I did all the work for these so I'm not cheating I'm just checking my answers.
(This got removed for some reason so this is my 2nd time posting this)
Geoff planted Dailies in his garden. They have bulbs that divide and reproduce underground. In the first year, they produced 6 bulbs. Second year, roduced 12, third year produced 24. how many bulbs should Geoff expect inn the 8th year?
A. 48
B. 384
C. 768
D. 1536
What are the first four terms of the squence represented by the expression n(n-2)-3?
A. -5, -2, 1, 4
B. -4, -3, 0, 5
C-3, 0, 3, 6
D-2, 0, 2, 4
I did post my answers but for some reason it got taken off even though I didn't say anything that wasn't related to math. Nobody even responded to it.
I also forgot to put what my answers were in the question lol
1. 768
2. B
To solve the first problem, we need to identify the pattern in how the number of bulbs is increasing each year. From the information given, we can observe that the number of bulbs is doubling each year. In other words, each year the number of bulbs is multiplied by 2.
Given that in the first year there are 6 bulbs, we can use the equation 6 * 2^(n-1) to represent the number of bulbs in the nth year, where n is the number of years.
To find the number of bulbs in the 8th year, we substitute n = 8 into the equation:
6 * 2^(8-1) = 6 * 2^7 = 6 * 128 = 768
Therefore, the answer is C. Geoff should expect 768 bulbs in the 8th year.
For the second problem, we need to evaluate the expression n(n-2)-3 for the first four terms.
Substituting n = 1 into the expression:
1(1-2)-3 = -2-3 = -5
Substituting n = 2 into the expression:
2(2-2)-3 = 2(0)-3 = 0-3 = -3
Substituting n = 3 into the expression:
3(3-2)-3 = 3(1)-3 = 3-3 = 0
Substituting n = 4 into the expression:
4(4-2)-3 = 4(2)-3 = 8-3 = 5
Therefore, the first four terms of the sequence represented by the expression are -5, -3, 0, 5.
The correct answer is A.