Answer:
I believe its C
Explanation:
I'm sorry if its wrong...
Answer: The FP-s are stored as sign (1.)111111111111 - number of bits.
Also 24 bits resolution there can be 23 zeros
Explanation:
The –1·2²⁴ might be stored as -1 · 1.00000000000000000000000(1 ←the 2⁰), it depends on how the FFP “engine” manages this, it may also be code specific a n+1–n does return 1 but 1–n+n does not. you should carry out a test for a specific compiler/computer
This is what javascript does with double (k+=1; n-=1) e.g
the (k+=2; n-=2) should be used to pass ±0x20000000000000
25! 27+24=51 if you divide 51 by two, you get 25.5
Because we can’t have a 0.5th of a child, you can round down to the nearest whole number. Leaving you with 25. Hope this helped :)
(modulus division works well for these kinds of questions)
In computer science, a 2–3 tree is a tree data structure, where every node with children (internal node) has either two children (2-node) and one data element or three children (3-nodes) and two data elements. According to Knuth, "a B-tree of order 3 is a 2-3 tree."
Answer:
Explanation:
This is a project I already submitted, but I never received feedback. All my upcoming assignments will be based on this, so I wanted to make sure it is correct. I got this program to work as far as calculating the information, but I was hoping someone could please let me know if it meets the parameters/requirements?
I did attempt it. It works and adds the items exactly how the example showed in the video. However, I wanted to make sure my code is solid, and not just a mishmash or working because I got lucky, and if there is a better more solid way to do this I want to make sure I can. I do want feedback so I can learn and get better and I am trying my hardest. I always right out all the code I see here and try it and learn what it does piece by piece, because it helps me learn and improve. I also want to make sure it is going to work once I start the code for the next half of the requirements.
i cant really help they dont let us put codes on here
