Computer Systems  In your implementation of the ALU there is no need to detect or handle overflow.  However, it's a good mental exercise to think about when/how an overflow might occur.  When adding numbers with different signs, overflow cannot occur because the sum must be no larger than one of the numbers. What happens when the signs of the numbers are the same? What happens in the case of subtraction? How do we detect when an overflow does occur?

Suggest one further small extension or improvement of your own to the modified flashcard program. Outline what the extension does and briefly say what additional sub-problem(s) would need to be added to the initial decomposition. Note that you are only required to describe your further extension and do not need to implement it in code. The maximum word count is 150 words, including the added sub-problem(s).

Thank you, but I'm just not able to follow what is being done. It looks to me like the first minterm was left out of the initial step with no explanation as to why, and then it says MXY' + M'XY + MXY = (MX + M' + M)XY', but if I multiply out (MX + M' + M)XY', I get MXY'+M'XY'+MXY'. The M'XY' is not in the left side; so I don't see how this works. Then, later on, a complement law is mentioned, which does not exist in the chart, and there is no law/identity in the chart that matches going from XY' to X'+Y'. I apologize for being dumb here, but I'm just not following how you're getting from one expression to another. Could you please dumb it down even more for me. Thank you.

In MIPS assembly, implement integer division using rounding (rather than truncation). This is accomplished by taking the remainder of the division and dividing it by the original divisor. If the new quotient is higher than or equal to one, multiply it by one.Otherwise, leave the original quotient alone.In MIPS assembly, implement integer division using rounding (rather than truncation). This is accomplished by taking the remainder of the division and dividing it by the original divisor. If the new quotient is higher than or equal to one, multiply it by one.Otherwise, leave the original quotient alone.In MIPS assembly, implement integer division using rounding (rather than truncation). This is accomplished by taking the remainder of the division and dividing it by the original divisor. If the new quotient is higher than or equal to one, multiply it by one.Otherwise, leave the original quotient alone.In MIPS assembly, implement integer division using rounding (rather than truncation).…

I attempted to do MOST of my HW bymyself, but I need IMMEDIATE help doin these last 3 questions.....

Outline Hobb's algorithm and centering algorithms as they relate to the phenomenon you're referencing.

Not regular division... it's binary.