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Thank you SOOOO MUCH!! This helped Enormously.  I get the problem 100% now, and I appreciate you spending the time to type all this out. I appreciate it!!

Thank you!

Look under your first post, some 15 questions below this one !!

This response had something to do with computers and programming, matrices and vectors (all of which are too advanced for me.) I already saw it, however I did not get ANY of it, so I came here. :)

The red area is the difference in the area of the sector less  the area of the isosceles triangle  with the radius as its equal sides

Area of sector  =   pi * r^2 * (68.9  / 360)  =   pi *(9.28)^2* (68.9  / 360)     (1)

Area  of triangle  = (1/2) (r^2 )  sin (68.9°)  =  (1/2) (9.28)^2 sin (68.9°)   (2)

So.....the red area = (1)  - (2)  =

[ pi *(9.28)^2* (68.9  / 360) ] - [  (1/2) (9.28)^2 sin (68.9°)  ]  =

        51.78     -   40.17     ≈

            11.6  cm^2

No it is right on cphill answer 

13pi/18 * 180/pi = 13 * 180 / 18 = 130°        

Arc  L = 18.4 in

r = { [ (360° * 18.4) /130° ] / pi} / 2 

The digit 9 appears 100 times, between 1 and 500, if you count the repeats such as 99.

https://web2.0calc.com/questions/plzzzzz-help-cphill-or-anyone

p=1; q=1;r=1;s=1;p=1;m=0;n=p+q+r+s;if(n==348 and 373 and 544, goto8, goto next);m=(a*b+b*c+c*d);max(m)==p;print max(m);next:a++;if(a<55, goto5, 0);a=1;b++;if(b<55, goto5, 0);a=1;b=1;c++;if(c<55, goto5,0);a=1;b=1;c=1;d++;if(d<55, goto5, discard=0;

OUTPUT: r = 31 and s = 8 and p = 34 and q = 8

F  =  (9/5) ( C ) + 32   subtract  32  from both sides

F - 32  =  (9/5) C          multiply both sides  by (5/9)

(5/9) [ F - 32 ]  = C

(5/9) [ 41 - 32 ] =  C

(5/9) (9)  = C

5°  =  C

I apologize for my lack of intelligence, but what is a?

Bill    1800 (1+.034)³     After 3 years

Sue   2300 (1+.024)³             Edit: this will be the amount in the account after 3 years....subtract original amounts to get interst earned !!

Bill's interst if rate changes   1800(1.034)² (1.04)¹  - 1800 = .....

Sue's interst was  2300(1.024)³ - 2300 =.....

We can use Heron's  formula here

s = the semi-perimeter   = [9 + 4 +7 ]  / 2 =  20/2  =  10

The Area  is

√ [ s  (s - a) (s - b) (s - c)  ]         where a,b,c  are  the sides of the triangle

So we have

√ [ 10 * ( 10-9) (10 -4) (10 - 7) ]  =

√ [  10 * 1 * 6 * 3  ]  =

√ 180  =

√ [36 * 5 ] = 

6√5   units^2

We  have  something like this

                                B

                25                        25

A                           30                            C

Using the Law of Cosines

30^2  = 25^2 + 25^2  - 2(25*25)  cos ABC

cos (ABC) =   [ 30^2 - 25^2 - 25^2 ] / [ [ -2 (25 * 25) ]  

cos (ABC)  =   7/25

So.....  sin (ABC)  =     sqrt (25^2 - 7^2]  / 25  =    sqrt [ 625 - 49 ] /25  = 

sqrt [ 576 ] / 25  =

24  / 25

on part a i dont know

Area of  sector  =  (1/2) r^2  ( theta in radians)

So we  have

Area of sector  = (1/2) * 7^2  *  ( 4pi/15)     =   98 pi

                                                                        _____    [ in^2 ]

                                                                          15

yes Part B is wrong. Bill has more interest. That was a mistake on my part

No prob  !!!!

i got 3 out of 5 i dont know whats missing

Formula  for arc length

Arc length  =  r * (theta in radians)    

So we  have that

18 . 4   = r  *  (13pi  /18)        multiply both sides by  (18/  [ 13 pi])

18.4 *  18 / [13pi ]     =  r  ≈  8.1  inches

This is a repost, I'd suggest scrolling down the list of answers. I've answered this problem(but my answer for part B is wrong, but someone has corrected me)

I really want to thank you from the bottom of my heart! thanks man.

I definitely second EP's comment! This site isn't for just giving you the answer, it's for showing you solutions and ways to think about these problems so you can improve! You have to realize that posting so excessively without giving second thought to the solutions is just a short term solution; it doesn't help you improve at all.