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Stu
Username
Stu
Score
1313
Membership
Stats
Questions
73
Answers
292
81 Questions
302 Answers
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+1313
Is this image conveying accurately?
is it accurate?
where did the 60 v come from?
do you have to draw on amp leaving battery to be correct? *fig 2.
why when take away the v source, do we leave 28 ohm when it it cannot carry current? is it just that when we cant
read more ..
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Stu
Oct 26, 2014
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help with circuits
can anyone make then selc available to call or skype with to assist me with some cir uit fundementals tonight for an exam in the morning?
LancelotLink
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Stu
Oct 26, 2014
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Mesh/Kirchhoffs theorem.
can someone define the values in the equation one and two stated in the image please.
Alan
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Stu
Oct 7, 2014
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It was blue, but now it isnt. Oh, it's the same I'm sure. fb is facebook, where I got a response. Just told it shouldn't be long and it was out of service just to be maintained or updates made.
Stu
Feb 17, 2014
#2
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Yep, except your post is blue. lol. I got a reply on fb from the guy who said changes were occurring.
Stu
Feb 17, 2014
#8
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[size=150]Tiger Maths: Algebra explained and illustrated. [/size]
The site is simple, well explained and with illustrations. Covering all the key algebra concepts and formula with a plethora of examples answered.
http://www.tiger-algebra.com/
Awesome site for those people learning Algebra.
Stu
Feb 17, 2014
#7
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[size=150]Nrich: Awesome pictorial based learning site: learn your way through what you want to know easily.[/size]
It is worth a stop for all age groups.
There is a few ways to find information on the site. I recommend the search option, or topics list. But you can also click on an appropriate learning level and go to early secondary learning for example. The site has had a lot of effort put into it. So don't let it go to waste. Get what you want to know at your finger tips.
Nrich:
http://nrich.maths.org/frontpage <-- highlight and right click, then go to.
This above Nrich site is a part of " uc-rgb.jpg Copyright © 1997 - 2014. University of Cambridge. All rights reserved.
NRICH is part of the family of activities in the Millennium Mathematics Project."
It has a good layout, great search engine and a wide variety of topics all with pictures to help understand your way through the learning.
You'l only find good answers, tools or other useful stuff you might be looking for. Not much missing, if there is try the first link in this post to www.mathportal.org.
Stu
Feb 15, 2014
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makes sense. I got it in the first bit. Just rationalising it in an improper fashion previously. Thanks.
Stu
Feb 15, 2014
#12
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Ok Rom makes sense. I'l put in the effort.
Stu
Feb 15, 2014
#4
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Thanks, yes I knew in my mind the whole fraction was rooted, could see it didnt look right, but skipped a step and - proof Im still learning. Sorry op for mistake. Thought I had this down pat.
Stu
Feb 15, 2014
#8
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I was on the right track Melody. Pat on the back for me. I don't know that it says will yet though. just going on what was written.
But i agree with you, that looks just like the answer I had to the same question in high school 14 years ago.
Stu
Feb 15, 2014
#2
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Here is an explaination.
http://www.mathportal.org/algebra/solving-equations/solving-linear-equations.php
Good luck, post in this thread again if stuck.
Stu
Feb 15, 2014
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.I think you can just transpose it to find the value for x. Therefore what you do to one side to change an individual identity (working with it's opposite to remove it) must be done to the other side. What sides? Each side of the equals, the left hand side and right hand side. Hope this helps. The answer is x is equal to 5/4 try to get to that.
____________
Thanks
Stu,
Learning E&E
Stu
Feb 15, 2014
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