Answer:
Let the number of digits be n and the number of elements in set be s.
<h3>When n = 1</h3>
- The set contains 1-digit numbers, 1 through 9,
- The set consists of 10 - 1 = 9 numbers.
<h3>When n = 2</h3>
- The set contains 2-digit numbers, 10 through 99,
- The set contains 100 - 10 = 90 numbers.
<h3>When n = 3</h3>
- The set contains 3-digit numbers, 100 through 999,
- The set contains 1000 - 100 = 900 numbers.
The pattern we see helps us determine the relationship between s and n as follows.
When set contains n-digit numbers, the set contains:
- s = 10ⁿ - 10ⁿ⁻¹ = 10ⁿ⁻¹(10 - 1) = 9*10ⁿ⁻¹ elements
We have s known, substitute it into equation above and solve for n:
- 900000000 = 9*10ⁿ¹
- 100000000 = 10ⁿ⁻¹
- 10⁸ = 10ⁿ⁻¹
- n - 1 = 8
- n = 9
The numbers in the set s are 9-digit long.
5 years, you have to make 6% a decimal so it would be 0.06. You would then multiply 2000 and 0.06 and get 120. You keep adding 120 to 2,000 until you reach 2,600. Hope this helped!
Answer:
one solutions
Step-by-step explanation:
graph both points on a graph if they cross you have one solution, if they never meet you have no solution and if it's one line you have infinite solutions
