Provide the first six terms of the following sequence: t1=81 tn=13tn-1n ≥ 2

81, 27, 9, 3, 1, 1/3

There are six terms listed. The first five are whole numbers and the sixth term is a fraction.

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Explanation:

The equation t1 = 81 means that the first term "t" is 81. The second term is t2 = 27 because we multiply 1/3 by 81, or divide 81 over 3, and we end up with 81/3 = 27 as the second term.

We know to multiply by 1/3 because of the equation tn = (1/3)*t_(n-1) where the "n-1" portion is smaller than 't'. We can write it like this

To get the next term, we multiply by 1/3 again: 27*(1/3) = 27/3 = 9 giving us the third term

And repeat until we have six terms

9*(1/3) = 9/3 = 3 is the fourth term

3*(1/3) = 3/3 = 1 is the fifth term

1*(1/3) = 1/3 is the sixth term

So that is how I got the first six terms to be 81, 27, 9, 3, 1, 1/3

answer: d. $20

step-by-step explanation:

just took the test

d. 8

step-by-step explanation:

the given radical expression is:

we can rewrite this as:

we express as square numbers to get;

the exponents will now cancel out to give.

the correct answer is d