Suppose that the population of a town is described by P=0.16t^2=7.2t=100, where P is the population in thousands and t is the time in years, (
with t=o representing the year 2000).
A. what will the population be in 2010?
B. what was the population in 1995?
C. when will the population reach 52,000?
For(a) I got P=188,OOO but i do not get the answers for Band C, pLS HELP ME.
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To find the population in 2010, we need to substitute t=10 into the equation P=0.16t^2+7.2t+100:
P = 0.16(10)^2 + 7.2(10) + 100
P = 0.16(100) + 72 + 100
P = 16 + 72 + 100
P = 188
So the population in 2010 will be 188,000 (in thousands).
To find the population in 1995, we need to substitute t=-5 into the equation:
P = 0.16(-5)^2 + 7.2(-5) + 100
P = 0.16(25) - 36 + 100
P = 4 - 36 + 100
P = 68
So the population in 1995 was 68,000 (in thousands).
To find when the population will reach 52,000, we need to solve the equation P = 0.16t^2 + 7.2t + 100 for t:
0.16t^2 + 7.2t + 100 = 52
0.16t^2 + 7.2t + 100 - 52 = 0
0.16t^2 + 7.2t + 48 = 0
To solve this quadratic equation, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 0.16, b = 7.2, and c = 48. Plugging these values into the quadratic formula:
t = (-7.2 ± √(7.2^2 - 4(0.16)(48))) / (2(0.16))
Calculating the expression under the square root:
t = (-7.2 ± √(51.84 - 3.84)) / 0.32
t = (-7.2 ± √48) / 0.32
Now we have two possible values for t:
t = (-7.2 + √48) / 0.32 ≈ 8.46
t = (-7.2 - √48) / 0.32 ≈ -15.46
Since time cannot be negative, we discard the negative value and conclude that the population will reach 52,000 (in thousands) approximately 8.46 years after the year 2000.
To find the answers to parts B and C, we need to use the given population equation: P = 0.16t^2 + 7.2t + 100, where P is the population in thousands and t is the time in years since 2000.
Let's solve each part step by step:
A. Population in 2010:
To find the population in 2010, we need to substitute t = 10 into the population equation:
P = 0.16(10)^2 + 7.2(10) + 100
P = 0.16(100) + 72 + 100
P = 16 + 72 + 100
P = 188
So, the population in 2010 will be 188,000.
B. Population in 1995:
To find the population in 1995, we need to determine the value of t that corresponds to the year 1995. Since t represents the time in years since 2000, we need to find the difference between the years 2000 and 1995.
Difference in years = 2000 - 1995 = 5
Substitute t = -5 into the population equation to find the population in 1995:
P = 0.16(-5)^2 + 7.2(-5) + 100
P = 0.16(25) - 36 + 100
P = 4 - 36 + 100
P = 68
So, the population in 1995 was 68,000.
C. Population reaching 52,000:
To find when the population will reach 52,000, we need to set the population equation equal to 52:
0.16t^2 + 7.2t + 100 = 52
Rearranging the equation and setting it equal to zero:
0.16t^2 + 7.2t + 100 - 52 = 0
0.16t^2 + 7.2t + 48 = 0
To solve the quadratic equation, we can either factor it or use the quadratic formula. In this case, factoring is not straightforward, so let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac))/(2a)
Using the values: a = 0.16, b = 7.2, c = 48:
t = (-7.2 ± √(7.2^2 - 4(0.16)(48)))/(2(0.16))
t = (-7.2 ± √(51.84 - 3.072))/(0.32)
t = (-7.2 ± √48.768)/(0.32)
Calculating the square root and simplifying further:
t = (-7.2 ± 6.9825)/(0.32)
Now we have two possible values for t, which represent the possible times when the population reaches 52,000:
1. t = (-7.2 + 6.9825)/(0.32) ≈ 0.24375
2. t = (-7.2 - 6.9825)/(0.32) ≈ -22.51875
Since time (t) cannot be negative in this context, we discard the second solution.
Therefore, the population will reach 52,000 approximately after 0.24375 years (or approximately 0.24375 * 365 days).