CPhill

A_sector  =  (1/2) r^2  (theta in rads)  =  (1/2) (5)^2 * (1.5)  = 

25 * 1.5 / 2   =

18.75 in^2

Jess...I'm never entirely sure how these are to be written

But......I think the full answer is  :   (x,y) ⇒  (x, y - 11)

Transformation  rule :   (x, y - 11)  ⇒  (6, 8 - 11)

Scale factor  =   Radius of  Circle 2 / Radius of Circle 1  =  2/8  =  1/4

Your answer is good, Supermanaccz....it would cover all possibilities.....mine was just using a basic identity.......both are OK

Hectictar's answer is about as good as it gets.....!!!!

Check your input, again......

Arc length, S...is given  by

S  =  r (theta  in radians)

So

S  = 28  (3pi/4)      =  28 (3/4)*pi =  [ 21 pi ]   inches

We jhave the identity....

cos (A)  = sin (90- A)

So

cos (63)  = sin (90- 63)

cos (63)  = sin (27).....so....x  = 27  (degrees)

We  can write this in exponential form

[5^(1/3) ] ^x  =  125

(5)^(x/3)  = (5)^3

Since the bases are the same.....we can solve for the exponents

x/3  = 3         multiply both sides by 3

x  =

9        and that's the evaluation for the original expression

Well...let's see....

15310 (1.2)*1   = 18372 ....nope....that's not correct  !!!!

Let's use the points

(0, 15310)  (1,18002)  (2, 21548)  ( 3, 25642)     and this utility to generate a quadratic function :

http://keisan.casio.com/exec/system/14059932254941

We get the function

y = 350.5x^2  + 2402.7x  + 15294.7

It appears that the last answer is pretty close to this....so...."D"

cos A  =   AC  / AB

And  AB  can be found as  √ [ 15^2 + 36^2 ] = √ [ 225 + 1296]  = √1521  = 39

So.....cos A  =    15 /39   =   5/13