(ar^b) ^4 = 16r^20 where a and b are positive integers work our a and b​

a = 2

b = 5

Step-by-step explanation:

Given :

(ar^b)^4 = 16r^20 ; a and b are positive integers :

Opening the bracket :

a^4r^4b = 16r^20

a^4 = 16 - - - - - (1)

r^4b = r^20 - - - (2)

a^4 = 16

Take the 4th root of both sides :

(a^4)^(1/4) = 16^1/4

a = 2

From (2)

r^4b = r^20

4b = 20

Divide both sides by 4

4b/4 = 20/4

b = 5

Hence ;

a = 2

b = 5

δabc ≅ δdcb by sas because

db ≅ ac . . given (side)

bc ≅ bc . . reflexive property (side)

∠dbc ≅ ∠ acb . . alternate interior angles where transversal bc crosses parallel lines bd and ac (included angle)