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I count 5....but I'm not sure.

Because chords CE and BD intersect within the circle, we can use the intersecting chords theorem to find minor arc BC

mDQE =  (1/2) ( arc DE + arc BC )

40 = (1/2) ( 30 + arc BC)

80  =  30 + arc BC

50° =  minor arc BC

And arc  BAE =

360 -  minor arc BC - minor  arc CD -  minor arc DE  =

360 - 50 - 70 - 30 =

360 - ( 150) =

210° 

And since PC is a tangent meeting the chord CE....the measure of angle PCE =  (1/2) ( minor arc BC and arc BAE ) = (1/2) ( 50 + 210) = (1/2)(260) = 130°  = mPCQ 

And angle CQP = 40°

So angle CPQ =  180 - mPCQ - mCQP =  180 -  130 - 40 =  10°

And angle CPE is twice this = 20°

So....angle CEP =  180 - mPCE - mCPE = 180 - 130 - 20 = 30°

And because angle CEP is an inscribed angle, then  minor arc CA is twice this = 60°

So

arc AE  =  360 -  minor arc CA - minor arc CD - minor arc  DE =

360 - 60 - 70 - 30 =

360 - [ 60 + 70 + 30 ] =

360 - [ 160 ] =

200°

This does not not look to be possible....I think the problem is that  minor arc DE is given as 30°....however....by the intersecting chords theorem, minor arc BC is computed to be greater than this

But..... minor arc DE looks to be much greater than minor  arc BC (???)

You are obviously trying to solve some financial problem, but your question is so incoherent that you would be much better off posting your FULL question and we will attempt to help you. Good luck.

Basically, you "type" in the math you want using the buttons at the top, then click the "Get Latex" button to get it in Latex. You can then input the latex in the forum.

I do use these myself, so I don't actually have to learn LAtex

You will have a   -5x^2  +42x^2   and a  -x^2   =  +36x^2            soooooo    36

I do AoPs, and we get a lot of problems, but it helps!

It can be shown that the max area is when each side of the rectangle = 200/4 = 50ft

So...the rectangle is a square

So.....the max area is 50^2  =   2500 ft^2

Do you guys deal with a bunch of AoPS problems like this?

Ok thanks.

Ok, thanks.

Thank you, CPhill! 

Notice that tertre's answer of  83 and 85  is correct

85^2 - 83^2  =

7225  - 6889   =

336.....

Good job, tertre!!!!

85^2-83^2=336, so it is true..

Or, (a+b)(a-b)=(85+83)(85-83)=168*2=336.

The larger of the two squares

??? I'll check my work again, are they asking for the 85 or 7225...

The two vertices we can look at are   3 + 5i  and  -1 - i

Taking the difference  between these complex numbers produces a complex vector  = 

(3 + 5i) - ( -1 - i) =  4 + 6i

The length of the diagonal is the length of this vector =

√[ 4^2 + 6^2 ]  =  √ [ 16 + 36 ]  =  √ 52  =  2√13 units

Checking that this is correct  the other two vertices are   4 + 0i    and - 2 + 4i

Take the difference of these two complex numbers and we get

6 - 4i

And the length of the diagonal is  √ [ 6^2 + (-4)^2 ]  = √52 = 2√13 units

So....true!!!

I am sorry for the inconvenience, but your answer, 7225, is incorrect. If you can, please redo this problem or ask for help. This question is way too hard for me. AoPS is a pain in the ____. I want to know if you guys get many questions from that website. 

We have \(x\) as the smaller odd number and \(x+2\) as the larger odd number.

Translating this into an equation, we have \((x+2)^2-x^2=336\).

Expanding \((x+2)^2-x^2=4x+4\).

Rewriting, we attain \(4x+4=336, 4x=332\) and \(x=83.\)

So,  \(x+2=85\) or \(\boxed{7225}.\)

We can let    f(x) = x^4    and g(x) = x^5

So

f(x^3)  =  [  x^3 ]^4 =  x^12

and

g(x^2)  =  [ x^2 ]^5  =  x^10

So

f(x^3) * g (x^2)  =

x^12 * x^10  =

x^22  ⇒    degree 22

No big deal.....all that Christmas food has made my mind hazy, too....LOL!!!!

Yes, you are correct CPhill. Misread the problem!

-16t^2 + 80t + 21

The  t  that maximizes this is given by :

-80 / [ 2 * -16]  =  80 / 32  =  5/2

Putting this back into the function, we get that the max height is

-16(5/2)^2 + 80(5/2) + 21  =

-16(25/4) + 200 + 21  =

-100 + 200 + 21 =

121 ft

Hint: The numbers when closer together have the maximum area.

Thanks!