Answer:
No
Reasoning:
If something is a perfect cube, it is able to be put under a cube root (![\sqrt[3]{..}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B..%7D)
) and will result in an integer (a non-decimal number > 0, basically).
So let's calculate ![\sqrt[3]{48}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B48%7D)
, and see if the result is an integer.
= 3.634.......
As you can see, the result is not an integer, therefore 48 is not a perfect cube.
45.019 i believe space filler
Answer:
58 degrees
Step-by-step explanation:
Answer:
1xgr7`0ryu932
Step-by-step explanation:
Answer: 12k + 5
Hope this helps:)
