CPhill

Thanks, Melody, for that one !!!

I assume by "length," you mean "height"

The volume of a cone is (1/3) of that of its associated cylinder (a cylinder of the same radius and height)

So

V = (1/3)(r^2)(h) = (1/3)(4^2)(6) = 32 cu in.

If this number is divisible by 2 and 5 it must be a multiple of 10.

And the only thee digit number which fits the bill is 290. It's not divisible by 3 and the sum of its digits  = 11.

(3 - 2i) /(4 - 5i)

Multiplyimg by the conjugate of (4 - 5i) = (4 + 5i)  in the numerator and denominator, we have

(15 + 7i + 10) / (16 + 25) =

(22 + 7i) / 41 =

(22 /41) + (7i / 41) =

(1/41)(22 + 7i)

(z₁)^2 = (1 - i√3)^2 = (1 - 2i√3 - 3) = (- 2 - 2i√3) = -2(1 + 2i√3)

(z₂)^2 = (√3 + i)^2 = (3 +2i√3 - 1) =   ( 2 + 2i√3) = 2( 1 + 2i√3)

So ......

[(z₁)^2] / [(z₂)^2] = [ -2(1 + 2i√3)] / [2(1 + 2i√3)]  = -2/2 = -1  =  -1 + 0i  = z₃

So

tan Θ = (b/a)  →  tan^-1 (b/a) = Θ  = arg z₃

And

tan^-1 (0/-1) = pi

So

arg z₃ = pi

Well, actually, the "formula" is:

A = (b*h) / 2      .....so we have.....

72 = (18 * h) / 2      multiply both sides by 2

144 = 18 * h            divide both sides by 18

144/18 = h = 72/9 = 8 inches

But we're not taking (1122) * (1122). We're taking 1122 and raising to the power of 1122 factorial....!!! (If I read the question correctly...)

I don't know how to determine this !!!  Whatever this number is, it is HUGE!!!

Here's another proof:

Let's suppose that there are two positive, relatively-prime integers, call them a and b, such that a/b  = √5 with a > b.

Multiply both sides by b, so we have

a = b√5

Now square both sides

a² = 5b²

And note that we can write 5b² as 4b² + b²    Now, subtract b² from both sides....so we have...

a² - b² = 4b²

Now, note that the right side is even, and suppose that a and b are both even. But, this is impossible since they are relatively prime and thus, have no factors in common.

And neither can one be odd an the other even, because the left side would be odd, but the right side is even.

The only possibility left is that both are odd. But this can't be so for the following reason: Factor the left side as (a + b) (a - b). And if a and b are odd, then (a + b) is even and so is (a - b) - (remember that a > b, so a - b > 0).

So we have

(a + b) ( a - b) = 4b²    Now divide both sides by 4 ....... and we can write

[(a + b)/2] [(a - b)/2]  = b²

Now, if (a-b)/2 is odd, then (a+b)/2 is even ..... and vice-versa. To see this,  let  (a - b)/2 = 2k + 1.     Then a - b  = 2(2k + 1)  ....  Add b to both sides..... a = 2(2k + 1) + b....Now add b again to both sides and we have  a + b =  2(2k +1) + 2b.......Now, divide by 2 on both sides, and we have (a+b)/2 = (2k + 1) + b, and both sides are even. Likewise, it can be proved that if (a +b)/ 2 is odd, then (a-b)/2 is even.

But, an even times an odd is always even...... but b² is odd.

Thus......a and b aren't both even or both odd, and neither are they of different parity. and nothing else can be possible. Thus, our assumption that there exists two positive, relatively-prime integers such that a/b = √5 must be false. And thus, √5 must be irrational.

12 min= .2 degrees

So we have

cot(85.2°) = 0.08397 = .0840 rounded

Actually, this theorem was proved by an English mathematician, Andrew Wiles, in the mid 1990's. It was one of the most famous "unsolved" problems in mathematics - remaining unproven for 358 years. Interestingly, Wiles himself almost gave up on his effort to prove it!!!

Here are some links to the theorem and to Wiles himself.