# en/0g613yeWAELN.xml.gz
# sco/0g613yeWAELN.xml.gz

(src)="1"> We need to calculate 9 . 005 minus 3 . 6 , or we could view it as 9 and 5 thousandths minus 3 and 6 tenths .
(trg)="1"> We need tae calculate 9 . 005 minus 3 . 6 , or we coud seeit aes 9 n 5 thoosants minus 3 n 6 tents .
(src)="2"> Whenever you do a subtracting decimals problem , the most important thing , and this is true when you 're adding decimals as well , is you have to line up the decimals .
(trg)="2"> Whaniver ye dae ae subtractin deceemals proablem , the maist important thing , n this is true whan ye 'r eikin deceemals n aw , is that ye hae tae line the deceemals up .
(src)="3"> So this is 9 . 005 minus 3 . 6 .
(trg)="3"> Sae 9 . 005 minus 3 . 6 .
(src)="4"> So we 've lined up the decimals , and now we 're ready to subtract .
(trg)="4"> Sae we 'v lined the decemals up , n nou we 'r readie tae subtract .
(trg)="5"> Nou we can subtract .
(src)="5"> Now we can subtract .
(trg)="6"> Sae we stert up here .
(src)="6"> So we start up here .
(src)="7"> We have 5 minus nothing .
(trg)="7"> We hae 5 minus nawthing .
(src)="8"> You can imagine this 3 . 6 , or this 3 and 6 tenths .
(trg)="8"> Ye coud imagen 3 . 6 , or 3 n 6 tents .
(src)="9"> We could add two zeroes right here , and it would be the same thing as 3 and 600 thousandths , which is the same thing as 6 tenths .
(trg)="9"> We coud eik twa zeros richt here , n it wid be the same thing aes 3 n 600 thoosants , the same aes 6 tents .
(src)="10"> And when you look at it that way , you 'd say , OK , 5 minus 0 is nothing , and you just write a 5 right there .
(trg)="10"> N whan ye luik at it that waa , ye 'd say , " O . K . , 5 minus 0 is nawthing , n ye juist sceeve ae 5 here " .
(src)="11"> Or you could have said , if there 's nothing there , it would have been 5 minus nothing is 5 .
(trg)="11"> Or ye coud 'v said , gif thaur 's nawthin there ,
(trg)="12"> It woud hae been 5 minus nawthing is 5 .
(src)="12"> Then you have 0 minus 0 , which is just 0 .
(trg)="13"> Than ye hae 0 minus 0 , n that 's 0 .
(src)="13"> And then you have a 0 minus 6 .
(trg)="14"> N than ye hae ae 0 minus 6 .
(src)="14"> And you can 't subtract 6 from 0 .
(trg)="15"> N ye canna sutract 6 fae 0 .
(src)="15"> So we need to get something into this space right here , and what we essentially are going to do is regroup .
(trg)="16"> Sae we need tae get sommit intae this space here , n whit we 'r baseeclie gaun tae dae is tae regroop .
(src)="16"> We 're going to take one 1 from the 9 , so let 's do that .
(trg)="17"> We 'r gaun tae tak ae 1 fae the 9 , sae lat 's dae that .
(src)="17"> So let 's take one 1 from the 9 , so it becomes an 8 .
(trg)="18"> Sae lats tak ae 1 fae the 9 , sae it becomes aen 8 .
(src)="18"> And we need to do something with that one 1 .
(trg)="19"> N we need tae dae sommit wi that 1 .
(src)="19"> We 're going to put it in the tenths place .
(trg)="20"> We 'r gaun tae put it in the tents steid .
(src)="20"> Now remember , one whole is equal to 10 tenths .
(trg)="21"> Mynd ye , yin hale is the sam aes 10 tents .
(src)="21"> This is the tenths place .
(trg)="22"> This is the tents steid .
(src)="22"> So then this will become 10 .
(trg)="23"> Sae than this wil become 10 .
(src)="23"> Sometimes it 's taught that you 're borrowing the 1 , but you 're really taking it , and you 're actually taking 10 from the place to your left .
(trg)="24"> Somtimes it 's said that ye 'r borroin the 1 , but ye 'r realie takin it , n ye 'r realie takin 10 fae the steid oan ye 'r cair .
(src)="24"> So one whole is 10 tenths , we 're in the tenths place .
(trg)="25"> Sae yin hale is 10 tents , we 'r in the tents steid .
(src)="25"> So you have 10 minus 6 .
(trg)="26"> Sae ye hae 10 minus 6 .
(src)="26"> Let me switch colors .
(trg)="27"> Lat me switch colours .
(src)="27"> 10 minus 6 is 4 .
(trg)="28"> 10 minus 6 is 4 .
(src)="28"> You have your decimal right there , and then you have 8 minus 3 is 5 .
(trg)="29"> Ye hae ye 'r deceemal richt there , n than ye hae 8 minus 3 is 5 .
(src)="29"> So 9 . 005 minus 3 . 6 is 5 . 405 .
(trg)="30"> Sae 9 . 005 minus 3 . 6 is 5 . 405 .

# en/7O4zTUHeOK8w.xml.gz
# sco/7O4zTUHeOK8w.xml.gz

(src)="1"> on Saturday william 's parents gave birth to twins and names them nadia and vanessa when they were first born .
(trg)="1"> On Satuday , Williams paurents gave birth tae twins n named thaim Nadia n Vanessa .
(trg)="2"> Whan thay were first born ,
(src)="2"> Nadia weighed 7 . 27 pounds and was 21 . 5 inches tall .
(trg)="3"> Nadia weiched 7 . 27 poonds n wis 21 . 5 inches taw , n Vanessa weiched 8 . 34 poonds .
(src)="3"> And Vanessa weighed 8 . 34 pounds . how much did the babies weigh in total so they tell us Nadia weighed 7 . 27 and Vanessa weighed 8 . 34 we have to add these up and really they just gave us this length of Nadia at birth is really a distraction so see that we don 't mindlessly add any numbers we see so this is really unnecessary information just to distract us so then we need to add Nadia 's birth weight plus Vanessa 's so it is 7 . 27 plus 8 . 34 and it is always important that we line up the decimal i like to do the decimal . so it is 8 . 34 and well 'I just add these these two up . so 7 plus 4 . and this is really 7 hundredths . plus 4 hundredths is 11 hundredths witch is the name thing as 1 hundredths and 1 tenth .
(trg)="4"> Whit did the bairns weich aw up ?
(trg)="5"> Sae thay tell us that Nadia weiched 7 . 27 , n Vanessa weiched 8 . 34 , we hae tae eik thir up , n realie , thay juist gave us Nadia 's langth at birth aes ae distraction ,
(trg)="6"> Sae mynd that we dinna myndlesslie eik onie nummers that we see .
(src)="4"> 1 tenth plus 2 tentsh plus 3 tenths is 6 tenths . we got our decimal sign right over there . and then 7 plus eight is fifteen . or you could even say it is 5 ones and one ten .
(trg)="11"> N this is the sam thing aes 1 hunnerts n 1 tent .
(trg)="12"> 1 tent plus 2 tents plus 3 tents is 6 tents .
(trg)="13"> We hae oor deceemal sign richt here , n than 7 plus 8 is 15 .
(src)="5"> and we 're done they weighed 15 . 61 pounds in total
(trg)="14"> Or ye coud it 's 5 yins n the ae ten .
(trg)="15"> N we 'r duin , thay weiched 15 . 61 poonds aw up .

# en/EQdIMg7Tquya.xml.gz
# sco/EQdIMg7Tquya.xml.gz

(src)="1"> May the strongest bot win !
(trg)="1"> Mey the strangest bot win !
(src)="2"> The match is over !
(trg)="2"> The maitch is ower !

# en/IpFzKHbQjcy5.xml.gz
# sco/IpFzKHbQjcy5.xml.gz

# en/eBjajVzw24wm.xml.gz
# sco/eBjajVzw24wm.xml.gz

(src)="1"> In this video I want to do a bunch of example problems that show up on standardized exams and definitely will help you with our divisibility module because it 's asking questions like this
(trg)="1"> In this video Ah want tae dae ae heap o exaumple proablems
(trg)="2"> That shaw up oan staunnardised exams ,
(trg)="3"> N will deefinitlie help ye wi oor diveeabeelitie module ,
(src)="2"> All numbers , and this is just one of the examples ,
(trg)="5"> Aw nummers , n this is but aen exaumple ,
(src)="3"> All numbers divisible by both 12 and 20 are also divisible by and the trick here is to realize that if a number is both divisible by 12 and 20 it has to be divisible by each of these guy 's prime factors
(trg)="6"> Aw nummers diveesable bi baith 12 n 20 ar dvieesable bi
(trg)="7"> N the nack here is tae see that gif ae nummer is diveesable bi baith 12 n 20
(trg)="8"> Than it haes tae be diveesable bi the prime facters baith thir nummers .
(src)="4"> So let 's take their prime factorization .
(trg)="9"> Sae lat 's tak thair prime facterisation .
(src)="5"> The prime factorization of 12 is 2 time 6 6 isn 't prime yet , so 6 is 2 times 3 ,
(trg)="10"> The prime facterisation o 12 is 2 times 6 , 6 is no ae prime , sae 6 is 2 times 3 ,
(src)="6"> So that is prime so any number divisible by 12 needs to be divisible by 2 times 2 times 3 .
(trg)="11"> Sae that 's prime .
(trg)="12"> Sae onie nummer diveesable bi 12 needs tae be diveesable bi 2 times 2 times 3 .
(src)="7"> So it 's prime factorization needs to have a 2 times a 2 times a 3 in it any number that 's divisible by 12
(trg)="13"> Sae it 's prime facterisation needs tae hae ae 2 times ae 2 times ae 3 in it .
(trg)="14"> Onie nummer that 's diveesable bi 12 .
(src)="8"> Now , any number that 's divisible by 20 , needs to be divisible by
(trg)="15"> Nou , onie nummer diveesable bi 20 , needs tae be diveesable bi ,
(src)="9"> Let 's take it 's prime factorization 2 times 10 , 10 is 2 times 5 so any number divisible by 20 , needs to also be divisible by 2 times 2 times 5 or another way of thinking about it , it needs to have two 2 's , and a 5 in it 's prime factorization
(trg)="16"> Lat 's tak it 's prime facterisation , 2 times 10 , n 10 is 2 times 5 .
(trg)="17"> Sae onie nummer that 's diveesable bi 20 , needs tae be diveesable bi 2 times 2 times 5 .
(trg)="18"> Or anither waa tae think o it ,
(src)="10"> Now if you 're divisible by both , you need to have two 2 's , a 3 , and a 5 . two 2 's and a 3 for 12 , and then two 2 's and a 5 for 20 and you can verify this for yourself if this is divisible by both
(trg)="20"> Nou , gif yer diveesable bi baith , than ye need tae hae twa 2´s , ae 3 , n ae 5 .
(trg)="21"> Twa 2´s n ae 3 fer 12 , n than twa 2´s n ae 5 fer 20 .
(trg)="22"> N ye can conferm this fer yersel , gif this diveesable bi baith ,
(src)="11"> Obviously , if you divide it by 20 , is the same thing as dividing it by 2 times 2 times 5
(trg)="23"> Obviooslie , gif ye divide bi 20 , it 's the sam aes dividin bi 2 times 2 times 5 .
(src)="12"> So you 're going to have , the 2 's are going to cancel out , the 5 's are going to cancel out your just going to have a 3 leftover , so it 's clearly divisible by 20 and if you were to divide it by 12 , you 'd divide it by 2 times 2 times 3 this is the same thing as 12 and so these guys would cancel out , and you would just have a 5 leftover so it 's clearly divisible by both , and this number right here is 60 it 's 4 times 3 , which is 12 , times 5 .
(trg)="24"> Sae ye 'r gaun tae hae ,
(trg)="25"> The 2´s will cancel oot , n the 5´s will cancel oot .
(trg)="26"> Ye 'r juist gaun tae hae ae 3 leftower , sae it 's clearlie diveesable bi 20 .
(src)="13"> It 's 60
(trg)="31"> It 's 4 times 3 , this is 12 , times 5 is 60 .
(src)="14"> This right here is actually the least common multiple of 12 and 20
(trg)="32"> This here is actualie the least common multiple o 12 n 20 ,
(trg)="33"> This isna the yinlie nummer that 's diveesable bi 12 n 20 ,
(trg)="34"> Ye coud multiplie this nummer bi ae heap o ither facters ,