How many times as many circles will there be at stage 21 compared to stage 20
1 answer:
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Answer:
1. 13, 18, 23, 28, 33, 38, 43, 48, 53
<em>Conjecture:</em> Each term is 5 more than the previous term.
2. 512, 256, 128, 64, 32, 16, 8, 4, 2
<em>Conjecture:</em> Each term is half of the previous term.
3. 1, 8, 27, 64, 125, 216, 343, 512, 729
<em>Conjecture:</em> Each term is a cube. To find the nth term, cube n.
4. 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31
<em>Conjecture:</em> Each term is the next consecutive prime number.
5. <u><em>Correct as is</em></u>
6. 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89
<em>Conjecture:</em> Each term is the sum of the 2 previous numbers.
Step-by-step explanation:
1.
Given: 13, 18, 23, 28
<em>We add 5 to each number</em>
13 + 5 = 18
18 + 5 = 23
23 + 5 = 28
28 + 5 = 33
33 + 5 = 38
38 + 5 = 43
43 + 5 = 48
48 + 5 = 53
2.
Given : 512, 256, 128, 64
<em>Divide each number by 2</em>
512 ÷ 2 =256
256 ÷ 2 = 128
128 ÷ 2 = 64
64 ÷ 2 = 32
32 ÷ 2 = 16
16 ÷ 2 = 8
8 ÷ 2 = 4
4 ÷ 2 = 2
3.
Given: 1, 8, 27, 64
<em>We have to cube n:</em>
<u><em>n³ = consecutive number cubed</em></u>
<u><em></em></u>
1³ = 1 or 1 × 1 × 1
2³ = 8 or 2 × 2 × 2
3³ = 27 or 3 × 3 × 3
4³ = 64 or 4 × 4 × 4
5³ = 125 or 5 × 5 × 5
6³ = 216 or 6 × 6 × 6
7³ = 343 or 7 × 7 × 7
8³ = 512 or 8 × 8 × 8
9³ = 729 or 9 × 9 × 9
4.
Consecutive prime number: <u><em>Consecutive prime numbers are those that have no gaps or prime numbers between them.</em></u>
Look at the picture in the link:
6.
Given: 1, 1, 2, 3, 5, 8
<em>We must add the two previous numbers to get the next term:</em>
1 + 1 = 2
1 + 2 = 3
2 + 3 = 5
3 + 5 = 8
5 + 8 = 13
8 + 13 = 21
13 + 21 = 34
21 = 34 = 55
34 + 55 = 89
Step-by-step explanation: