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Q: How does this sequence finish 3 9 81?

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81

81

Divide by 3

3-1-.33 repeating

243

81/9 = 9

81

It is 3 followed by 1

81 Each number is 3 x the previous number. So, 3 x 27 = 81.

81/9 = 9

The general term for the sequence 3, 9, 27, 81, 243, . . . is:

First divide 3 into 81 81 ÷ 3 = 27 so 81 = 3 × 27 Next divide 3 into 27 27 ÷ 3 = 9 so 27 = 3 × 9 and hence 81 =3 × 27 = 3 × 3 × 9 But 9 = 3 × 3 and thus 81 = 3 × 3 × 9 = 3 × 3 × 3 × 3 = 34

9 * 9 = 81 and3*3*3*3 = 81

81/12 = 8 times with a remaider of 9

9 243 - 81 = 162 81 - 27 = 54 162 = 54 * 3 54 = 18 * 3 27 - 18 = 9

The answer is 1,3,9,27,81. 81*1=81 3*27=81 9*9=81 27*3=81 81*1=81

1, 3, 9, 27, 81. 1 x 81 = 81, 3 x 27 = 81, 9 x 9 = 81.

The first four terms are 3 9 27 81 and 729 is the 6th term.

9*9 = 81 square cm

1 times 81 3 times 27 9 times 9

This sequence is the squaring of the numbers 1, 3, 5, and 7. The next number is 9^2 = 81.

Yes, because 9x9=81 9 divides into 3 three times 3x3=9 So if 9 is a factor of 81, then so must 3. 81/3= 27

1, 3, 9, 27, 81, -1, -3, -9, -27, -81

1, 3, 9, 27, 81: 1 x 81, 3 x 27, 9 x 9.

27