1/9 on a calculator is 0.11111111
I wonder if you mean to write ![\sqrt[3]{256}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B256%7D)
in place of 
...
If you meant what you wrote, then we have
If you meant to write (the cube root of 256), then we could go on to have
![\sqrt[3]{256}=\sqrt[3]{16^2}=\sqrt[3]{(4^2)^2}=\sqrt[3]{4^4}=\sqrt[3]{4^3\cdot4}=4\sqrt[3]4](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B256%7D%3D%5Csqrt%5B3%5D%7B16%5E2%7D%3D%5Csqrt%5B3%5D%7B%284%5E2%29%5E2%7D%3D%5Csqrt%5B3%5D%7B4%5E4%7D%3D%5Csqrt%5B3%5D%7B4%5E3%5Ccdot4%7D%3D4%5Csqrt%5B3%5D4)
Answer:
514 days
Step-by-step explanation:
We use the formula (r+n-1)/r(n-1)
There are 1540 combinations. Then, we'll divide them by three because were not just finding the amount of combinations, but the time it'd take to play them all.
We get 513.333(cont.)
thats not really possible, so we're gonna go with 514 instead, rounding up. Normally we'd round down, because the number behind the decimal is below 5, but we really want to make sure we get every combination in there.
Hope this is helpful!
Step-by-step explanation:
f(x) = x² + 1, where x >= 0.
Let y be f(x).
=> y = x² + 1
=> x² = y - 1
=> x = √(y - 1) because x is non-negative.
Hence f^-1(x) = √(x - 1), where x >= 1.
