McGraw Hill Glencoe Algebra 2, 2012
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McGraw Hill Glencoe Algebra 2, 2012 View details
6. Rational Exponents
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Exercise 52 Page 427

Practice makes perfect
a To find the number of deer after 4 12 years we will substitute 4 12 in the given formula. Our next step will be to rewrite this mixed number as a fraction. Let's start.
D=100* 2^(t2)
D=100* 2^(4 122)
D=100* 2^(. 92 /2.)
D=100*2^(94)
Next, we will rewrite the fraction in the exponent using the radicals.
D=100*2^(94)

a^(mn)=sqrt(a^m)

D=100*sqrt(2^9)
D=100*sqrt(512)
D=100*4.7568...
D=475.68...
D≈ 475
Notice that in this case we are rounding to the nearest integer that is not greater than the rounding number. After 4 12 years, there will be approximately 475 deer in this population.
b We are asked to make a table that charts this population of deer every year for the next five years. Therefore, we should substitute each integer between 1 and 5 in the given formula and evaluate. Let's start with t=1.
D=100*2^(t2)
D=100*2^(12)
D=100*sqrt(2)
D=141.4213...
D≈ 141
After the first year there are 141 deer in the population. We will do the same with the rest of the time period.
Year, t D=100*2^(t2) The number of deer, D
1 D=100*2^(12) ≈ 141
2 D=100*2^(22) 200
3 D=100*2^(32) ≈ 282
4 D=100*2^(42) 400
5 D=100*2^(52) ≈ 565
c To make a graph showing the changes in the number of deer in this population, we will use the table we made in Part B. Since we know the population at the beginning is 100 deer, the starting point of this graph will be at ( 0, 100). Let's plot the points we found and connect them with a smooth curve.
d Looking at the graph made in Part C, we can see that the number of deer seems to increase faster and faster each year. Thinking critically about animal populations, we know that this is not reasonable. There is always some limiting factor, such as available food, water, or territory.