The answers is a because the radius must be 5
Answer:
Use multitape Turing machine to simulate doubly infinite one
Explanation:
It is obvious that Turing machine with doubly infinite tape can simulate ordinary TM. For the other direction, note that 2-tape Turing machine is essentially itself a Turing machine with doubly (double) infinite tape. When it reaches the left-hand side end of first tape, it switches to the second one, and vice versa.
You can factor -80 as
So, we have
The square root of a product is the product of the square roots:
Since 
and 
, we have
Hello there,
The "prime factor's" of 1,386 would be "2*3*3*7*11.
When we multiply those together. We get 1,386.
Hope this helps.
~Jurgen
![x^ \frac{m}{n} = \sqrt[n]{x^m}](https://tex.z-dn.net/?f=x%5E%20%5Cfrac%7Bm%7D%7Bn%7D%20%3D%20%5Csqrt%5Bn%5D%7Bx%5Em%7D%20)
![((7x^3)(y^2))^ \frac{2}{7} = \sqrt[7]{((7x^3)(y^2))^2}](https://tex.z-dn.net/?f=%20%28%287x%5E3%29%28y%5E2%29%29%5E%20%5Cfrac%7B2%7D%7B7%7D%20%3D%20%5Csqrt%5B7%5D%7B%28%287x%5E3%29%28y%5E2%29%29%5E2%7D%20)
=![\sqrt[7]{((7x^3)^2(y^2)^2)}](https://tex.z-dn.net/?f=%20%5Csqrt%5B7%5D%7B%28%287x%5E3%29%5E2%28y%5E2%29%5E2%29%7D%20)
=![\sqrt[7]{((7^2x^6)(y^4))}](https://tex.z-dn.net/?f=%20%5Csqrt%5B7%5D%7B%28%287%5E2x%5E6%29%28y%5E4%29%29%7D%20)
=![\sqrt[7]{((49x^6)(y^4))}](https://tex.z-dn.net/?f=%20%5Csqrt%5B7%5D%7B%28%2849x%5E6%29%28y%5E4%29%29%7D%20)
=![\sqrt[7]{49x^6y^4}](https://tex.z-dn.net/?f=%20%5Csqrt%5B7%5D%7B49x%5E6y%5E4%7D%20)
