# en/3B5gglzOKC3q.xml.gz
# oc/3B5gglzOKC3q.xml.gz

(src)="1"> Denis is vacationing in China and he wants to spend 30 dollars on a new sweater .
(trg)="1"> Denis pren de vacanças en China e vòl despensar 30 $ per un tricòt novèl .
(src)="2"> The sweater he likes costs 197 Chinese yuan .
(trg)="2"> Lo tricòt que li agrada còsta 197 yuan chineses .
(src)="3"> One US dollar can be converted into 6 Chinese Juan .
(trg)="3"> Un dolar american se pòt convertir contra 6 yuan chineses .
(src)="4"> Denis will have blank Chinese yuan if he converts his 30 US dollars .
(trg)="4"> Denis aurà _______ yuan chineses se convertís sos 30 dolars americans .
(src)="5"> Let 's think about this .
(trg)="5"> Sosquem .
(src)="6"> He 's going to take 30 dollars and the conversion rate is 6 yuan per dollar so he 's going to have 30 dollars times 6 yuan per dollar , he 's going to have 30 times 6 yuan .
(trg)="6"> Va prene 30 $ e lo taus de conversion , lo taus de conversion es 6 yuan per dolar .
(trg)="7"> Va doncas aver 30 dolars còps 6, 1 per dolar , 30 còps 6 yuan .
(src)="7"> And 30 x 6 , well that 's the same thing as 3 x 6 x 10 or 180 .
(trg)="8"> 30 x 6 , es coma 3 x 6 x 10 o 180 .
(src)="8"> So he 's going to have 180 Chinese yuan .
(trg)="9"> Va doncas aver 180 yuan chineses .
(src)="9"> Now does he have enough money to buy the sweater ?
(trg)="10"> Ara , a pro argent per se crompar lo tricòt ?
(src)="10"> Well the sweater costs 197 yuan .
(src)="11"> So no , he does not have enough money to buy the sweater .
(trg)="11"> Lo tricòt còsta 197 yuan , doncas non , a pas pro argent per se crompar lo tricòt .

# en/4ard0nYxZkur.xml.gz
# oc/4ard0nYxZkur.xml.gz

(src)="1"> Use a number line to compare 11 . 5 and 11 . 7 .
(trg)="1"> Utiliza una benda numerica per comparar 11, 5 e 11, 7 .
(src)="2"> So let 's draw a number line here .
(trg)="2"> Dessenhem una benda numerica aicí .
(src)="3"> And I am going to focus between 11 and 12 because that 's where our two numbers are sitting .
(trg)="3"> Me vau concentrar sus la partida entre 11 e 12 , es ailà que se tròban nòstres dos nombres .
(src)="4"> They are 11 and then something else , some number of tenths .
(trg)="4"> Son 11 e quicòm mai , un cert nombre de disièmas .
(src)="5"> So this right here is 11 .
(trg)="5"> Aquí avèm 11 .
(src)="6"> This right here would be 12 .
(trg)="6"> Aquí avèm 12 .
(src)="7"> And then let me draw the tenths .
(trg)="7"> Ara , dessenhi los disièmas .
(src)="8"> So this would be , smack- dab in between , so that would be eleven and five tenths , or that would be 11 . 5 .
(trg)="8"> Aquí , exactament al mièg , doncas onze e cinc disièmas ( o 11, 5 ) seriá aquí .
(src)="9"> Well , I 've already done the first part .
(trg)="9"> Bon , avèm ja fach la mitat .
(src)="10"> I 've figured out where 11 . 5 is .
(src)="11"> It 's smack- dab in between 11 and 12 .
(trg)="10"> Ai trobat ont es 11, 5 : exactament al mièg entre 11 e 12 .
(src)="12"> It 's eleven and five tenths .
(trg)="11"> Es onze e cinc disièmas .
(src)="13"> But let me find everything else .
(src)="14"> Let me mark everything else on this number line .
(trg)="12"> Ara , vau marcar tot sus la benda numerica .
(src)="15"> So that 's 1 tenth , 2 tenths , 3 tenths , 4 tenths , 5 tenths , 6 tenths , 7 tenths , 8 tenths , 9 tenths and then 10 tenths right on the 12 .
(trg)="13"> Aquí , i a 1 disièma , 2 disièmas , 3 disièmas , 4 disièmas , 5 disièmas 6/ 10 , 7/ 10 , 8/ 10 , 9/ 10 e ara 10/ 10 sul 12 .
(src)="16"> It 's not completely drawn to scale .
(src)="17"> I 'm hand- drawing it as good as I can .
(trg)="14"> Es pas dessenhat a l' escala , dessenhi a man levada coma pòdi .
(src)="18"> So where is 11 . 7 going to be ?
(trg)="15"> Ont va èsser 11, 7 ?
(src)="19"> Well this is 11 . 5 .
(trg)="16"> Bon , aquí i a 11, 5 .
(src)="20"> This is 11 . 6 .
(trg)="17"> Aquí es 11, 6 .
(src)="21"> This is 11 . 7 .
(trg)="18"> Aquí es 11, 7 .
(src)="22"> Eleven and seven tenths .
(trg)="19"> Onze e sèt disièmas .
(src)="23"> 1 tenth , 2 tenths , 3 tenths , 4 tenths , 5 tenths , 6 tenths , 7 tenths .
(trg)="20"> 1/ 10 , 2/ 10 , 3/ 10 , 4/ 10 , 5/ 10 , 6/ 10 , 7/ 10 .
(src)="24"> This is 11 . 7 .
(trg)="21"> Aquí es 11, 7 .
(src)="25"> And the way we 've drawn our number line , we are increasing as we go to the right .
(trg)="22"> Sus la benda numerica , los nombres son mai bèls quand anam a drecha .
(src)="26"> 11 . 7 is to the right of 11 . 5 .
(trg)="23"> 11, 7 es a drecha de 11, 5 .
(src)="27"> It 's clearly greater than 11 . 5 .
(trg)="24"> Es clarament mai bèl que 11, 5 .
(src)="28"> 11 . 7 & gt ; 11 . 5
(trg)="25"> 11, 7 & gt ; 11, 5 .
(src)="29"> And really , seriously , you didn 't have to draw a number line to figure that out .
(trg)="26"> Seriosament , aviás pas besonh de dessenhar una benda numerica per t' en rendre compte .
(src)="30"> They 're both 11 and something else .
(trg)="27"> Los dos son 11 e quicòm mai .
(src)="31"> This is 11 and 5 tenths .
(trg)="28"> Es 11 e 5 disièmas .
(src)="32"> This is 11 and 7 tenths .
(trg)="29"> Es 11 e 7 disièmas .
(src)="33"> So , clearly , this one is going to be greater .
(trg)="30"> Doncas , aquel va èsser mai bèl .
(src)="34"> Both have 11 , but this has 7 tenths as opposed to 5 tenths .
(trg)="31"> Los dos an 11 mas aquel a 7 disièmas e el n 'a 5 ( disièmas ) .

# en/66PyFALvfF3R.xml.gz
# oc/66PyFALvfF3R.xml.gz

(src)="1"> We 're asked to select which fractions add together to make 25 over 22 , or 25/ 22 .
(trg)="1"> Devèm causir quinas fraccions podèm addicionar per obténer 25 sus 22 , o 25/ 22 .
(src)="2"> You can use as many fractions as you need .
(trg)="2"> Utiliza tantas fraccions coma vòls .
(src)="3"> Put all unused fractions into the trash can .
(trg)="3"> Bota las fraccions qu' utilizas pas a l' escobilièr .
(src)="4"> So let 's think about how we could do this .
(trg)="4"> Ara , soscam a cossí podèm far aquò .
(src)="5"> So I actually want to use the largest fraction first , just so I can get pretty close .
(trg)="5"> Vòli utilizar primièr la fraccion mai bèla , atal , vau arribar tot prèp .
(src)="6"> So I 'm going to make 16/ 22 .
(trg)="6"> Vau doncas prene 16/ 22 .
(src)="7"> And let 's see , if I add 8/ 22 to that , then that 's going to get me to -- let 's see , 16 plus 8 , that gets me to 24/ 22 .
(trg)="7"> E anam veire , se i ajusti 8/ 22 , me va menar a ... , anam veire 16 + 8 , me mena a 24/ 22 .
(src)="8"> This is 16/ 22 plus 8/ 22 is going to get me 24/ 22 .
(trg)="8"> 16/ 22 + 8/ 22 me va menar a 24/ 22 .
(src)="9"> And if I get one more , that gets me to 25/ 22 .
(trg)="9"> Doncas se preni un de mai ( 1/ 22 de mai ) , me mena a 25/ 22 .
(src)="10"> And then I could put these unused ones down here in the trash can .
(trg)="10"> E pòdi botar las fraccions pas utilizadas a l' escobilièr .
(src)="11"> So I 'm going to put these unused ones down here .
(trg)="11"> Vau doncas botar las fraccions pas utilizadas aquí .
(src)="12"> And let 's actually check our answer .
(trg)="12"> Verificam nòstra responsa .
(src)="13"> Got it right .
(trg)="13"> Es plan !
(src)="14"> And there 's multiple ways that we could have actually done this .
(trg)="14"> I a mantun biais d' aver la bona responsa .
(src)="15"> Actually , it 's not clear that there 's multiple ways that we could have done this .
(trg)="15"> Pas segur que i aja mantun biais d' obténer aquò .
(src)="16"> Let 's see -- is there any other way ?
(trg)="16"> Anam veire , i a un autre biais ?
(src)="17"> Yeah , because even if we did the 2 and the 4 here , we would have to get to -- let 's see , 8 would get us to -- this is 2/ 22 plus 4/ 22 is 6/ 22 , plus 8/ 22 is going to be 14/ 22 , yeah .
(trg)="17"> Òc , perque quitament s' aviam pres lo 2 e lo 4 aquí , arribariam a ... anam veire ... 8 nos menariá a ... aquò 's 2/ 22 + 4/ 22 , fa 6/ 22 + 8/ 22 va far 14/ 22 .
(src)="18"> And then if you put 16/ 22 , there 's going to be too many .
(trg)="18"> E se ajustas 16/ 22 , fa va tròp .
(src)="19"> So actually , the way we did it was the way that you 've got to do it .
(trg)="19"> Doncas çò qu' avèm fach èra lo sol biais per capitar .
(src)="20"> So the 2 and the 4 I 'm going to put in the trash can .
(trg)="20"> Lo 2 e lo 4 , los vau botar dins l' escobilièr .
(src)="21"> And 16 of something plus 1 of something plus 8 of something is going to be 25 of that something .
(trg)="21"> E 16 de quicòm + 1 de quicòm mai + 8 de quicòm va far 25 d' aquel quicòm .
(src)="22"> And in this case , the something that we 're talking about are 22nds .
(trg)="22"> Dins aqueste cas , lo quicòm que sèm a parlar son de / 22 .

# en/7chKngEvF7Zc.xml.gz
# oc/7chKngEvF7Zc.xml.gz

(src)="1"> PROBLEM :
(src)="2"> " Where is 30 on the number line ? "
(trg)="1"> Problèma : ont es 30 sus la benda numerica ?
(src)="3"> So on the number line right over here , we see that as we go from 0 , the first slash isn 't 1 .
(src)="4"> It 's 3 .
(trg)="2"> Sus la benda numerica aquí , vesèm que partissèm de 0 e que la primièra graduacion es pas 0 : es 3 .
(src)="5"> So with each slash we move to the right , the numbers go up by 3 .
(trg)="3"> Doncas cada graduacion nos fa avançar de 3 .
(src)="6"> So let 's see if we can figure out where 30 is .
(trg)="4"> Anam veire si trobam ont es 30 .
(src)="7"> So this slash is marked as 3 .
(trg)="5"> Sus aquesta graduacion , i a 3 .
(src)="8"> So this is going to be 3 more .
(src)="9"> So it 's 6 .
(trg)="6"> Aquò va doncas èsser 3 de mai :
(src)="10"> Then 9 , 12 , 15 , 18 , 21 , 24 , 27 , and 30 .
(trg)="8"> Puèi 9 , 12 , 15 , 18 , 21 , 24 , 27 e 30 .
(src)="11"> Another way you could have thought about this is since each of these marks is 3 , to get to 30 , we have to step through 10 of these marks .
(trg)="9"> Un autre biais de trobar : coma cada graduacion val 3 , nos cal comptar 10 graduacions .
(src)="12"> So we go [ COUNTlNG ]
(trg)="10"> I anam :
(src)="13"> 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 .
(trg)="11"> 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 .
(src)="14"> Let 's do a few more .
(trg)="12"> Ne fasèm d' autres .
(src)="15"> PROBLEM :
(src)="16"> " Where is 24 on the number line ? "
(trg)="13"> Problèma : ont es 24 sus la benda numerica ?
(src)="17"> Well , once again , each of these marks is 3 .
(trg)="14"> Un còp de mai , cada graduacion val 3 .
(src)="18"> So it 's going to be 3 , 6 , 9 , 12 , 15 , 18 , 21 , and 24 .
(trg)="15"> Va èsser , 3 , 6 , 9 , 12 , 15 , 18 , 21 e 24 .
(src)="19"> Let 's do one more .
(trg)="16"> Un autre ...
(src)="20"> So in this problem , each mark is 4 .
(trg)="17"> Dins aqueste problèma , cada graduacion val 4 .
(src)="21"> So we 're going to go 4 , 8, 12 , 16 , and 20 .
(trg)="18"> Va far 4 , 8 , 12 , 16 e 20 .
(src)="22"> Another way you could think about it is that if each step is 4 , you have to go up 5 times to get up to 20 .
(trg)="19"> Un autre biais de capitar es : una graduacion = 4 , ne cal doncas comptar 5 per anar a 20 .
(src)="23"> So this is 1 2 3 , 4 , 5 .
(trg)="20"> 1 , 2 , 3 , 4 , 5 .
(src)="24"> Either way , this should be the right answer .
(trg)="21"> Cada còp , nos mena a la bona responsa .

# en/ARIAKj5VlMW3.xml.gz
# oc/ARIAKj5VlMW3.xml.gz

(src)="1"> Esperanto is a language you can use for everything .
(trg)="1"> L' Esperanto es una lenga adaptada a tot .
(src)="2"> ESPERANTO
(trg)="2"> ESPERANTO
(src)="3"> IS A LANGUAGE
(trg)="3"> ES UNA LENGA
(src)="4"> SUlTABLE FOR
(trg)="4"> ADAPTADA PER
(src)="5"> INTERNATlONAL COMMUNlCATlON
(trg)="5"> LA COMUNICACION INTERNACIONAU
(src)="6"> TRAVEL
(trg)="6"> LOS VIATGES
(src)="7"> INTERNET
(trg)="7"> INTERNÈT
(src)="8"> LANGUAGE FESTlVALS
(trg)="8"> LOS FESTENAUS DE LENGA
(src)="9"> INTERCULTURAL LEARNlNG
(trg)="9"> L' ENSENHAMENT INTER- CULTURAU
(src)="10"> CONCERTS
(trg)="10"> LOS CONCÈRTES
(src)="11"> GAMES
(trg)="11"> JOGAR
(src)="12"> SClENCE
(trg)="12"> LA SCIÉNCIA
(src)="13"> THEATRE
(trg)="13"> LO TEATRE
(src)="14"> MAGAZlNES
(trg)="14"> LAS REVISTAS
(src)="15"> MAKlNG NEW FRlENDS
(trg)="15"> TROBAR AMICS NAVÈTHS
(src)="16"> BOOKS
(trg)="16"> LOS LIBERS