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anyanavicka [17]
3 years ago
7

What value is equivalent to: (8*4*2/8*7)^2 x (8^0/7^-3)^3 x 7^-9???

Mathematics
1 answer:
ankoles [38]3 years ago
6 0

(8*4*2/8*7)^2 x (8^0/7^-3)^3 x 7^-9

Simplify the equations:

(8*4*2/8*7)^2 = (64/56)^2 = (8/7)^2

(8^0/7^-3)^3 = (1/7^-3)^3 = 7^9

Now you have :

(8/7)^2 x 7^9 x 7^-9

Multiply 7^9 x 7^-9 by adding the exponents to get 7^0, which = 1

Now you have:

(8/7)^2 x 1

(8/7)^2 = (8x8) / (7x7) = 64/49

64/49 x 1 = 64/49

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Answer:

The 29th term is

<h2>243</h2>

Step-by-step explanation:

The above sequence is an arithmetic sequence

For an nth term in an arithmetic sequence

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where n is the number of terms

a is the first term

d is the common difference

From the question

a = - 121

d = -108 -- 121 = - 108 + 121 = 13

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The 29th term of the sequence is

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= -121 + 28(13)

= -121 + 364

The final answer is

<h2>243</h2>

Hope this helps you

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