Answer:
ye me too i cant answer thay
Im confused. Its not equal to any of the choices
Unless C is the cube root of 32^5.
[t3] \sqrt[n]{x} [32^5]
Answer:

Step-by-step explanation:
as with all functions f(t), the average rate of change on [a.b] is

It could be B but I want to say D because it relates more to the question.
Answer:
Step-by-step explanation:
By applying tangent rule in the given right triangle AOB,
tan(30°) = 


By applying tangent rule in the given right triangle BOC,
tan(60°) = 
OC = BO(√3)
OA + OC = AC

2√3(BO) = 60
BO = 10√3
OC = BO(√3)
OC = (10√3)(√3)
OC = 30
By applying tangent rule in right triangle DOC,
tan(60°) = 
OD = OC(√3)
OD = 30√3
Since, BD = BO + OD
BD = 10√3 + 30√3
BD = 40√3
≈ 69.3