Pearson Algebra 1 Common Core, 2011
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Pearson Algebra 1 Common Core, 2011 View details
Chapter Review
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Exercise 71 Page 477

Practice makes perfect
a Let's take a look at the given function.

p(x)=50* 2^xHere, p(x) is the bacteria population after x 30-min periods. To calculate how many bacteria there will be after 2h, we have to start by determining how many 30-min periods there are in 2hours. 2h=120min 120min÷ 30min= 4 There are four 30-min periods in 2hours. Therefore, the bacteria population after 2h is equal to p( 4), the bacteria population after four 30-min periods. Let's substitute x= 4 into the given function and simplify!

p(x)=50* 2^x
p( 4)=50* 2^4
p(4)=50* 16
p(4)=800

There will be 800 bacteria after 2h.

b Let's take a look at the given function.

p(x)=50* 2^x Here, p(x) is the bacteria population after x 30-min periods. To calculate how many bacteria there will be after 1day, we have to start by determining how many 30-min periods there are in 1day. First, note the following. 1h=60min 60min÷ 30min=2 There are two 30-min periods in 1h. Now, recall that 1day is 24hours. 24* 2= 48 There are 48 30-min periods in 1day. Therefore, the bacteria population after 1day is equal to p( 48), the bacteria population after 48 30-min periods. Let's substitute x= 48 into the given function and simplify!

p(x)=50* 2^x
p( 48)=50* 2^(48)
p(48)≈ 1.4* 10^(16)

There will be about 1.4* 10^(16) bacteria after 1day.