GYanggg

Method 1: 

To find the unit digit of the expression, we need to find a pattern. 

After writing out a few more numbers, we see:

\(1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\cdot10...\cdot2012\cdot2013\)

Since multiplying any integer with 10, the result ends in 0,

the unit digit of this expression must be zero. 

Method 2:

Of course, you could multiply the whole expression and calculate the value of 2013!. 

2013!

= 2842867247059648714320882908184358975298425936765406743563017882090099387783580447012935263746562

335479064550804211732063022834149625421992014144514533688428601587813569194324989561917151782797873

121680666949119475158973400795970620477769770891025873157955892369624843826017526081184552011934374

10509731578204812073608581513295488735471553922625629140543458920383377989900058870151031893904324

786467813499293437953323054452848703591411894569362652829036859143720665327952287750921278821073795

758246971107141102165987055971413647452259971850140909083491892667926811689827067309820345136189931

70855874333269695793623349197860089943442174193730066430857215683894230520597646816671734574453822

639829410734166055686591247872808311907999627073361328419227145279682743490293656700038542243000003

54656476434833237074182399956043794238102817592695696938568409658563414032484092565521956662668178

04130797077932032476566168993055491553005065129612241809221756958758551301616031692650277494508565

460020142819737041175520752950885102631597178638938715655511417824993651493965622682149873599776821

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27228801998127490992546175714650004281060557576018602691444874370276588961454729805801934914327058

639037910641855479898434777131761105855951851430766729127680477612727441963181686830328681323514755

07005255980785550480809320586257442330649968338872970362487582453860064408263458172262252063541389

40472165440197844386043987814485357207839608993158387923679839401916385370678987012450565846378062

37613648468018169610742424883398829055732890553015662215573197318332952206241281889197604097179798

088111804842744903817885667786672979352296770176797979680935970290662288862203487507979401519096847

866678683866374334376409418963307236435361004707445941188594507325969642135133107415655296396624159

611926210964204186787681300922165759386829935051758089290240223443538483111578632645575815706664545

129018761043826893436452374686888469782790455363052163042518632450978515011042593378534376399702189

20273056745879634273623885977170895848007403898018718056610467971822515583288587585365491735918541

851143874245054399927740846012278308508172250269186018235162874294246905205271515797568854333409581

770621535680031307344634119307920989359990306607019552656310949483911595447438683550126438222044423

66574187301773842976392960318817492317270290321216739518314342760193902063057941457483831918266124

856180712708873479072562585163511439010136904713443685275196453218976196831564862256983178474297470

717874357825058000841842254607086820958709964285390681344929611021650748906543867856832645308948315

63132549697850639084675795153902191945156201050737533340297121449992322830373504638815285378819787

554773575857714346271171627666997528270237625481717740484365952887157175985255192895785552274978335

022571528293668119818363728825071916771945692681017826150901126099177461980958354110093788032189674

99188857015469067059756448734088835527244682814249255050856850527306308568919331798552245212702085

62951496568791453972719099564639553188855510265294703433263335170190878675189768260844753417897774

605623321349011902193140551488084053156588942044109445461823324089443588958669455041738856562689822

933876705027721348392092785430536231155624184547493347003183551786284141683397325217471718551661094

77719366226966982601829031348339568014756885616782225353731679589010642963026728648636517643953417

22332329923130642392597499850698224808818858610577949452559361090645680123768915090996283714284645

18499556483198459740683706440681874790291565697629012736855369437572800033629881008632192695545890

165897605607139837425754932756265266177004015134335575298431975102791240204949642576808809116443446

03495723958279131978382862453869345658201667502410202572628757335499674300863432592256152060213732

58766077086853543964535758107538070039372783281779176173268921617552162373199770869845016818789242

170921555048050613527266247691008484193008401103135422070066418155209427253547525169099964052804460

89149387826652320289706871854037767039903259795085325094930026669064944815410134739486230574390677

16354970459715223485500393463339996144504094685302690606100003782316415853025878871684344019606419

775272528889669587663352056426642112160119700550423526808274366819908984680988815165676455989729650

20594215802444175390346271358292514861528156681872687363261748664778410513181823674285103194306067

045099545803147235396638706228707189288952637962203047115037086421526867398070632958542701199245133

382396133624499203582231312010840884212052056354447679961154306001820509943943259757985331301915430

07073726382576912304230842839818060193895673180137969400931470686507837694128076839580668577787352

528579884315928111505284991546171783935459962376675014391494104156202279442380953797629681894096893

415368905476382033883187685926067674083948155038913063463236521164557031729533512019335884749555275

63220129086963377342371823922961841428147215856281545446983868005765000965710748347784855863778139

511179159754686588299301850513114088473086011993200873704826365083841064351095378526835498963203426

65003300891762896539159219941617766093956216205158201836090300940905254000256692488424335980452653

96159843083730318095959670247829067563318797906385452400640000000000000000000000000000000000000000

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000000000000000000000000000000000000000000000000000000000000000000000.

Indeed, the result ends in a zero, a lot of them. 

I hope this helped,

Gavin. 

The first step in order to tackle this problem is to draw another square enclosing square DEFG.

First, we prove that \(WXYZ\) is actually a square: 

\( \because\overline{ED}=\overline{DG}=\overline{GF}=\overline{FE}\because\angle{Z}=\angle{Y}=\angle{YXW}=\angle{ZWX}\because\angle{DEZ}+\angle{WEF}=90º,\angle{DEZ}+\angle{ZDE}=90º\Rightarrow\angle{ZDE}=\angle{WEF}.\)

\(\text{Using the same reasoning, we get:} \angle{ZDE}=\angle{WEF}=\angle{DGY}=\angle{GFX}.\)
\(\therefore\text{By AAS congruency:} \triangle{ZDE}\cong\triangle{YGD}\cong\triangle{XFG}\cong\triangle{WEF}.\)

From this, we get

\(\overline{ZE}+\overline{EW}=\overline{ZD}+\overline{DY}=\overline{YG}+\overline{GX}=\overline{FX}+\overline{FW}, \)which simplifies to \(\overline{ZW}=\overline{ZY}=\overline{YX}=\overline{XW}.\)

Therefore \(WXYZ \) is a square.

Since \(\triangle ABC\) is equilateral,\( \angle B=60º. \because \triangle BEW\) is a 30-60-90 triangle, \(\frac{\overline{EW}}{\overline{BW}}=\sqrt3.  \text{Same goes with } \triangle GXC, \frac{\overline{GX}}{\overline{XC}}=\sqrt3.\)
\(\text{If }\overline{EW}=x \text{ and } \overline{GX}=y,\text{ we get }\overline{BW}=\frac{x}{\sqrt3} \text{ and } \overline{XC}=\frac{y}{\sqrt3}.\)

If the equilateral triangle's side length is \(a, a=\overline{BW}+\overline{WF}+\overline{FX}+\overline{XC}=\frac{x}{\sqrt3}+y+x+\frac{y}{\sqrt3}.\)

After simplifying, we get \(x+y=\frac{3-\sqrt3}{2}a.\)

\(\because \overline{WX}=x+y \therefore x+y=\overline{DH}.\)

Since in any case, \(\overline{DH}=\frac{3-\sqrt3}{2}a, \) the length remains consistent.

\(\text{We plug in }-4\text{ for }x\text{ in the function }t(x) = |{-3+2x}|.\\ t(-4) = |{-3+2\cdot(-4)}|=11.\\ \text{Then, we do the same for }11.\\ t(11) = |{-3+2\cdot(11)}|=\boxed{19}.\)

I hope this helped,

Gavin

\(\text{For any three real numbers a, b, and c, with } b \neq c,\\\text{the operation P is defined by } P(a, b, c) = \frac{a}{b-c}.\\ \text{What is } P\left(P(1,2,3), P(2,3,1), P(3,1,2)\right)?\\ P(1,2,3)=\frac1{2-3}=-1\\ P(2,3,1)=\frac2{3-1}=1\\ P(3,1,2)=\frac3{1-2}=-3\\ P(-1,1,-3)=\frac{-1}{1-(-3)}=-\frac14\\ \text{I hope this helped,}\\ \text{Gavin.}\)

Hey Guest!

\(\text{volume of the cube} - \text{volume of the cylindrical section} = \text{volume of the remaining solid.}  \)

The volume of the cube: \(\text{length} * \text{width} * \text{height} = 4^3 = 64\)

Volume of the cylinder: \(π * \text{radius}^2 * \text{height} = 22π * 4 = 16π\)

\(64-16π=13.73251754\cdots\)

I hope this helped,

Gavin. 

Find all real numbers that satisfy the equation: \(\frac{1}{r^3+7}-7=\frac{-r^3}{r^3+7}\).

\(\frac{1}{r^3+7}-7=\frac{-r^3}{r^3+7}\\ \frac{1}{r^3+7}-\frac{-r^3}{r^3+7}=7\\ \frac{1+r^3}{r^3+7}=7\\ 1+r^3=7r^3+49\\ 6r^3=-48\\ r^3=-8\\ r=\boxed{-2} \)

I hope this helped,

Gavin

Hey guest!

I found my careless mistake, when I determined the probabilty for case 1. 

I fixed it and the answer should be right now!

Hey guest!

We can split this probabilty problem into two cases:

Case 1: Markov gets 3-6 on the die for the first roll and 1-2 on the die for the second roll. 

Case 2: Markov gets 1-2 on the die for the first roll and flips a head on the second turn. 

Then, we find the probabilty of each case and then find the sum. 

Case 1: 

There are six possible rolls on the die. We need to get 3, 4, 5, or 6 on the first roll, and 1 or 2 on the second. 

\(\frac23\cdot\frac13=\frac29\)

Case 2: 

There are six possible rolls on the die and two outcomes for the coin. 

\(\frac13\cdot\frac12=\frac16\)

Then, we find the sum of the cases to reach the desired result. 

\(\frac29+\frac16=\boxed{\frac7{18}}\)

I hope this helped,

Gavin

However, the most irritating of all, are guests who hide behind their anonymity and spew BS that absolutely does not help the growth of the questioner. 

Saying "I hope this helped" is just simply a positive signature of mine, concluding my solution, much better than starting your answer off with "this is easy". I think this, we can both agree on. 

So, Guest, instead of taking your time to write these futile responses, please go learn some math, or how to use the comma, or something else that will contribute to the overall mentality of the user of this website. 

I hope this helped,

Gavin. 

Yes, kevbamboo, I well understand that there are multiple ways of approaching this problem. 

Your way completely skips steps a, b, c, and d. 

If you want to get technical, I can use 4 different methods to calculate the sum, and start my solution off with "this is easy". 

However, the method the questioner requests is obviously the one presented in my solution. 

Steps a, b, c, and d are just parts of the proof and investigation of the method.