CPhill

When in doubt....just look at the answers...b and d would produce positive first terms....but the given problem has a negative first term.....

So...it appears it's a two-horse race....

Expand each remaining one and see what happens....

I'm unfamiliar with the notation used in the last one, so I can't comment on that one

On the -5i one....think about where the terminal point of this angle lies.....it's on the y axis, 5 units "down" from the origin in the "imaginary" direction {pardon the non-math terms, here}

So, the correct angle should be -pi.2, not pi/2......

I don't think that "Fermat's Last Theorem" got as much attention as this one did...!!!

{But..it took a lot longer to prove....over 350 years....}

Regardless...it was a good one to play around with, Mathematician.....many different approaches by our "professionals"

And even some bumbling and stumbling by me, too!!!

This boils down to finding the side length of an equiateral triangle inscribed in a circle. Of course, the area is maximized when the side lengths are maximized....!!!

Applying the Calculus to our given problem where r = 4.5 and x = s

We have

A = .5x(4.5 + √(20.25 - .25x^2))

A = 2.25x + .5x√(20.25 - .25x^2)

dA/dx = 2.25 +.5(20.25 - .25x^2)^(.5) - (.125x^2)(20.25 - .25x^2)^(-.5) = 0

18 + 4(20.25 - .25x^2)^(.5) - x^2(20.25 - .25x^2)^(-.5) = 0  rearrange

18 + 4(20.25 - .25x^2)^(.5) = x^2(20.25 - .25x^2)^(-.5)

18 (20.25 - .25x^2)^(.5) + 4 (20.25 - .25x^2) = x^2

18(20.25- .25x^2)^(.5) = x^2 - 4 (20.25 - .25x^2)     square both sides

324(20.25- .25x^2) = x^4 - 8 (20.25 - .25x^2)x^2 + 16(20.25-.25x^2)^2

6561 - 81x^2 = x^4 - 162x^2 + 2x^4 + 16(6561/16 - 61/8x^2 + 1/16x^4)

6561 - 81x^2 = 4x^4 - 324x^2  +6561

4x^4 - 243x^2 = 0   factor

x^2 (4x^2 - 243) = 0      reject x =0

4x^2 - 243 - 0

4x^2 = 243

x^2 = 243/4

x = √(81 * 3) / 2 =   (9/2)√3  = 4.5√3 = s

And this result can be generalized to r√3 for any given radius

Let x be the number of dollars, rhen 2x will be the number of quarters....

So

x + .25(2x) = 96    simplify

x +.50x = 96

1.5x = 96

(3/2)x = 96

x = 64

And 2x = 128 = the number of quarters

The discriminant = b^2 - 4ac.... so we have

13^2 - 4(11)(-28) = > 0    so this tells us we have real roots

But...maybe I'm missing something.....the discriminant doesn't tell us anything about whether or not f(x) ever = -100....it just tells us whether or not we may have "real" zeroes

We'd need to find the vertex for this

The x coordinate for the vertex is given by -b/2a  = -13/2(11)  = -13/22

And putting this back into the function.....we have the y value at the vertex = about -31.84

And since this parabola turns "up," (-13/22, -31.84) is the lowest point on the graph.....so f(x) never = -100....

Don't worry, saseflower, I misread it too, at first....!!!

I wish I knew how to construct "3-D" objects with it.....I haven't found the trick for that!!!

From A and C, drop a perpendicular to BD cutting EF at G and H and BD at I and J, respectively.

Then, by similar triangles  AEG = ABI and CFH = CDJ

But, since E anf F are midpoints of AB and CD, and all parts of similar triangles are in the same ratios,  EG = 1/2 of BI and FH = 1/2 of JD. Thus (EG + FH) = 1/2 of (BI + JD).

And AC = GH = IJ    ...thus GH = 1/2(AC + IJ)

Therefore EF = (EG + FH) + GH  =  1/2(BI + JD) + 1/2(AC + IJ) = 1/2(BI + IJ + JD + AC)

But (BI + IJ + JD)  = (BD)

So we have

EF = 1/2(BD + AC) = 1/2(AC + BD)

Let's work on the 'x" terms, first

Notice that

1x + 99x = 100x

And

3x + 97x = 100x

So, it appears that adding all the x terms pair-wise like this will produce 100x. And we have 25 of them...so 25 x 100x = 2500x

Likewise

2 + 100, 4 + 98, 6 + 96.... all produce 102......and we have 25 pairs of these. So 102 x 25 = 2550

So, the "series" sums to   2550 + 2500x