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AS Computing @ Aquinas College
Приєднався 2 чер 2014
Learning VB.Net for the AQA AS Computing Course
Brackets and Rearrangement
Boolean algebra simplification. Looking at moving brackets and rearranging terms.
Переглядів: 492
Відео
The Distributive Law
Переглядів 19 тис.9 років тому
Describes the distributive law. Multiplying out brackets and factoring.
The Absorption Law
Переглядів 84 тис.9 років тому
This video describes the absorption law used to simplify Boolean expressions.
Boolean Identities
Переглядів 1,4 тис.9 років тому
This video describes the Boolean identities needed for the AQA Computing AS exam.
1. Number Systems
Переглядів 8310 років тому
I created this video with the UA-cam Video Editor (ua-cam.com/users/editor)
Number Systems
Переглядів 2810 років тому
An introduction to number systems. Covers Base 10 and Base 2.
2. Binary Conversions
Переглядів 7110 років тому
Describes how to convert between binary and denary numbers.
asmr
What about this 48:6(4+4) do I do 48:48 or do I do 8(8)
Fantastic!
Thanks, my teacher poorly explained this law.
man i study in germany they complicate things so much. thanks for the simple and relaxing explanation. cheers mate
thaNK you
Amazing proof for absorption law! I love the idea of reverse distributive property
Thank you very much
The way you teach us 😍brilliant, I wish you still post videos, I can't thank you enough.
04:36 there is a mistake. te result should be =1(B'+A)=B'+A
Thank you for the help. The absorption law is one of those laws where you must just recognize it.
sirrrr you're amaazingggg
so there should be only two operators to apply the absorption law ?
You're very soft spoken
Thank you so much.
A.(A+B)=A.A+A.B=(A+AB)=A(1+B)=A(1)=A
but we can also apply absorption law on ii why?
2 points of absorption law made the whole game easy
Class ended before my teacher got to this law, thank you for the video. Was a great help on my homework!
(x+y)(x+z) = x^2 +xz +xy +yz. How is that equal to x+(yz) ?
X2=x And take x common. 1 + something = 1 Apply them
@@aryamallick5426 so (x+y)(x+z) = x + xz + yx + yz = x + x(z+y) + yz = x(1+z+y) +yz //(1+z+y)=1 since true or anything between true or false is always true =x + yz //I was literally trying to find the answer for like hours , thanks for helping me out
bro forgot the video is about boolean algebra
Thanks sir you just save me hahaha. Thank you very much
Lovely voice
Why are so many people grateful for this video when it doesn't provide any proof that would allow us to understand this law?
Looks very useful but I can't hear anything sadly
Thank you! This was helpful!
1:35 i do not understand this
=x+(yz)
@@thankyouthankyou1172 Yea, that does not make sense to me either
Great vid. Thanks!
What law would Not C + (C+B) be?
C+B
@4:35 the final answer must be 1*~B+A = ~B+A
i'd like to know the theory behind this. i know i can do this in a truth table but i still don't understand the theory cause my mind looks at it in an algeraic way
It's so hard to find theory videos on logic - I can answer the questions on the test ok but I don't really understand what I'm doing and why it works. If anyone finds a good resource I'd love to know
Same qst
AND finds the overlap between between 2 parts and OR finds the total coverage. So A.(A+B) is the overlap between A and (A+B) combined. And the only thing A and (A+B) have in common is A, which is therefore the answer. With A+(A.B), look at (A.B) first, the result of (A.B) must be less than or equal to A and must totally be inside the section A. Therefore you have A+(Something that is included in the set of A) which is similar to doing A+A which = A
So taking ur example on i) would it = A? And C' + (C. A) = C'. ?
The answer would be C + A.
Use a truth table to check your answer.
C' + (C. A) = C'+A
a series of presentations well done
Can you do your videos with better volume please!!!
thank u sir
It was really useful actually! thank you very much
Does this law work when there is more than 2 terms in the bracket? Thanks
no, you'll have to make it 2 items first.
Interesting sound at the start haha!
Thanks
Thanks for the video definitely helped.
Thanks for the educational lesson on absorption law, that was actually quite a nice and clear explanation.
Thanks. Was wrecking my brain but I understand now :)
1.16 has a mistake it should be = x+ (y.z)
Awesome, thank you!
I just want to say thank you for making this series of tutorials; they were great!
THANK YOU SO MUCH!
please released more video for absorption law. I can not understand this vidio
this has been great tutorial series so far. could you please tell me where can i find rest of the video tutorials, sir?
sound completely stops at 7:40
awesome!
The sound went out half way through. Otherwise, I did like the video because he broke down the Boolean identities into 8 rules in 3 categories. That is how I learn. Thank you!