[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 400
You are missing trading opportunities:
- Free trading apps
- Over 8,000 signals for copying
- Economic news for exploring financial markets
Registration
Log in
You agree to website policy and terms of use
If you do not have an account, please register
The logarithm on any basis is calculated using the school formula:
loga(b)= logc(b) / logc(a)
Now you can take the base of natural logarithms as c.
P.S. Alas, I failed to make logarithm bases with subscripts: the <sub> tag won't work for some reason.
The logarithm on any base is calculated using the school formula:
loga(b)= logc(b) / logc(a)
The base of natural logarithms can now be taken as c.
P.S. Alas, I failed to make logarithm bases with subscripts: the <sub> tag won't work for some reason.
Thank you very much!
Algebra. grade 9 problem book.
The line y=-2x+b touches the circle x^2+y^2=5 at a point with negative abscissa. Determine the coordinates of the tangent point.
Sergei, what's the punchline of the problem?
In ninth grade, people know how to solve quadratic equations. It's a hassle, of course, but it's solvable.
Substitute y from the straight line equation into the circle equation and require the uniqueness of the solution to the quadratic equation (concerned!):
x^2 + (2x-b)^2 = 5
5x^2 - 4xb + b^2 - 5 = 0 (*)
The discriminant is zero: D = 16bb - 4*5*(bb-5) = 0
Hence 100 = 4bb.
Hence |b| = 5.
Draw a picture of a circle and a line. It turns out that the tangent at the point with negative abscissa can be only when b is negative. Hence b=-5.
Hence (*) becomes: 5x^2 +20x + 20 = 0
x = -2, hence y=-2x+b = 4-5 = -1.
Point (-2;-1).
P.S. Well, yes, it takes about three minutes. That's if you write it out carefully and neatly.
We have nine pieces of paper with numbers and algebraic signs drawn on them:
101-102=1. Obviously, the identity is not true. We only need to move one piece of paper somewhere (remove it, turn it upside down) to make the identity true.
For example:
101-10=12.
Any options?
Alexei. Thank you. I hadn't guessed to equate the discriminant to zero. Got the solution.
I realized it's not for 9th grade. I found another solution. It's nicer. Without the discriminant.
We have nine pieces of paper with numbers and algebraic signs drawn on them:
101-102=1. Obviously, the identity is not true. We only need to move one piece of paper somewhere (remove it, turn it upside down) for the identity to be true.
For example:
101-10=12.
Any options?
101-10^2=1
101-10^2=1
The sign "^" is missing in the original data.
This seems to be the solution:
101 = 102 - 1,
but it doesn't fit the "just one paper" condition either.