Explanation:
A sequence is a list of numbers.
A <em>geometric</em> sequence is a list of numbers such that the ratio of each number to the one before it is the same. The common ratio can be any non-zero value.
<u>Examples</u>
- 1, 2, 4, 8, ... common ratio is 2
- 27, 9, 3, 1, ... common ratio is 1/3
- 6, -24, 96, -384, ... common ratio is -4
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<u>General Term</u>
Terms of a sequence are numbered starting with 1. We sometimes use the symbol a(n) or an to refer to the n-th term. The general term of a geometric sequence, a(n), can be described by the formula ...
a(n) = a(1)×r^(n-1) . . . . . n-th term of a geometric sequence
where a(1) is the first term, and r is the common ratio. The above example sequences have the formulas ...
- a(n) = 2^(n -1)
- a(n) = 27×(1/3)^(n -1)
- a(n) = 6×(-4)^(n -1)
You can see that these formulas are exponential in nature.
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<u>Sum of Terms</u>
Another useful formula for geometric sequences is the formula for the sum of n terms.
S(n) = a(1)×(r^n -1)/(r -1) . . . . . sum of n terms of a geometric sequence
When |r| < 1, the sum converges as n approaches infinity. The infinite sum is ...
S = a(1)/(1-r)
Answer:
1. b ∈ B 2. ∀ a ∈ N; 2a ∈ Z 3. N ⊂ Z ⊂ Q ⊂ R 4. J ≤ J⁻¹ : J ∈ Z⁻
Step-by-step explanation:
1. Let b be the number and B be the set, so mathematically, it is written as
b ∈ B.
2. Let a be an element of natural number N and 2a be an even number. Since 2a is in the set of integers Z, we write
∀ a ∈ N; 2a ∈ Z
3. Let N represent the set of natural numbers, Z represent the set of integers, Q represent the set of rational numbers, and R represent the set of rational numbers.
Since each set is a subset of the latter set, we write
N ⊂ Z ⊂ Q ⊂ R .
4. Let J be the negative integer which is an element if negative integers. Let the set of negative integers be represented by Z⁻. Since J is less than or equal to its inverse, we write
J ≤ J⁻¹ : J ∈ Z⁻
First, let’s all acknowledge that whoever comes up with problems like this WANTS kids to hate math...smh
I’m sure there is a prettier way to solve this, but here’s what I did:
8(2.25) + 3(22.50) =
18 + 67.50 = 85.50 per “set” of balls/jerseys
400/85.50 = 4.678 = number of “sets” he can buy. Round down to 4 so we have room for tax.
85.5 x 4 “sets”= $342
Tax on 342 is 0.06 x 342 = 20.52
$342 + 20.52 = $362.52 spent
Basketballs = 4 sets x 8 balls per set= 32
Jerseys = 4 sets x 3 jerseys per set= 12
32 basketballs, 12 jerseys, $362.52 spent
Answer:
Step-by-step explanation:
2(5)(-5)-(5)^2
=−75
Answer:
Step-by-step explanation:
well it would be about if it got random winners it would be 24759
hope this helps
