What is the area of the shaded region in the given circle in terms of pi and in simplest form? A) (120pi + 6 sqrt 3) m^2

B) (96pi + 36 sqrt 3) m^2

C) (120pi + 36 sqrt 3) m^2

D) (96pi + 6 sqrt 3) m^2

answer: cos [   arcsin (3/5) - arccos (1/2) ] =

step-by-step explanation: × = 3/5 = 0.6--> x = 36.87 and x = 180 - 36.87 = 143.13 deg

cos y = 1/2 --> y = +- 60 deg a. cos (x - y) = cos (143.13 - 60) = cos 83,13  

=

0.12

b. cos (36.87 - 60) = cos (-23.23) = 0.92

hence, the required result is

step-by-step explanation:

suppose we have two matrices a and b

we will assume the two matrices

we have to compute b-a which means subtracting corresponding elements of a from b.

-

all you have to do is multiply them both to see how many out comes you can put:

3 x 2 = 6

you can make 6 diferent combinations of crusts and toppings.

~batmans wife dun dun