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1. I had no idea that posting a few questions is a problem.
Had you not promised almost two months ago that you would post no more than three questions per day? (https://www.mathisfunforum.com/viewtopic.php?pid=434726#p434726)
Did you not, just three days ago, issue a rule restricting the number of questions others may post in one day? (https://www.mathisfunforum.com/viewtopic.php?pid=435960#p435960)
2. No need to threaten banning me from the website.
Was there a need for you to threaten banning me from the website? (https://www.mathisfunforum.com/viewtopic.php?pid=435630#p435630)
Was there a need, two and a half years ago, to demand a ban on someone who wasn't even here? (https://www.mathisfunforum.com/viewtopic.php?id=26386)
3. Member amnkb never complained to me about my reply to his answers.
Did amnkb not apologize many times for having somehow offended you, asking you please to explain what s/he was doing wrong? Did you not then criticise amnkb for "saying sorry all the time"? (https://www.mathisfunforum.com/viewtopic.php?pid=435912#p435912)
Should not one attempt in general to subscribe to the standards one imposes on others? ("...with what judgment ye judge, ye shall be judged: and with what measure ye mete, it shall be measured to you again.")
Aside to Admin Bob: You seem to have developed, during your many orbits about our Sun, deep reserves of patience and kindness. Your character is one I aspire to.
Popularity of a thread, based at least in part on the subject line; or popularity of a user, based at least in part on the user's number of posts; is not quite the same as a "like" button that can allow users to, say, thank a specific helper for specific good advice.
It appears that FluxBB does not have "like" buttons as options. It seems that the package is also fairly out of date. Perhaps the back-end script could be updated?
How do you [evaluate or simplify] a square root raised to another square root?
What is sqrt{2}^(sqrt{2})?
Use technology to obtain a decimal approximation. That is, use your calculator or an online resource such as Wolfram Alpha:
https://www.wolframalpha.com/input?i2d=true&i=Power%5B%5C%2840%29Sqrt%5B2%5D%5C%2841%29%2CSqrt%5B2%5D%5D
What is the best way to solve tricky or lengthy word problems?
Take your time, and be clear on your reasoning. Otherwise, it's just a matter of practice, practice, practice.
Why does the radicand come out?
See here:
sqrt{(a + b)^2} = (a + b)
Why is that the case?
Squaring and square-rooting are inverse operations (assuming the argument of the square root is non-negative). They undo each other. It is the same reason that 1 + 1 - 1 = 1 or (2*3)/3 = 2.
Complete the square.
ax^2 + ba + c = 0, where a > 1.
If the middle term is "ba", then there is no middle term, so the square-completion process is nothing more than isolating the one square:
Since anything can be viewed as having a zero added to it, then:
If the middle term is meant to be "bax", then try following the steps shown here: https://www.mathisfunforum.com/viewtopic.php?pid=435501#p435501
This business of rounding part way through a problem can lead to very inaccurate answers. here's an example:
Given x^2 = 56 work out x^5
On a calculator x = 7.483314774...
If I call that 7.5 and work out x^5 I get 23730.46875
If I use the full value I get 23467.67513.....
You can see that using the rounded up number and multiplying five times leads to an answer that is quite a bit too big.
Yes! Amen! Wait to round until the end!
I remember a question posed in one of my classes as an undergraduate:
Young people and old people face many of the same problems, often to do with discrimination. Why do old people fight it harder?
The answer: You can grow out of being young; you can't grow out of being old.
Best wishes on the job search and, if you glean any advice that you think might help fellow seniors, please share!
I apologize for my failure at formatting in the above. I though [ imath ] tags would work. Kindly please ignore the "\qquad" part at the beginning of each line. D'oh!
Suppose that m and n are positive integers with m > n.
If a = (m^2 - n^2), b = 2mn, and c = (m^2 + n^2), show that a, b, and c are the lengths of the sides of a right triangle.Note: This formula can be used to find the sides of a right triangle that are integers such as 3, 4, 5; 5, 12, 13; and so on. Such triplets of integers are called Pythagorean triplets.
NOTE: Looking for the set up only.
I suspect that they're expecting you to notice how the given side-length expressions hint of connections to perfect-square trinomiais.
If you add two of the above expressions, you get the third of the above expressions. Which pair works?
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