Expanding the limit, we get (x^2+2x∆x+∆x^2-2x-2∆x+1-x^2+2x-1)/<span>∆x
Crossing the 1s , the 2xs, and the x^2s out, we get
(2x</span>∆x+∆x^2-2∆x)/<span>∆x
Dividing the </span><span>∆x, we get
2x+</span><span>∆x-2.
Making the limit of </span><span>∆x=0, we get 2x-2.</span>
Answer:
1/5
Step-by-step explanation:
1 pack of paper/ 5 groups
or all three groups
3/5
50 tens
5000 thousands
50000 ten thousand
50 tens
5
Hope this helps
Answer:
Step-by-step explanation:
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.
