{a},\:{b},\:{c}\:\in\:\mathbb{N}\:;\:\mathrm{2}^{\:{a}} +\:\mathrm{4}^{\:{b}} +\:\mathrm{8}^{\:{c}} =\:\mathrm{328} \\ $$
Question Number 176928 Answers: 1 Comments: 0
$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{smallest}\:\mathrm{three}\:\mathrm{digit}\:\mathrm{number} \\ $$$$\mathrm{divisible}\:\mathrm{by}\:\frac{\mathrm{2}}{\mathrm{3}},\:\frac{\mathrm{7}}{\mathrm{9}},\:\frac{\mathrm{4}}{\mathrm{13}}\:\mathrm{without}\:\mathrm{a}\:\mathrm{remainder}\:? \\ $$
Question Number 176925 Answers: 1 Comments: 0
Question Number 176917 Answers: 0 Comments: 0
Question Number 176915 Answers: 1 Comments: 0
Question Number 176913 Answers: 1 Comments: 1
$${Determiner}\:{la}\:{hauteur}\:\mathrm{D}{E}\left({r}+{x}\right)\:{en}\:{fonction}\:{de}\:{r} \\ $$$${r}=\mathrm{OOC}=\mathrm{BH}\:\:\:\:\:\mathrm{BF}=\mathrm{20} \\ $$$$\mathrm{pour}\:\mathrm{que}\:\mathrm{distance}\left(\mathrm{AB}+\mathrm{BC}+\mathrm{CD}+\mathrm{DE}+\mathrm{EF}\:\:\mathrm{soit}\:\right. \\ $$$$\mathrm{sgale}\:\mathrm{AC}+\mathrm{arcCDF} \\ $$
Question Number 176898 Answers: 1 Comments: 0
Question Number 176746 Answers: 5 Comments: 0
$${a}_{{n}+\mathrm{2}} −\mathrm{5}{a}_{{n}+\mathrm{1}} +\mathrm{6}{a}_{{n}} =\mathrm{3}{n}+\mathrm{5}^{{n}} \\ $$$${a}_{\mathrm{1}} =\mathrm{1},\:{a}_{\mathrm{2}} =\mathrm{0} \\ $$$${find}\:{a}_{{n}} \\ $$
Question Number 176741 Answers: 1 Comments: 1
$$\: \\ $$$$\:{In}\:{the}\:{following}\:{figure},\:{if}\:\:\:\frac{{S}_{\mathrm{1}} \:+{S}_{\mathrm{2}} }{\:\sqrt{{S}_{\mathrm{3}} }}\:=\:\frac{\mathrm{6}}{\:\sqrt{\pi}} \\ $$$$\:{find}\:{the}\:{area}\:{of}\:{the}\:{circular}\:{crown}\: \\ $$$$\: \\ $$
Question Number 176727 Answers: 0 Comments: 0
$$\mathrm{In}\:\:\:\bigtriangleup\mathrm{ABC}\:\:\:\mathrm{the}\:\mathrm{following}\:\mathrm{relationship} \\ $$$$\mathrm{holds}: \\ $$$$\mathrm{6r}\:\:\underset{\boldsymbol{\mathrm{cyc}}} {\sum}\:\frac{\mathrm{r}_{\boldsymbol{\mathrm{a}}} }{\mathrm{s}\:+\:\mathrm{n}_{\boldsymbol{\mathrm{a}}} }\:\:+\:\:\underset{\boldsymbol{\mathrm{cyc}}} {\sum}\:\mathrm{n}_{\boldsymbol{\mathrm{a}}} \:\geqslant\:\mathrm{3s} \\ $$
Question Number 176723 Answers: 0 Comments: 0
$${donner}\:{la}\:{forme}\:{trigonometrique}\left({i}−\mathrm{1}\right)^{\mathrm{5}} /\left({i}−\mathrm{4}\right)^{\mathrm{4}} \\ $$
Question Number 176721 Answers: 2 Comments: 0
Question Number 176722 Answers: 0 Comments: 1
$$\frac{\sqrt{\mathrm{11}}+\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{11}}+\mathrm{11}}\:=\:\frac{\left(\sqrt{\mathrm{11}}+\mathrm{1}\right)\left(\mathrm{2}\sqrt{\mathrm{11}}−\mathrm{11}\right)}{\left(\mathrm{2}\sqrt{\mathrm{11}}+\mathrm{11}\right)\left(\mathrm{2}\sqrt{\mathrm{11}}−\mathrm{11}\right)}\:=\: \\ $$$$\frac{\mathrm{2}\sqrt{\mathrm{11}}−\mathrm{11}+\mathrm{2}×\mathrm{11}+\mathrm{11}\sqrt{\mathrm{11}}}{\mathrm{2}×\mathrm{11}−\mathrm{121}}\:=\:\frac{\mathrm{13}\sqrt{\mathrm{11}}+\mathrm{11}}{−\mathrm{99}} \\ $$$$\boldsymbol{{A}}{ns} \\ $$
Question Number 176719 Answers: 0 Comments: 0
$$ \\ $$
Question Number 176718 Answers: 1 Comments: 1
$${x}^{\mathrm{3}} +{y}^{\mathrm{3}} ={z}^{\mathrm{2}} \\ $$$${x}^{\mathrm{3}} +{z}^{\mathrm{3}} ={y}^{\mathrm{2}} \\ $$$${y}^{\mathrm{3}} +{z}^{\mathrm{3}} ={x}^{\mathrm{2}} \\ $$$$\mathrm{obviously}\:{x}={y}={z}=\mathrm{0}\vee\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{trying}\:\mathrm{to}\:\mathrm{totally}\:\mathrm{solve}\:\mathrm{it} \\ $$$$\mathrm{let}\:{y}={px}\wedge{z}={qx} \\ $$$$\left({p}^{\mathrm{3}} +\mathrm{1}\right){x}^{\mathrm{3}} ={q}^{\mathrm{2}} {x}^{\mathrm{2}} \\ $$$$\left({q}^{\mathrm{3}} +\mathrm{1}\right){x}^{\mathrm{3}} ={p}^{\mathrm{2}} {x}^{\mathrm{2}} \\ $$$$\left({p}^{\mathrm{3}} +{q}^{\mathrm{3}} \right){x}^{\mathrm{3}} ={x}^{\mathrm{2}} \\ $$$$\Rightarrow\:{x}=\mathrm{0}\:\Rightarrow\:{y}=\mathrm{0}\wedge{z}=\mathrm{0}\:\bigstar \\ $$$${x}=\frac{{q}^{\mathrm{2}} }{{p}^{\mathrm{3}} +\mathrm{1}} \\ $$$${x}=\frac{{p}^{\mathrm{2}} }{{q}^{\mathrm{3}} +\mathrm{1}} \\ $$$${x}=\frac{\mathrm{1}}{{p}^{\mathrm{3}} +{q}^{\mathrm{3}} } \\ $$$$\Rightarrow \\ $$$$\frac{{p}^{\mathrm{2}} }{{q}^{\mathrm{3}} +\mathrm{1}}=\frac{{q}^{\mathrm{2}} }{{p}^{\mathrm{3}} +\mathrm{1}} \\ $$$$\frac{{p}^{\mathrm{2}} }{{q}^{\mathrm{3}} +\mathrm{1}}=\frac{\mathrm{1}}{{p}^{\mathrm{3}} +{q}^{\mathrm{3}} } \\ $$$$========== \\ $$$${p}^{\mathrm{5}} +{p}^{\mathrm{2}} −{q}^{\mathrm{5}} −{q}^{\mathrm{2}} =\mathrm{0} \\ $$$${p}^{\mathrm{5}} +{p}^{\mathrm{2}} {q}^{\mathrm{3}} −{q}^{\mathrm{3}} −\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{subtracting}\:\mathrm{both} \\ $$$${p}^{\mathrm{2}} \left({q}^{\mathrm{3}} −\mathrm{1}\right)+{q}^{\mathrm{5}} −{q}^{\mathrm{3}} +{q}^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$$\left({q}−\mathrm{1}\right)\left({p}^{\mathrm{2}} \left({q}^{\mathrm{2}} +{q}+\mathrm{1}\right)+{q}^{\mathrm{4}} +{q}^{\mathrm{3}} +{q}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\Rightarrow\:{q}=\mathrm{1}\:\Rightarrow\:{p}^{\mathrm{5}} +{p}^{\mathrm{2}} −\mathrm{2}=\mathrm{0} \\ $$$$\:\:\:\:\:\left({p}−\mathrm{1}\right)\left({p}^{\mathrm{4}} +{p}^{\mathrm{3}} +\mathrm{2}{p}+\mathrm{2}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\Rightarrow\:{p}=\mathrm{1}\:\Rightarrow\:{x}={y}={z}=\frac{\mathrm{1}}{\mathrm{2}}\:\bigstar \\ $$$$\:\:\:\:\:{p}^{\mathrm{4}} +{p}^{\mathrm{3}} +\mathrm{2}{p}+\mathrm{2}=\mathrm{0} \\ $$$$\:\:\:\:\:\mathrm{no}\:\mathrm{useable}\:\mathrm{exact}\:\mathrm{solutions} \\ $$$$\:\:\:\:\:{p}\approx−.\mathrm{975564}\pm.\mathrm{528237i} \\ $$$$\:\:\:\:\:\Rightarrow\:{x}\approx.\mathrm{33635}\mp.\mathrm{515329i} \\ $$$$\:\:\:\:\:\wedge\:{y}\approx−.\mathrm{0559113}\pm.\mathrm{680406i} \\ $$$$\:\:\:\:\:\wedge\:{z}={x}\:\bigstar \\ $$$$\:\:\:\:\:{p}\approx.\mathrm{475564}\pm\mathrm{1}.\mathrm{18273i} \\ $$$$\:\:\:\:\:\Rightarrow\:{x}\approx−.\mathrm{586346}\pm.\mathrm{562464i} \\ $$$$\:\:\:\:\:\wedge\:{y}\approx−.\mathrm{944089}\mp.\mathrm{426001i} \\ $$$$\:\:\:\:\:\wedge\:{z}={x}\:\bigstar \\ $$$${p}^{\mathrm{2}} \left({q}^{\mathrm{2}} +{q}+\mathrm{1}\right)+{q}^{\mathrm{4}} +{q}^{\mathrm{3}} +{q}+\mathrm{1}=\mathrm{0} \\ $$$${p}^{\mathrm{2}} =−\frac{\left({q}+\mathrm{1}\right)^{\mathrm{2}} \left({q}^{\mathrm{2}} −{q}+\mathrm{1}\right)}{{q}^{\mathrm{2}} +{q}+\mathrm{1}} \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{be}\:\mathrm{sure}\:\mathrm{that}\:\left({q}\neq−\mathrm{1}\Rightarrow{p}=\mathrm{0}\right) \\ $$$$\Rightarrow\:{p}^{\mathrm{2}} <\mathrm{0}\:\Rightarrow\:{p}=\pm\sqrt{\frac{{q}^{\mathrm{2}} −{q}+\mathrm{1}}{{q}^{\mathrm{2}} +{q}+\mathrm{1}}}\left({q}+\mathrm{1}\right)\mathrm{i} \\ $$$$...\mathrm{I}'\mathrm{ll}\:\mathrm{continue}\:\mathrm{later} \\ $$
Question Number 176716 Answers: 1 Comments: 0
Question Number 176710 Answers: 3 Comments: 0
Question Number 176708 Answers: 1 Comments: 0
$$\:{a}_{{n}+\mathrm{2}} −\mathrm{3}{a}_{{n}+\mathrm{1}} +\mathrm{2}{a}_{{n}} ={n}+\mathrm{3}^{{n}} \: \\ $$$${please}\:{find}\:{a}_{{n}} \\ $$$$ \\ $$
Question Number 176692 Answers: 1 Comments: 1
Question Number 176684 Answers: 1 Comments: 2
Question Number 176679 Answers: 2 Comments: 0
$$ \\ $$$$\:\:\:\:\:\:\:\:{Eeasy}\:\:{integral}.... \\ $$$$\:\:\:\:\:\:\boldsymbol{\Omega}\:=\:\int_{−\int_{\mathrm{0}} ^{\:\infty} {e}^{\:−{x}^{\:\mathrm{2}} } {dx}} ^{\:\int_{\mathrm{0}} ^{\:\infty} {e}^{\:−{x}^{\:\mathrm{2}} } {dx}} {sin}^{\:\mathrm{2}} \left({t}\right).{ln}^{\:\mathrm{3}} \left(\:{t}\:+\:\sqrt{\mathrm{1}+{t}^{\:\mathrm{2}} }\right){dt}\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:−−−{m}.{n}−−− \\ $$
Question Number 176676 Answers: 3 Comments: 0
$$\mathrm{If}\:\:\mathrm{x}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} }=\mathrm{1}, \\ $$$$\mathrm{prove}\:\mathrm{that} \\ $$$$\mathrm{x}^{\mathrm{5}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{5}} }=−\left(\mathrm{x}^{\mathrm{4}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{4}} }\right) \\ $$
Question Number 176672 Answers: 1 Comments: 0
Pg 320 Pg 321 Pg 322 Pg 323 Pg 324 Pg 325 Pg 326 Pg 327 Pg 328 Pg 329
Terms of Service
Privacy Policy
Contact: info@tinkutara.com