The Purplemath Forums

Looks great so far! (MFJ)

The mean of a set of eight numbers is 3 and the mean of a different set of twelve numbers is x . Given that the mean of the combined set of twenty numbers is 9, calculate x . My thinking: Since the average of the 8 numbers is 3, and you find this average by dividing the total by 8, then that total ...

Thank you!

The triangles say r = (11h)/21, and I get the same answer.

When you're doing truth tables, why does "P implies Q" evaluate to "true" if P is false? How are "P implies Q" and "P is true only if Q is true" "equivalent"?

And maybe no "Caps Lock"? Because SHOUTING seems rude, ya know?

I guess that kind of makes sense, but that "only if" thing is still weird!

Prove by induction that, for any natural umber n, 6 divides n^3 - n. Let n = 1. Then n^3 - n = 1 - 1 = 0, and 6(0) = 0, so 6 divides n^3 - n. Let n = k, and assume 6 divides k^3 - k, so 6m = k^3 - k for some natural number m. Let n = k + 1. Then (k + 1)^3 - (k + 1) = k^3 + 3k^2 + 3k + 1 - k - 1 = k^...

a) Give an example of a one-to-one function f:N->N (the function f, from the natural numbers to the natural numbers) that does not map N onto N. Can I use f(n) = 2n, because that uses all of the n's in N, and each f(n) has only one n going to it, but it doesn't "cover" all of N? b) Given a...

If k is odd, so k = 2p + 1 for some natural p, then k^2 = (2p + 1)^2 = 4p^2 + 4p + 1 and k^2 + k = (4p^2 + 4p + 1) + (2p + 1) = 4p^2 + 6p + 2 = 2(2p^2 + 3p + 1). This means k^2 + k is still divisible by 2, so 3(k^2 + k) is divisible by 6, so the whole thing is divisible by 6.

Thank you!