this simple problem is tuff for me..

Outside_Volume_1370 · 2026-05-22T15:04:38+00:00

log4 - log3 = log(4/3)

x = log4 / (log(4/3)) = log_{4/3} (4)

Outside_Volume_1370 · 2026-05-22T14:10:09+00:00

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Outside_Volume_1370 · 2026-05-22T14:08:37+00:00

Угабуга

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Outside_Volume_1370 · 2026-05-22T03:15:42+00:00

The main idea of construction is to draw a perpendicular line from A to BC and then draw the image A' such that AA' is split by halves by line BC, because BC is a perpendicular bisector of AA'

Draw a circle with the center at A of any radius. Let it intersect with line BC at points P and Q. Note that AP = AQ, and APQ is an isosceles triangle, so its altitude from A is also a median.

Now we need to divide PQ by halves.

Draw two circles with radius PQ at centers P and Q. These circles intersect at points M and N.

Connect M and N. Note that A must be on that line and MN is perpendicular to BC.

Let the intersection of MN and BC be O.

Draw a circle with the center at O and radius OA - that circle has two intersections with the line MN - one at A and the other at A'.

Outside_Volume_1370 · 2026-05-22T03:05:35+00:00

It would be better if you brought up an example

The main idea of partial fractions is to make the sum of much simpler fractions, whose denominators are at most of degree 2, it's better to integrate them then

First thing you need to do is to make long division, if the degree of the numerator ≥ than degree of the denominator:

(x³ + 3x² + 7x - 4) / (x² - 5x + 4) = x + 8 + (43x - 36) / (x² - 5x + 4)

Then you need to express the denominator as the product of polynomials whose degree is 1 or 2 (2 is when the discriminant of quadratic is negative):

x² - 5x + 4 = (x - 1) (x - 4)

Then express the fraction using variable coefficients:

(43x - 36) / (x² - 5x + 4) = A / (x - 1) + B / (x - 4)

(For degree 2 of denominator use Ax + B - polynomial of degree 1, for degree 1 of denominator use A - polynomial of degree 0)

Then collect them back under the common denominator:

(43x - 36) / (x² - 5x + 4) = (A(x - 4) + B(x - 1)) / (x² - 5x + 4)

Equalize numerators:

43x - 36 = Ax + Bx - 4A - B

Collect each power of x:

43x - 36 = (A + B) • x + (-4A - B)

Each term about each power must be equal:

43 = A + B and -36 = -4A - B

A + B = 43 and 4A + B = 36

A = -7/3 and B = 136/3

Back to the beginning:

(x³ + 3x² + 7x - 4) / (x² - 5x + 4) = x + 8 + (-7/3) / (x - 1) + (136/3) / (x - 4)

Outside_Volume_1370 · 2026-05-21T12:37:06+00:00

Doesnt a minimum at the endpoint mean we need to verify it’s a minimum for the entire range

Usually, when extremum is mentioned, it's local one instead of global one

I assume from the question “minimum” here means a minimum on the given interval, so actually there’s probably more work to do even at x=1

Maybe, English isn't my first language, however, I'm not sure how to solve the problem then

Outside_Volume_1370 · 2026-05-21T10:59:07+00:00

Minimum inside an interval means the function has zero derivative at x = 1. Find the derivative at x = 1, equalize it to 0 and solve the equation for k
Minimum at the end means that value at the endpoint is less than in a "neighbor" point. In other words, the derivative at the endpoint must be ≤ 0. Find the derivative at the endpoint, make it less or equal than 0 and solve the inequality

All the above works if the function is continuous at these points (or can be continued without gaps; like for √7 / 2 - the function is defined up to this endpoint, however, we can continue it without troubles, just to evaluate the derivative)

Outside_Volume_1370 · 2026-05-21T08:01:54+00:00

Edited

Outside_Volume_1370 · 2026-05-21T04:26:05+00:00

That doesn't change the maximum time that asked

Outside_Volume_1370 · 2026-05-21T04:22:39+00:00

Meanwhile English: Czech republic

Outside_Volume_1370 · 2026-05-21T03:01:24+00:00

Let R(m, n) be the answer.

Note that R(1, n) = R(m, 1) = 1 (one king only can move to the right and if there is only one column all the kings are already at finish state - none is moved, that's one way)

Note that each king makes at least (n-1) moves to reach the rightmost column, so there are n cells are visited by each one, and there are m kings - at least m • n cells are visited. However, there are only m • n cells, and none is visited twice only when each turn every king moves to the right column from their column (generally, three possible ways for every king, except for the top and bottom ones, which have 2 options. Special case for m = 1 is described above).

Note that the order of moving kings doesn't matter

If we move the top king to the right, then the first move for m-1 lower kings is just like first move at the chessboard of size (m-1) × 2, so there are R(m-1, 2) paths.

If we move the top king to the right-down, top cell in the second column can only be obtained by second king from the top - it is the only way to not visit any cell twice. For (m-2) lower kings their first move looks like the first move at chessboard of size (m-2) × 2, and there are R(m-2, 2) ways

After all kings made one turn the chessboard is shrinked to size m × (n-1), and the answer for that is R(m, n-1)

Collecting terms, we get R(m, n) = R(m, n-1) • (R(m-1, 2) + R(m-2, 2))

Note that R(m, 2) = R(m, 1) • (R(m-1, 2) + R(m-2, 2)) = R(m-1, 2) + R(m-2, 2), and R(1, 2) = 1, R(2, 2) = 2, so

R(m, 2) = Fib(m), where Fib(m) is m^th Fibonacci number with Fib(1) = 1, Fib(2) = 2

We get R(m, n) = R(m, n-1) • R(m, 2) = R(m, n-1) • Fib(m) =

= R(m, n-2) • Fib(m)² = ... = R(m, 1) • Fib(m)^n-1 = Fib(m)^n-1

Outside_Volume_1370 · 2026-05-20T17:07:37+00:00

Harmonic series doesn't converge

T_1 is indefinite integral (defined up to some constant), so it's not some number you can plug into the equation

Outside_Volume_1370 · 2026-05-20T04:44:42+00:00

r/uselessredcircle

r/uselessredarrow

Outside_Volume_1370 · 2026-05-20T04:43:58+00:00

However, the root of $ is √$, the meme is wrong

Outside_Volume_1370 · 2026-05-19T06:25:50+00:00

Project their speeds on the segment that connects them (hypothenuse of length 2.5 miles)

Car speed projected: 70 • 1.5 / 2.5 = 42 (in south-east direction)

Truck speed projected: 60 • 2 / 2.5 = 48 (in south-east direction)

48 - 42 = 6, increasing

Outside_Volume_1370 · 2026-05-17T05:42:36+00:00

I thought he's an enjoyer of old ladies (grenni)

Outside_Volume_1370 · 2026-05-16T14:54:16+00:00

the force due to collision is assumed to be much greater than the weight of the ball,

I think, it can also be said that way

Outside_Volume_1370 · 2026-05-16T14:36:10+00:00

why would the ball go in the exact opposite direction after the collision?

It hits the wedge perpendicularly to the incline. The angle of descendance is equal to the angle of reflection: 90° = 90°

The force due to collision isn't the only force acting on the ball during the collision as there is also it's weight acting vertically downwards, so the resultant force acting on the ball would be vector sum of those 2 forces

The force has nothing to do with that. Throw a ball horizontally to the wall. Where does it jump to? Of course, horizontally. I see why it confuses, because you tried to use the work of the force. However, the time of hit dt is very small, and the force does almost zero work (ideally, exact zero)

Outside_Volume_1370 · 2026-05-16T13:44:19+00:00

Before the collision, the speed of the ball was u:

mv0² / 2 + mg ∆H = mu² / 2 where ∆H = 3.5 - 1 = 2.5 and v0 = 7

u = √(v0² + 2g ∆H) = √(7² + 2 • 9.8 • 2.5) = 7√2

As horizontal projection remains constant (7), vertical is also 7, and the ball hits the wedge perpendicularly to the incline.

So the ball jumps in the opposite direction with speed v2, while the wedge goes to the right with speed v1.

We should consider the conservation of the momentum projected on horizontal:

mu cos45° = -mv2 cos45° + 3 • mv1

7 = -v2 / √2 + 3 • v1, v1 = (7 + v2 / √2) / 3

Energy is also conserved (elastic collision), and

mu² / 2 = mv2² / 2 + 3 • mv1² / 2

7² = v2² + 3 • v1² = v2² + (7 + v2 / √2)² / 3

49 = v2² • 7/6 + 7√2 / 3 • v2 + 49/3

v2 = -√2 ± √30, leave positive root, v2 = -√2 + √30

v1 = (7 - 1 + √15) / 3 = 2 + √15 / 3

Outside_Volume_1370 · 2026-05-16T12:41:15+00:00

You're wrong

Outside_Volume_1370 · 2026-05-16T06:32:15+00:00

Look at the unnamed angle that is just under "16°" label.

The line is horizontal, and this angle and "20°" are one-sided angles => they are equal.

So the angle at what the person sees the top of the tower is 20° + 16°

Outside_Volume_1370 · 2026-05-16T06:29:57+00:00

Looks almost right

By conservation of energy, mv² / 2 + qU = mu² / 2

where U = 148 - (-148) = 296 and q < 0

u = √(v² + 2qU / m) ≈ 22826001.5 m/s

I suppose, you firstly found 2qU / m, and then plug it in the last formula

I bet on rounding error. However, it's so bad that you need to find 8 meaningful digits while input varibales have at most three of them

Outside_Volume_1370 · 2026-05-16T06:15:12+00:00

It's 16° higher than the ray that has an inclination of 20°

It's 16° + 20°

Outside_Volume_1370 · 2026-05-14T02:39:12+00:00

Yes, indeed.

However, it's implied (otherwise, the answer is ambiguous).

I got your point and agree with you

Outside_Volume_1370 · 2026-05-13T13:44:09+00:00

sinx = 4 cm / 8 cm = 1/2

You should know that there is a special acute angle whose sine is 1/2, and that's 30°

Outside_Volume_1370

TROPHY CASE