Core Connections Integrated II, 2015
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Core Connections Integrated II, 2015 View details
3. Section 1.3
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Exercise 103 Page 54

Practice makes perfect
a From the exercise, we know that there was a total of 109 000 votes cast. If we let B and C represent the number of votes for the two candidates, we can write the following equation for the total number of votes cast.
B+C=106 000We also know that candidate C received 15 000 fewer votes than candidate B. Therefore, if we subtract 15 000 from B, we would get an equation describing the number of votes for candidate C. C=B-15 000 If we substitute the expression for C into the first equation, we can solve for B.
B+C=106 000
B+( B-15 000)=106 000
B+B-15 000=106 000
Solve for B
2B-15 000=106 000
2B=121 000
B=60 500
Candidate B received 60 500 votes.
b From the exercise we know that the price of coffee beans has increased by 10 % per year, which means the cost is increasing exponentially. This means we can describe the price with an exponential function.
y=ab^xIn this equation, a is the initial value and b is the rate of change. From the exercise we know that the current cost is $14, which means a= 14. We also know that the price increases by 10 % each year, which means b= 1.1. y= 14( 1.1)^x To find the cost two years ago, we should substitute x=-2 into the equation and simplify.
y=14(1.1)^x
y=14(1.1)^(- 2)
Evaluate right-hand side
y=14(0.82644...)
y=11.57016...
y≈ 11.57
Two years ago the cost was about $11.57.
c Like in Part B, we have a percentage change, which means we can model the situation with an exponential model.
y=ab^xFrom the exercise we know that the current cost is $29 000, which means a= 29 000. We also know that the price depreciates by 11 % annually, which means b= 0.89 y= 29 000( 0.89)^x To find the car's value in four years, we should substitute x=4 into the equation and simplify.
y=29 000(0.89)^x
y=29 000(0.89)^4
Evaluate
y=29 000(0.627422...)
y=18 195.24989...
y≈ 18 195
In four years the value will be about $18 195.
d Since both women write problems at a constant rate, we can describe the number of problems they write with a linear equation.
y=mx+b In this equation, m is the slope and b is the y-intercept. Since Ms. B started with 10 problems and writes 6 problems an hour, we know that b= 10 and m= 6. Ms. B y= 6x+ 10 Also, we know that Ms. D writes 10 problems an hour but does not start of with any problems. This means m= 10 and b= 0. Notice that since the y-intercept is 0, we do not have to write that out. Ms. D y= 10x By equating the functions we can solve for the number of hours it takes, x, before they have written the same number of problems.
y=6x+10
10x=6x+10
4x=10
x=2.5
In 2.5 hours they will have written the same number of exercises.