Every Saturday, Justin gets an allowance of $10 from his parents. During every week since Justin began getting an allowance, he spends $8 and saves the remaining $2 of his allowance before he receives his next allowance. The sequence shown represents the amount of money Justin has right before and right after he gets his allowance for six consecutive weeks, starting with $0.
0, 10, 2, 12, 4, 14, 6, 16, 8, 18, 10, 20
Explain why this sequence is or is not a function?
A.
This sequence is not a function because the amount $10 appears in the sequence more than once.
B.
This sequence is not a function because the sequence does not form a line.
C.
This sequence is a function because each input (odd terms in sequence) maps to one output (even terms in sequence).
D.
This sequence is a function because each output (even terms in sequence) maps to one input (odd terms in sequence).
the whole question is poorly stated.
each term in the sequence is an output: the savings balance.
so is it D
is it D
1. What are the next three terms in the sequence?
–1, 9, 19, 29, …
38, 37, 32
40, 51, 62
39, 49, 59
38, 47, 56
The sequence is increasing by 10 in each term.
Therefore, the next three terms are 39, 49, 59.
So, the answer is (C) 39, 49, 59.
2. Geoff planted dahlias in his garden. Dahlias have bulbs that divide and reproduce underground. In the first year, Geoff’s garden produced 8 bulbs. In the second year, it produced 16 bulbs, and in the third year it produced 32 bulbs. If this pattern continues, how many bulbs should Geoff expect in the sixth year?
64 bulbs
512 bulbs
128 bulbs
256 bulbs
The pattern of the number of bulbs is that it is doubling each year.
So, in the fourth year, the number of bulbs will be 64 (32 x 2).
In the fifth year, the number of bulbs will be 128 (64 x 2).
And, in the sixth year, the number of bulbs will be 256 (128 x 2).
Therefore, Geoff should expect 256 bulbs in the sixth year.
So, the answer is (D) 256 bulbs.
3. What are the first four terms of the sequence represented by the expression n(n – 2) – 3?
–5, –2, 1, 4
–4, –3, 0, 5
–3, 0, 3, 6
–2, 0, 2, 4
If we substitute n = 1, 2, 3, and 4 in the expression, we get:
n(n – 2) – 3 = (1)(–1) – 3 = –4
n(n – 2) – 3 = (2)(0) – 3 = –3
n(n – 2) – 3 = (3)(1) – 3 = 0
n(n – 2) – 3 = (4)(2) – 3 = 5
Therefore, the first four terms of the sequence are –4, –3, 0, 5.
So, the answer is (B) –4, –3, 0, 5.
4. A car traveling at 23 mi/h accelerates to 46 mi/h in 5 seconds. It maintains that speed for 5 seconds and then slows to a stop in 5 seconds. Which graph shows the car’s speed over time?
(Image: A number of points are shown on a graph. The points are connected by line segments.)
(Image: A number of points are shown on a graph. The points are connected by line segments.)
(Image: A number of points are shown on a graph. The points are connected by line segments.)
(Image: A number of points are shown on a graph. The points are connected by line segments.)