CPhill

OK...we can use

(-∞ < x < -11) U (3.5 < x < ∞)

x(x+7.5) > 38.5        simplifying, we have

x^2 + 7.5x > 38.5    subtract 38.5 from both sides

x^2 + 7.5x - 38.5 > 0

Setting the polynomial = 0 and factoring, we have

(a -7)(a-3) = 0  ..... so a = 7 or a = 3

And for all a > 7, the polynomial is postive.

So, the greatest "a" that makes the polynomial ≤ 0 is when a = 7.

Note that...for any given perimeter, the area is at a max when L = W.

So let L=W and so that LW = max area

Now, if we decrease W by 1 and increase L by 1, we have

(W - 1) (L +1) = WL + (W-L) -1  = max area + 0 - 1  = max area - 1

And if  we decrease W by 2 and increase L by 2, we have

(W - 2)(L+2) = WL + 2(W-L) - 4  = max area + 0 - 4 = max area - 4

And with successive increases/decreases of n = 3, 4, 5.......etc., the max area is decreased by n^2. So, the smallest of these decreases occurs between n = 0 and n = 1. Or, looking at it from another perspective, inceasing the Width from (W -1) to W and decreasing the Length from (L+1) to L ( where W = L), results in x = 1.

n^2 - 9n + 18

We only need to find out where the "zeroes" of this polynomial are.

So setting it to 0 and factoring, we have

(n - 6) (n-3) = 0

So.......n=6 and n=3

Note that when n = 5, the polynomial's value = -2  and when n = 7, the polynomial's value is 4. And for all integers, n > 7, this polynomial is > 0.

So the largest integer that makes this polynomial negative is when n = 5. Note that n = 6 just makes the polynomial = 0, as we found  !!!!!

This one is interesting....it may be a little difficult to picture what is going on....here is a diagram.....

Note that we have a set of lines drawn through B. And, drawing a line from C to each line through B such that they are perpendicular at their intersection generates a set of right triangles. And, as Anonymous has pointed out, the locus of all possible points at the perpendiculars generates a circle with a radius of BC/2. And if we let BC = the distance from (20,14) to (18,0), then BC = √200. And the area of the circle is pi*(√200/2)²  = pi*(200/4) = 50pi = 157.07. And the greatest integer not exceeding this is 157.

Thank you for this great answer, Anonymous......I couldn't picture what was going on until I read your response and "sussed it out" for myself with a picture!!!

A^-n = 1/Aⁿ

So

(-6)^-4 = 1/6⁴    = 1/1296

Another way to see this is to note that that the sum of the first n even positive integers is n(n+1)

And the sum of the first n positive odd integers is n^2.

Therefore

n(n+1) - n^2   = n^2 + n - n^2  =  n

And since n = 102 in both cases, then that's the final sum.

Note that we're getting 150ml more than we usually do.

So 150/750 = 1/5 = 20% "free"........of course, I always want to ask them.....why don't you just pour the "free" part from 5 bottles into one??.....then, it shouldn't cost me anything...

I had one of thse days, yesterday....

My advice???......back away from the computer......SLOWLY