Week 6 - Positional Notation: Expanding decimals

Question 1

Hindu-Arabic positional numeration: Decimals expanded into standard form

2.4
36.02
4.571

Answer 1

2.4 = 2 ones + 4 tenths
36.02 = 3 tens + 6 ones + 0 tenths + 2 hundredths
4.571 = 4 ones + 5 tenths + 7 hundredths + 1 thousandth

Question 2

H-A positional numeration: Decimals expanded into standard forms as powers of 10
(using table)

2.4
36.02
4.571

Answer 2

2.4 = 2 ones + 4 tenths (expanded)
= (2x10^0) + (4x10^-1) (expanded into powers of 10)
= 2 + 4/10
= 2 4/10

36.02 = 3 tens + 6 ones + 0 tenths + 2 hundredths
= (3x10^1) + (6x10^0) + (0x10^-1) + (2x10^-2)
= 30 + 6 + 0 + 2/100
= 36 2/100

4.571 = 4 ones + 5 tenths + 7 hundredths + 1 thousandth
= (4x10^0) + (5x10^-1) + (7x10^-2) + (1x10^-3)
= 4 + 5/10 + 7/100 + 1/1000
= 4 571/1000

Question 3

H-A positional numeration: Given non-standard form, write the decimals

20 thousandths + 4 tens

50 tenths + 17 hundredths

Answer 3

20 thousandths + 4 tens
= 40.020

50 tenths + 17 hundredths
= 5.17
(10 tenths = 1 one, so 50 tenths = 5 ones)
(10 hundredths = 1 tenth, so 17th hundredths would = 1 tenth and 7 hundredths