[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 216
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(x^2 - x)=a;
Чему равен x, при известном а?
Too trivial, since all that is needed and sufficient here is to solve the quadratic equation:
x2 - x - a = 0;
Я уже чувствую, что еще пару постов в эту ветку, и я буду окончательно опозорен, но решение этого вопроса лежит исключительно в практической области. Хорошо, предположим что x=10, тогда а=90:
10^2 - 10=90;
Попробуем найти искомое значение (10) с помощью учебника по математики за 8 лкасс:
Были получены два значения и оба не правильных. В первом случае 9^2 - 9 = 72, а не 90, во втором случае значение верно по модулю но неверно по знаку. Спрашивается, почему не один из корней не равен +10? Еще раз прошу прощения за мой кретинизм, но уж если позориться то до конца:)
I don't know which textbook you meant, through solving quadratic equations like: x^2 - 1 - a = 0
We have two potential solutions:
x1 = (1 + sqrt(1 + 4*a)) / 2
x2 = (1 - sqrt(1 + 4*a)) / 2
Since a = 90, 1 + 4*a = 361. The square root of 361 equals 19. Add 1 to it, i.e. we get 20 and divide by 2. Thus x1 = 10. The answer is correct.
In the second case x2 = -9. The answer is also correct, because 81 + 9 = 90.
In short, go and learn the maths - it is tame.
In order to fulfill the condition of the problem we need to connect the ends of the red segments with lines, straight or broken - it doesn't matter, the main thing is that the connecting lines should not cross the black segments, as they have all been crossed once. Consider figure 1. We can connect 4 of 5 red segments inside it, so one of them has no continuation inside the piece. It means that the polyline we are looking for has one of its ends inside of 1. However, the same can be said for shapes 2 and 3, which would mean that the polyline has 3 ends, which is impossible.
This is cheating. The THIRD FIGURE has two ends and all 5 of 5 can be crossed?! It's 1 and 2 have one end. If I'm confused, make it clear: either I'm wrong, or these cards are spoiled. no maths .... or we'll be misunderstanding each other.
Yes, folks, it's getting quite boring without me. All that dry maths :)
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Here'sa question for you: There is one such capacitor, its capacity is 2200 pF, you need 2100 pF.
What is a simple way to reduce the capacitance of this capacitor? It is not possible to connect other capacitors to it.
Да, народ, совсем у вас без меня скучно стало. Одна "сухая" математика :)
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Вопрос на засыпку: Есть один вот такой конденсатор, его ёмкость - 2200 пФ, нужно - 2100 пФ.
Как простым способом уменьшить ёмкость этого конденсатора? Подключать к нему другие конденсаторы нельзя.
Drill a hole.
Дырку просверлить.
Not far from the truth. they make kerfs on them == reduced plate area == reduced capacity.
It's all about maths, it's all about logic. You must be referring to formulas.
I looked up the solution. It's about the same. The main thing is this: if none of the ends of the polyline lies inside one of the regions 1, 2, 3, the polyline must enter as many times as it exits, i.e. the number of its intersections with the region boundaries must be even. But here it is odd, and there are three such regions.
Drill a hole.
MetaDriver , vegetate, yes you're right, that's what I meant. Either drill the hole with a diamond drill bit, or grind out the flats, then wipe, wash, dry and insulate the condenser by dipping it in paint or varnish.
Now the task is more difficult:
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There is one such capacitor, its capacity is 4700 μF, you need 4600 μF.
How can I reduce the capacitance of this capacitor? You cannot connect any other capacitors to it.
Nor can I dismantle, grind or drill it.
It can break.