proyaop

Wow that was a great way to do it! Thank you so much Asinus! 

oh and hi

By vieta, the sum of the roots (a+b) is 5 (the b term * -1) and the product of the roots (ab) is -11 (the c term).

To clarify: $a + b = 5$, and $ab = -11$

Then, when we open up the parenthesis for (a+3)(b+3), we get ab + 3b + 3a + 9. We can simplify that new expression to:

$-11 + 3(5) + 9$

So, (a + 3)(b + 3) = 13.

Since it is equilateral, then AD bisects CB, so CD = 4. Using the pythagorean theorem, AD is 8^2 - 4^2 = AD^2

So AD = $4\sqrt{3}$

Wow! That was very detailed and complicated! Maybe leaving out undefined in the sum is the answer. And the answer to the problem is in the formatting of a fraction, so the answer to the problem shouldn't have any imaginary numbers.

Thank you... 

We can simplify this to 8000 divided by 15, getting 533.333333.

I got it! The answer is $({4\over5},{4\over3})$.

Here is how I got it:

Let S be the twice the area of the triangle. Then the side lengths are S, S/2, and S/h. Since the triangle is acute, these side lengths must satisfy these inequalities:

$S^2 + ({S\over2})^2>({S\over h})^2$

$({S\over2})^2+({S\over h})^2 >S^2$

$S^2 + ({S\over h})^2>({S\over2})^2$

We can divide each inequality by $S^2$ to simplify, and the third inequality will automatically be satisfied, so we are left with:

$1 + {1\over4} > {1\over h^2}$

${1\over4} + {1\over h^2} > 1$

Simplifying, we get h^2>4/5, and h^2<4/3, so the interval for h^2 is the (4/5, 4/3).

I chose all except C.

If two triangles share the same size they are congruent, and if they are the same name they are the same triangle.

But if the triangle is the same shape, then it could be a similiar triangle, not a congruent triangle. Unless the triangle has the same name, then it is still congruent and is the same triangle.

We can use complementary counting, and see what are the odds of 0 heads to be up. Since there are 3 coins, and there is a 50% chance to get heads or tails, the odds of having 0 heads is (1/2)^3 = 1 in 8. 

Then, subtracting 1/8 from the total probability (8/8), we get the chance for at least one head to be up as 7/8.

pryap