SOPHIA COLLEGE ALGEBRA MILESTONE 5 – ALL ANSWERS !!!

SOPHIA COLLEGE ALGEBRA MILESTONE 5 – ALL ANSWERS !!! You passed this Milestone 21 questions were answered correctly. 1 question was answered incorrectly. 1 Consider the function . What are the domain and range of this function? • • correct • • RATIONALE A Square root function has the domain restriction that the radicand (the value underneath the radical) cannot be negative. To find the specific domain, construct an inequality showing that the radicand must be greater than or equal to zero. The expression under the radical, , must be greater than or equal to zero. To solve this inequality, add to both sides to undo the subtraction of . This tell us that must be greater than or equal to . In other words, must be less than or equal to . We can write this inequality in the other direction. This is the domain of the function, which means all values must be less than or equal to . To find the range, consider the fact that it is not possible for the input of the function to be a negative number. For all x-values less than or equal to , the function will have non-negative values for y that only get bigger and bigger as x increases. The range is all values greater than or equal to zero. CONCEPT Finding the Domain and Range of Functions 2 Kevin examines the following data, which shows the balance in an investment account. Year Balance 1 $5,000.00 2 $5,100.00 3 $5,202.00 4 $5,306.04 5 $5,412.16 What is the formula for the geometric sequence represented by the data above? • • • • correct RATIONALE This is the general formula for a geometric sequence. We will use information in the table to find values for and . Let's start with finding , the value of the first term. The first term, which is , so will be replaced by in the formula. Next, let's find , the common ratio. To find , take the value of any term, and divide it by the value of the previous term to find the common ratio. For example, so . Finally, plug in values for and into the geometric sequence formula. This is the formula for the geometric sequence. CONCEPT Introduction to Geometric Sequences 3 Find the solution for in the equation . • • • • correct RATIONALE To solve this equation, begin by dividing both sides by to cancel the coefficient in front of the exponential. divided by is equal to . To undo the variable exponent, apply a logarithm to both sides. If you apply a log to one side, you must apply a log to other side. Next, apply the Power Property of Logs which states that exponents inside a logarithm can be expressed as outside scalar multiples of the logarithm. This means that can be expressed as . Now divide both sides by to isolate . To solve, you can use a calculator to evaluate and . is approximately equal to and is approximately equal to . Finally, divide to solve for . This is the solution to the equation . CONCEPT Solving Exponential Equations using Logarithms 4 Select the graph of . • correct • • • RATIONALE CONCEPT Graph of an Exponential Equation 5 Suppose $20,000 is deposited into an account paying 4.5% interest, compounded annually. How much money is in the account after four years if no withdrawals or additional deposits are made? • $24,350.52 • $23,850.37 correct • $24,100.00 • $23,600.00 RATIONALE This is the general equation for compounding interest, where is the principal balance, is the annual percentage rate (APR), is time (in years), and is the number of times per year interest is compounded. Use the information provided to plug in the values for each variable. In this case, the principal balance is , the APR is (remember to express the percentage as a decimal, ), the time is years, and it is compounded one time a year, so is . Next, evaluate the right side of the equation. Because the interest is compounded only once a year, the fraction and exponent are easy to simplify. Evaluate the addition in inside the parentheses. plus is equivalent to . Now, apply the exponent. to the power of is equal to . Finally, multiply this value by . The account will have a balance of $23,850.37. CONCEPT Compound Interest 6 Find the sum of the following geometric series. • • • • correct RATIONALE In the above series, the first term is so we will substitute for in the sum of an infinite geometric sequence formula. Next, let's determine , the common ratio. To find , divide the value of any term by the value of the term before it to find the common ratio. For example, so . We can now substitute the values in for and to solve for . To find the sum, we will first evaluate the denominator. minus simplifies to . Finally, evaluate the division to find the value of the sum. divided by equals , which is the sum of the infinite geometric sequence. CONCEPT Sum of an Infinite Geometric Sequence 7 Consider the function . Find the formula for the inverse function. • • • correct • RATIONALE To find the inverse of a function, you can write the function as , swap the variables and , and then rewrite the equation with on one side. First, start by swapping with . Here is the function written as an equation where Next, we will swap the variables, and . . Now that the variables are swapped, we will manipulate this equation to place on one side of the equation. We'll start by squaring both sides to undo the radical. When a square root is squared, the result is the expression under the radical. Next, we will subtract from both sides to undo the addition of . Multiply both sides by to undo the in front of . We have now isolated the variable to one side of the equation, which results in the inverse function. This is the inverse of . CONCEPT Finding the Inverse of a Function 8 Suppose $12,000 is deposited into an account paying 5.5% interest, compounded continuously. How much money is in the account after five years if no withdrawals or additional deposits are made? • $15,683.52 • $15,300.00 • $15,798.37 correct • $15,512.31 RATIONALE This is the general equation for compounding interest. is the principal (or beginning) balance, is the annual interest rate, and is time (in years). The number is approximately . Using the information provided, plug in the appropriate values into the formula. The account starts with , which gives us the value for . The account earns interest at a rate of or as a decimal. This gives us the value for . We want to know the value of the account after five years, so is . To evaluate, first simplify the exponent. times equals . Next, either use the button on your calculator or the approximation of and raise this value to the power. to the power of is equivalent to . Finally, multiply by . The account has a balance of $15,798.37 after 5 years. CONCEPT Continuously Compounding Interest 9 Solve the following logarithmic equation. • • correct • • RATIONALE Logarithms and exponents are inverse operations. We can use the base of the logarithm, , as a base to an exponent, and place the logarithmic expression as an exponent in the equation. We'll have to do this to both sides of the equation. Here, we used as a base number on both sides of the equation. When we do this, the logarithm and exponent will cancel each other out. On the left side, the logarithm and exponent cancel each other out, leaving only . On the right side, is equal to . Finally, divide both sides by to solve for . The solution to the equation is . CONCEPT Solving Logarithmic Equations using Exponents 10 Find the solution to the following equation. • • • • correct RATIONALE To solve this equation, first re-write the term on the right side so that both terms have a common base. Since is a power of , we can re-write as . The two terms now have the same base, . Use the properties of exponents to simplify by multiplying the exponents and . times is equal to . Because the bases are the same, we can focus on the exponents and set them equal to each other. Since the exponents are equivalent, we can simply solve for . Add to both sides to undo the on the right. We now have the term isolated on the left side. Next, divide both sides by to solve for x. Once we divide by , we have isolated x on the left side. However, we can still simplify this fraction. is equivalent to . The solution for is . CONCEPT Solving an Exponential Equation 11 George monitors the number of influenza infections reported in a certain neighborhood in a given week. The recent numbers are shown in this table: Week Number of People 0 40 1 48 2 58 3 69 According to his reports, the reported infections are growing at a rate of 20%. If the number of infections continues to grow exponentially, what will the number of infections be in week 8? • 172 people correct • 384 people • 256 people • 203 people RATIONALE In general, exponential growth is modeled using this equation. We will use information from the problem to find values to plug into this equation. The initial number of infections is , so this is the value for . The infection rate is , so this is our value for (remember to write it as a decimal). We want to know how many infections there will be in week 8, so we will use for the value for . We will need to solve for . Start by simplifying what's inside the parentheses. plus is . Next, take this value to the power of . to the power of is . Finally, multiply this by There will be 172 people infected in week 8. CONCEPT Exponential Growth 12 Simplify the rational expression by canceling common factors: • • correct • • RATIONALE In order to cancel like terms, we have to first factor both the numerator and denominator, and then look for common factors in the top and bottom of the fraction. Let's take a look at the numerator first. To factor this quadratic, we need to identify a pair of integers whose product is the constant term ( ) and whose sum is the coefficient of the x- term ( ). Two integers that multiply to and add to are and . This means that and make up the integer pair for this quadratic. The original numerator can be written as . Let's take a look at the denominator. Again, we need to identify a pair of integers whose product is the constant term ( ) and whose sum is the coefficient of the x-term ( ). Two integers that multiply to and add to are and . This means that and make up the integer pair for this quadratic. The original denominator can be written as . Now we can rewrite the original fraction with the new factored numerator and denominator. The original fraction is now rewritten with both numerator and denominator written in factored form. To simplify, we cancel factors that appear in both the top and the bottom. The factor can be canceled. The rational expression can be simplified to . CONCEPT Simplifying Rational Expressions 13 Consider the functions and . Find the value of . • • correct • • RATIONALE To evaluate this composite function, focus on the innermost function first. Evaluate first by plugging in for the variable in the function . Once has been replaced with , evaluate the expression. The function evaluates to . To evaluate , use the value of , which is , as the input for the function . Once has been replaced with , evaluate the expression. This tells us that is equal to . CONCEPT Function of a Function 14 Suppose , , and . Find the value of the following expression. • • correct • • RATIONALE This question involves several properties of logarithms. The Quotient Property of Logs states that division inside a logarithm can be expressed as subtraction of individual logarithms. This means we can express as . Next, the Product Property of Logs states that multiplication inside a logarithm can be expressed as addition of individual logarithms. This means we can express as . Then, the Power Property of Logs states that exponents inside a logarithm can be expressed as outside scalar multiples of the logarithm. CONCEPT Applying Properties of Logarithms 15 Write the following as a single rational expression. • • • correct • RATIONALE Just as with numeric fractions, we can subtract algebraic fractions. To do so, we need to rewrite the fractions so that they have a common denominator. One strategy is to multiply the two denominators together. But if we change the denominator, we have to also change the numerator to keep things equal. Multiply the numerator and denominator of the first fraction by the denominator of the second fraction; repeat for the second fraction. We now have an equivalent expression with common denominators. Before we subtract across the numerators, let's simplify each numerator. times is equal to and times is equal to . We can now subtract across the numerators. The numerator is and the denominator is . Finally, simplify the numerator. This is the simplified rational expression. CONCEPT Adding and Subtracting Rational Expressions 16 Suppose and . Find the value of . • • • correct • RATIONALE To evaluate , evaluate and separately, and then multiply the values together. Start by evaluating first by substituting for in the given function. is plugged in for all instances of in the function . Next, evaluate the exponent. squared is . Then, add and together. plus is . Repeat this process for by substituting for in the given function. is plugged in for all instances of in the function . Next, evaluate the multiplication. times is . Then, add and together. plus is . Finally, multiply the results of and together. We know that and . Multiply these values together. times is . CONCEPT Multiplying and Dividing Functions 17 Write the following as a single rational expression. • correct • • • RATIONALE Just as with numeric fractions, we can re-write division of algebraic fractions as multiplication and multiply across numerators and denominators. To re-write fraction division as multiplication, re-write the second fraction as its reciprocal (flipping the numerator and denominator). changes to and division changes to multiplication. We can now multiply across the numerators and denominators. times is equal to and times is equal to . Next, find any common factors in the numerator and denominator. Both the numerator and denominator have a factor of . We can cancel out these factors and simplify. Once all common factors have been canceled out in the numerator and denominator, write the fraction in simplest form. This is the the simplified fraction written as a single rational expression. CONCEPT Multiplying and Dividing Rational Expressions 18 Find the sum of the first 10 terms of the following geometric sequences: • • • • correct RATIONALE This is the formula to find the sum of a finite geometric sequence. We will use information from the given sequence to find values for , , and . Let's start by finding , the value of the first term. In the above sequence, the first term is . So we will substitute for in the sum of a geometric sequence formula. Next, let's determine the variable . To find , divide the value of any term by the value of the term before it to find the common difference. For example, so . Finally, we need to review how many terms we want to consider, which will be . We are asked to find the sum of the first 10 terms, so . We can now substitute values in for , , and to solve for . Once the values for , , and have been plugged into the sum formula, we can simplify the numerator. CONCEPT minus is and minus is . Then, divide the numerator and denominator. The negative values in the numerator and denominator cancel to result in a positive value of . Finally, multiply this by to find the sum. The sum of the first ten terms in the sequence is . Sum of a Finite Geometric Sequence 19 Evaluate the following expression using the properties of logarithms. • correct • • • RATIONALE raised equal to the next to the power of is , so is . We can repeat this process with term, . number, tells us that , equals . raised to some raised to the power of is , so is equal to . Repeat this one more time for the last term, . number, tells us that , equals raised to some raised to the power of is , so is Substitute the calculated in for the log expressions to evaluate. Once the values are substituted, evaluate to addition. The expression evaluates to 7. CONCEPT Introduction to Logarithms 20 An initial deposit of $1200 is put into an account that earns 5% interest, compounded annually. Each year, an additional deposit of $1200 is added to the account. What is the value of the account after the tenth deposit if no withdrawals or additional deposits are made? • $15,093.47 correct • $12,004.20 • $11,628.90 • $14,624.40 RATIONALE We can use this formula to find the sum of a finite geometric sequence. We will use information from the problem to find values for , r, and n. The initial deposit is ; this is the value for . To find , find the common ratio between two terms. For example, , so is . We want to know the balance of the account after 10 years, so is First, simplify the exponent. minus is and minus is Then, evaluate the division. CONCEPT divided by is . Finally, multiply by . The balance of the account after 10 years will be $15,093.47. Sum of a Geometric Sequence in the Real World 21 The population of a small town is decreasing at a rate of 4% each year. The following table shows a projection of the population, , after years. Year Population 0 7,900 1 7,584 2 7,281 3 6,989 4 6,710 If the population of the small town is currently 7,900 people, how many years will it take for the population to reach 4,300 people? • 13.9 years • 14.9 years correct • 15.5 years • 14.5 years RATIONALE minus is . Next, divide both sides by . We are now left with an exponential expression on the right side. To undo the variable exponent, take the log of both sides. Once we have taken the log of both sides, apply the Power Property of Logs, which allows exponents inside logs to be written as outside factors. Since is outside the log expression, divide both is now isolated. To evaluate, use a calculator to find the value of and . CONCEPT is equal to and is equal to . Finally, divide. It will take 14.9 years for the population to reach 7900. Exponential Decay 22 To collect data on the signal strengths in a neighborhood, Tracy must drive from house to house and take readings. She has a graduate student, Dave, to assist her. Tracy figures it would take her eight hours to complete the task working alone, and that it would take Dave 12 hours if he completed the task by himself. How long will it take Tracy and Dave to complete the task together? • 4.8 hours • 6.0 hours correct • 3.6 hours • 7.2 hours RATIONALE We can use the relationship between work, rate, and time to build equations to solve this problem. To begin, we need to identify the relationship for Tracy and Dave, separately. Tracy's rate is 1 task in 8 hours. We do not know the time it will take them to work on the project together, so we will just denote the time as . When we multiply the rate and time together, we get Tracy's work at (notice how the units for hours cancel). We can repeat this process for Dave. Dave's rate is 1 task in 12 hours. We do not know the time it will take them to work on the project together, so we will just denote the time as . When we multiply the rate and time together, we get Dave's work at . Now we can create a rational equation by adding Tracy and Dave's work together. The work is 1 task, so their combined work is equal to 1 task. In the rational equation, is the amount of time it takes Tracy and Dave to complete the task together. With rational equations, we can add them together once they have common denominators. A simple way to find common denominators is to multiply the denominators of the fractions together. Make sure that you are doing this to all terms to ensure they all end up with the same denominator. Next, evaluate the multiplication. Now we have three fractions with common denominators. Once we have this, we can solve for using the expressions in the numerator. To solve for , first combine like terms. plus is equal to . Finally divide both sides by to solve for . It will take Tracy and Dave 4.8 hours to complete the task together.