Answer:
2, 2, 3, 5, 7 that is the farthest you can go
Step-by-step explanation:
2,2,3,5,7 prime factorization of 420 but impossible to get 9 factors
Lawrence's is the correct answer. he put 3 plates on each box in both situations
Answer:
![6 \sqrt[3]{5}](https://tex.z-dn.net/?f=6%20%5Csqrt%5B3%5D%7B5%7D)
Step-by-step explanation:
For the problem,
, use rules for simplifying cube roots. Under the operations of multiplication and division, if the roots have the same index (here it is 3) you can combine them.
![\sqrt[3]{24} *\sqrt[3]{45} = \sqrt[3]{24*45}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B24%7D%20%2A%5Csqrt%5B3%5D%7B45%7D%20%3D%20%5Csqrt%5B3%5D%7B24%2A45%7D)
You can multiply it out completely, however to simplify after you'll need to pull out perfect cubes. Factor 24 and 45 into any perfect cube factors which multiply to each number. If none are there, then prime factors will do. You can group factors together such as 3*3*3 which is 27 and a perfect cube.
![\sqrt[3]{24*45} =\sqrt[3]{3*8*5*3*3} = 6 \sqrt[3]{5}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B24%2A45%7D%20%3D%5Csqrt%5B3%5D%7B3%2A8%2A5%2A3%2A3%7D%20%20%3D%206%20%5Csqrt%5B3%5D%7B5%7D)
Answer: 5/8
Step-by-step explanation: