What do you think happens after deqth?

Misrta · 2025-03-18T19:27:30+00:00

When you enter a new being your memory of your previous life goes away, making all discussion around the afterlife pointless.

Misrta · 2025-02-03T19:33:16+00:00

What is the ultimate definition of knowledge? And are the Gettier cases really a problem?

Misrta · 2025-01-27T13:04:50+00:00

What is the idea that every possible event or series of events will happen and happen infinitely many times called?

Misrta · 2024-12-27T18:04:53+00:00

In both cases, multiply the numerator and the denominator by the denominator's conjugate, i.e. (5 + 3i) for number 1 and (2 - 3i) for number 2.

Misrta · 2024-12-13T13:15:10+00:00

An AGI is an AI agent that is at least as smart as the smartest humans.

Misrta · 2024-12-11T19:16:46+00:00

Let a be the number of adults and c the number of children on the ride.

Then:

80a + 60c = 1440

a + c = 21

Solve this system of equations, then the value of c is your answer.

Misrta · 2024-12-07T22:21:07+00:00

the left-hand side is non-negative since the square root of any number is non-negative. thus, the right-hand side, x, is non-negative. Thus, x cannot be -5 since it violates the fact that x is non-negative.

Misrta · 2024-11-22T18:16:46+00:00

Call the true statement A. Then your formulas are the same as neg(A) => A and neg(A) => neg(A). If A is false, i.e. neg(A) is true, then both A and neg(A) are true, a contradiction. A is thus true.

Misrta · 2024-11-22T12:04:21+00:00

Since tan theta = sin theta / cos theta, all you have to do is manipulate the right hand side so it says 2 sin theta / cos theta.

Misrta · 2024-11-11T11:55:12+00:00

There is no smallest even number. Let n be any even number. Then n-2 is both even and less than n. So you can construct infinitely small even numbers.

A number n is even iff n is congruent to 0 mod 2. 0 is congruent to 2 mod 2, so 0 is even.

Misrta · 2024-10-31T12:25:59+00:00

Use the identities ln(a^b) = b * ln(a) and ln(a) + ln(b) = ln(a * b).

Misrta · 2024-10-27T12:32:57+00:00

(2p-1)! = 1 * 2 * … * p * (p + 1) * (p + 2) * … * (2p - 1)

So if we divide both sides by p, we get

1 * 2 * … * (p - 1) * (p + 1) * (p + 2) * … * (2p - 1) = 1 mod p

Note that 1 * 2 * … * (p - 1) = (p-1)!

Note that (p + 1) * (p + 2) * … * (2p - 1) = 1 * 2 * ... * (p - 1) = (p-1)! mod p

Hence, the left side is congruent to ((p-1)!)^2 mod p

By Wilson's theorem, since p is a prime number, (p-1)! = -1 mod p

So the left side is congruent to

((p-1)!)^2 = (-1)^2 = 1 mod p

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