1.
a) metres to centimetres :
multiply length by 100
b) metres to millimetres:
multiply length by 1000
c) kilograms to grams:
multiply the mass value by 1000
d) litres to millilitres :
multiply volume by 1000
2.
a) 3 m = 3× 100 = 300 cm
b) 28 cm = 28 × 10 = 280 mm
c) 2.4 km = 2.4 × 1000
= 24 × 10^-1 × 10^3
= 24 × 10^2 =2400 m
d) 485 mm =485 / 10
= 485 / 10 ^1
= 485 × 10 ^-1
= 48.5 cm
e) 35 cm = 35 / 100
= 35 /10^2
= 35 × 10 ^ -2
= 0.35 m
f) 2.4 m = 2.4 / 1000
= 24 × 10 ^-1 / 10^3
= 24 × 10^-1 × 10 ^-3
= 24 × 10 ^ -4
= 0.0024 km
g) 2495 mm = 2495 /1000
= 2495 /10^ 3
= 2495 × 10 ^-3
=2.495 m
Answer:
Solution and how to convert 8 / 21 into a percentage
0.38 times 100 = 38.1. That's all there is to it!
Step-by-step explanation:
hope this helps
now, there are 12 months in a year, so 18 months is really 18/12 of a year, thus
![~~~~~~ \textit{Simple Interest Earned Amount} \\\\ A=P(1+rt)\qquad \begin{cases} A=\textit{accumulated amount}\dotfill & \$4000\\ P=\textit{original amount deposited}\\ r=rate\to 5\%\to \frac{5}{100}\dotfill &0.05\\ t=years\to \frac{18}{12}\dotfill &\frac{3}{2} \end{cases} \\\\\\ 4000=P[1+(0.05)(\frac{3}{2})]\implies 4000=P(1.075) \\\\\\ \cfrac{4000}{1.075}=P\implies 3720.93\approx P](https://tex.z-dn.net/?f=~~~~~~%20%5Ctextit%7BSimple%20Interest%20Earned%20Amount%7D%20%5C%5C%5C%5C%20A%3DP%281%2Brt%29%5Cqquad%20%5Cbegin%7Bcases%7D%20A%3D%5Ctextit%7Baccumulated%20amount%7D%5Cdotfill%20%26%20%5C%244000%5C%5C%20P%3D%5Ctextit%7Boriginal%20amount%20deposited%7D%5C%5C%20r%3Drate%5Cto%205%5C%25%5Cto%20%5Cfrac%7B5%7D%7B100%7D%5Cdotfill%20%260.05%5C%5C%20t%3Dyears%5Cto%20%5Cfrac%7B18%7D%7B12%7D%5Cdotfill%20%26%5Cfrac%7B3%7D%7B2%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%204000%3DP%5B1%2B%280.05%29%28%5Cfrac%7B3%7D%7B2%7D%29%5D%5Cimplies%204000%3DP%281.075%29%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7B4000%7D%7B1.075%7D%3DP%5Cimplies%203720.93%5Capprox%20P)
"Product" signifies multiplication, and "half" means divide by two (or multiply by 1/2). So half the product of b and c is (bc)/2 or (1/2)(bc).
