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The Official Homework Help Thread

I need clarification... If I have 4x^2 and x is given as, say 3, which do I multiply first??

Remember the order of operations, PEMDAS
Parentheses (none), then exponents. So you would perform x^2 first and then multiply what you get from that to 4. Hope this helps :)
 
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There was no other information... What I wrote was the question

In that case the general simplification is |3/(ab)| ^^.

I think that some confusion might have arisen from the terminology used though in the problem. The problem you posted seems to be a problem of "simplifying" an expression as opposed to "solving" an equation (an equation has an equality sign). The terminology is accidentally interchanged all the time xP.
 
I got a bit confused when doing this in class and I have a midterm tomorrow. I doubt there is anyone here that can help me but it's all unique number system work. In this case we assume we're working in a 5 bit register:

"Use 2?s complement arithmetic to make the calculation 2610[SUB]10[/SUB] − 1710[SUB]10[/SUB] using a 5 bit
register."

Essentially, I believe a 5-bit binary register would consist of:

00000
to
11111

but the 5th digit place is a placeholder for even and odd numbers. 0 is positive and 1 is negative.

00101 would be equal to 5
and to find the negative counter part you find the first 1 from the right, skip it, and flip every digit past it, apparently.

11011 would be equal to -5 if I am doing it correctly.

Help
 
You're doing it correctly from what I can see, granted this isn't my specialty by any means!

Personally I think it's easier to understand if you assume you are working with N-1 digits in the register with a single bit at the start that controls the sign. For positive numbers you read it just like a four bit binary with "0" at the start. For negative numbers you read the first bit as "1" which means you subtract the next four bits from "10000" and then turn the result into a negative number.

That means that the complement of a number is always a flip of all bits + 1, so the new number would be a flip of all bits to the left of the "1" farthest to the right; which makes sense because subtracting a number from the same number + 1 in binary will always result in flips for all bits and then add 1 back, again flipping the bits from right to left until you reach the bit where the first "1" was originally because there is now a "0" in that bit to hold the "1" carried from the bit to the right.

For your example, 11011 is the complement of 00101 in a 5-bit register so it looks like you understand the concept! Hopefully this was somewhat helpful. :)
 
You're doing it correctly from what I can see, granted this isn't my specialty by any means!

Personally I think it's easier to understand if you assume you are working with N-1 digits in the register with a single bit at the start that controls the sign. For positive numbers you read it just like a four bit binary with "0" at the start. For negative numbers you read the first bit as "1" which means you subtract the next four bits from "10000" and then turn the result into a negative number.

That means that the complement of a number is always a flip of all bits + 1, so the new number would be a flip of all bits to the left of the "1" farthest to the right; which makes sense because subtracting a number from the same number + 1 in binary will always result in flips for all bits and then add 1 back, again flipping the bits from right to left until you reach the bit where the first "1" was originally because there is now a "0" in that bit to hold the "1" carried from the bit to the right.

For your example, 11011 is the complement of 00101 in a 5-bit register so it looks like you understand the concept! Hopefully this was somewhat helpful. :)

Thanks, this somewhat helps. How is it possible to write 27 in base 10 in a 5-bit register? It's confusing me?
I know the highest digit for any number register is 2^n - 1 where n is bit register # - 1.

PS: Are you in math/CS?
 
Thanks, this somewhat helps. How is it possible to write 27 in base 10 in a 5-bit register? It's confusing me?
I know the highest digit for any number register is 2^n - 1 where n is bit register # - 1.

PS: Are you in math/CS?

I'm actually not sure either. I was under the impression that 15 is the largest number you can represent in a single 5 bit register when using two's complement. Perhaps they intend you use more than one register?

It also might be possible that I just don't how to make large numbers on small bit registers since this definitely isn't my area of expertise. And no, I didn't major in CS but have been known to slang numbers on occasion. :p
 
wat is a ^3

third root 90 they mean

- - - Post Merge - - -

I'm actually not sure either. I was under the impression that 15 is the largest number you can represent in a single 5 bit register when using two's complement. Perhaps they intend you use more than one register?

It also might be possible that I just don't how to make large numbers on small bit registers since this definitely isn't my area of expertise. And no, I didn't major in CS but have been known to slang numbers on occasion. :p

I just got the answers online - it was a trick question haha
 
estimate ^3sqrt90 and i need an explanation on how to solve it

If ^3sqrt(90) is meant to be the third root of 90 like Trundle said, then if you want only a rough approximation you will first need to consider cubes of some numbers. 1^3 = 1, 2^3 = 8, 3^3 = 27, 4^3 = 64, 5^3 = 125, ...

Notice that 4^3 < 90 < 5^3. If you "third root" each side of the inequality you get that 4 < (90)^(1/3) < 5. So a rough approximation of the third root of 90 is any number between 4 and 5, perhaps 4.5 (The actual answer is about 4.48...).

Of course if you meant to say (sqrt(90))^(1/3), (90)^(3/2), etc... then the process follows similarly ^^.
 
What's a good way to remember the endings of masc plurals, fem plurals, etc. in French?
 
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