Answer: 
Step-by-step explanation:
Given: A cubic kilometer=
cubic centimeters
The volume of world’s oceans=
cubic kilometers of water.
⇒ The volume of world’s oceans=
cubic centimeters of water.
Volume of a bucket = 20,000 cubic centimeters of water.
The number of bucket-loads would it take to bucket out the world’s oceans

![\Rightarrow\ n=\frac{1.4\times10^{9+15}}{0.2\times10^5}......[a^n\times a^m=a^{m+n}]\\\Rightarrow\ n=7\times10^{24-5}.....[\frac{a^m}{a^n}=a^{m-n}]\\\Rightyarrow\ n=7\times10^{19}](https://tex.z-dn.net/?f=%5CRightarrow%5C%20n%3D%5Cfrac%7B1.4%5Ctimes10%5E%7B9%2B15%7D%7D%7B0.2%5Ctimes10%5E5%7D......%5Ba%5En%5Ctimes%20a%5Em%3Da%5E%7Bm%2Bn%7D%5D%5C%5C%5CRightarrow%5C%20n%3D7%5Ctimes10%5E%7B24-5%7D.....%5B%5Cfrac%7Ba%5Em%7D%7Ba%5En%7D%3Da%5E%7Bm-n%7D%5D%5C%5C%5CRightyarrow%5C%20n%3D7%5Ctimes10%5E%7B19%7D)
hence,
bucketloads would it take to bucket out the world’s oceans.
Answer:
<em>False</em>
Step-by-step explanation:
took the test
Answer:
20 units
Step-by-step explanation:
s = 8
Perimeter = s + s + 

= 8 + 8 + 4
= 20 units
Answer:
30 m/s
Step-by-step explanation:
<em>p = mv</em>
<em>momentum</em><em> </em><em>=</em><em> </em><em>mass</em><em> </em><em>×</em><em> </em><em>velocity</em>
<em>here</em><em> </em><em>the</em><em> </em><em>momentum</em><em> </em><em>is</em><em> </em><em>60kgm</em><em>/</em><em>s</em>
<em>mass</em><em> </em><em>is</em><em> </em><em>2kg</em>
<em>substitute</em><em> </em><em>in</em><em> </em><em>the</em><em> </em><em>equation</em>
<em>6</em><em>0</em><em> </em><em>=</em><em> </em><em>2v</em>
v= 30m/s
Answer: -1/2, -0.22, 0, 12%, 0.56
Step-by-step explanation:
-1/2 is equal to -0.50, making it the number with the least value.
-0.22 is closer to 0 than -0.50, meaning it is greater then -0.50 and less than 0.
0 is between the negative and positive numbers, giving it the spot that it has.
12% is equivalent to 0.12, meaning it is more than 0, and less than 0.56, which is the greatest number.
0.56 has more value than any other number in the problem, meaning it goes last in the order.