Simplify the following:
2^3 (4×36/12 + 1/4)
4×36/12 = (4×36)/12:
2^3 ((4×36)/12 + 1/4)
2^3 = 2×2^2:
2×2^2 ((4×36)/12 + 1/4)
2^2 = 4:
2×4 ((4×36)/12 + 1/4)
2×4 = 8:
8 ((4×36)/12 + 1/4)
36/12 = (12×3)/12 = 3:
8 (4×3 + 1/4)
4×3 = 12:
8 (12 + 1/4)
Put 1/4 + 12 over the common denominator 4. 1/4 + 12 = 1/4 + (4×12)/4:
8 1/4 + (4×12)/4
4×12 = 48:
8 (1/4 + 48/4)
1/4 + 48/4 = (1 + 48)/4:
8 (1 + 48)/4
1 + 48 = 49:
8×49/4
8×49/4 = (8×49)/4:
(8×49)/4
8/4 = (4×2)/4 = 2:
2×49
2×49 = 98:
Answer: 98
it is infintely many i think so
Answer:
The required recursive formula is:
Step-by-step explanation:
We are given a geometric sequence as:
6,-18,54,-162,.....
Clearly after looking at different terms of the sequence we could observe that the sequence is an geometric progression (G.P.) with common ratio= -3 denoted by r.
Let 
represents the nth term of the sequence.
This means that:
As the common ratio is -3.
so,
Hence, the required recursive formula for the geometric sequence is:
Answer:
Draw a 3-D of a terrain indicating different fault behaviors. Use the terrain below. Be sure that the entire terrain should cover at least three type of fault. Labcl the fault arcas and its part
