1. A town has a population of 100,000 now and the population is increasing 10% every year. What will be the population in 5 years?
2. How many poles will be in a pile of telephone poles if there are 30 in the bottom layer, 29 in the second, and so on, until there is one in the top layer?
3. How far up and down will a ball travel before stopping if it is dropped from a height f 12 m, and each rebound is 1/3 of the previous distance.
#1: 100000 * 1.1^5
#2: Sum of the first n integers is n(n+1)/2
#3 12 + 2*4/(1 - 1/3)
1. To find the population in 5 years, we need to calculate the increase in population for each year and then add it to the initial population.
First, let's find the increase in population for one year. We know the population is increasing by 10% each year, which means it will be multiplied by 1.10.
So, the population after one year will be 100,000 * 1.10 = 110,000.
Since we want to find the population after 5 years, we have to repeat this process four more times:
Year 1: 110,000 * 1.10 = 121,000
Year 2: 121,000 * 1.10 = 133,100
Year 3: 133,100 * 1.10 = 146,410
Year 4: 146,410 * 1.10 = 161,051
Finally, for Year 5:
Year 5: 161,051 * 1.10 = 177,156.1
Therefore, the population in 5 years will be approximately 177,156.
2. This problem involves finding the sum of an arithmetic sequence. To find the number of poles in the pile, we need to sum the number of poles in each layer.
We are given that in the bottom layer, there are 30 poles. In the second layer, there are 29 poles, and so on. The pattern suggests that we have a decreasing arithmetic sequence.
We can calculate the sum using the formula for the sum of an arithmetic sequence:
Sn = (n/2) * (a + L)
Where Sn is the sum, n is the number of terms, a is the first term, and L is the last term.
In this case, the number of terms is equal to the number of layers, which is 30.
Using the given information, we can substitute the variables into the formula:
Sn = (30/2) * (30 + 1)
Simplifying:
Sn = 15 * 31
Sn = 465
Therefore, there will be 465 poles in the pile.
3. To find how far the ball travels before stopping, we need to sum the geometric series formed by the rebounds.
Given that the ball is dropped from a height of 12 meters and each rebound is 1/3 of the previous distance, we can interpret this as a decreasing geometric sequence.
The formula for the sum of a finite geometric series is:
Sn = a * (1 - r^n) / (1 - r)
Where Sn is the sum, a is the first term, r is the common ratio, and n is the number of terms.
In this case, the first term is 12 meters, the common ratio is 1/3, and we need to find the number of terms.
Let's calculate the number of terms using the formula for the sum of an infinite geometric series:
Sn = a / (1 - r)
12 = a / (1 - 1/3)
12 = 3a / (2/3)
12 * (2/3) = 3a
8 = 3a
a = 8 / 3
Now we can calculate the sum of the series:
Sn = (8/3) * (1 - (1/3)^n) / (1 - 1/3)
To determine the number of terms, we need to find how many rebounds occur before the ball stops bouncing. Each rebound is 1/3 of the previous distance, so the height of each rebound is (1/3)(previous height).
Given that the first term is 12 meters, the height of the first rebound is (1/3)(12) = 4 meters. For each subsequent rebound, the height decreases by a factor of 1/3.
To determine when the ball stops, we need to find the point at which the height becomes negligible, meaning it is less than some threshold (for example, 0.01 meters).
Let's say the height threshold is 0.01 meters. We can set up an equation to solve for the number of terms:
(1/3)^n * (8/3) < 0.01
Simplifying the inequality:
(1/3)^n < 0.01 * (3/8)
(1/3)^n < 0.00375
To solve for n, we can use logarithms:
n * log(1/3) < log(0.00375)
n > log(0.00375) / log(1/3)
Using a calculator, we can find that n > 4.81. Since n represents the number of rebounds, we'll take the next whole number, n = 5.
Now, let's substitute the values into the formula for the sum of the geometric series:
Sn = (8/3) * (1 - (1/3)^5) / (1 - 1/3)
Calculating:
Sn ≈ (8/3) * (1 - 1/243) / (2/3)
Sn ≈ (8/3) * (242/243) / (2/3)
Sn ≈ (8/3) * (242/243) * (3/2)
Sn ≈ (8 * 242) / 243
Sn ≈ 1936 / 243
Therefore, the ball will travel approximately 7.97 meters up and down before stopping.