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Exercises 1 An arithmetic sequence is characterized by the fact that there is a common difference between consecutive terms and it can be written in the form: an=a1+(n−1)d, This is a linear function. If we think of an as the y values and n as the x values, then we can think of d as the slope, as a1 as the y-intercept. Our function then looks very similar to the slope-intercept form: y=b+(x−1)m. Additionally, in an arithmetic sequence, n must be whole numbers: First term: Second term: Third term: n=1n=2n=3 As with any function where the domain is limited to positive integers, the domain is discrete. In general, an arithmetic sequence looks similar to the graph shown below. | |

Exercises 2 Essentially, we are given four different questions and asked to find the solutions to each one. One of the four questions will provide a different answer than the other three. Let's look at each question individually and then compare.First question Find the slope of the linear function. To find the slope of the function, we need to choose two arbitrary points and use the slope formula. Let's use the first two, (2,10) and (3,13). m=x2−x1y2−y1Substitute (2,10) & (3,13)m=3−213−10Subtract termsm=131a=am=3 The slope of the linear function is 3.Second question Find the difference between consecutive terms of the arithmetic sequence. The x values of the coordinate pairs represent n, the number of the term within the sequence, n=1 is the first term, n=2 is the second term, and so on. Because the values for n are 2, 3, 4, and 5, these are all consecutive terms. We can find the difference by subtracting the y values of the ordered pairs. 13−10=316−13=319−16=3 The difference between each pair of consecutive terms is 3.Third question Find the difference between the terms a2 and a4. To find the difference between a2 and a4, first we need to find the value of these terms. a2 indicates the y-coordinate for the second term, which is 10. Similarly, a4 indicates the y-coordinate for the fourth term, which is 16. The difference is then: 16−10=6.Fourth question Find the common difference of the arithmetic sequence. The common difference is the number of units that separates each pair of consecutive terms in a arithmetic sequence. Because the values for n are 2, 3, 4, and 5, these are all consecutive terms. We can check the difference between each pair. 13−10=316−13=319−16=3 The difference is always 3. Therefore, the common difference d is 3.Conclusion The only question with a different answer was the third, the difference between the second and fourth terms. The slope of the linear function representing an arithmetic sequence is the same thing as the difference between consecutive terms which is just another way to describe the common difference. | |

Exercises 3 In an arithmetic sequence, we find each term by adding the common difference to the previous term. For the given sequence, we need to add 13 to the first term to find the second term. 2⟶+1315… We can continue in the same way to find any number of terms in the sequence. Here we've been asked to find the three terms that come after the first term. This means that we will add 13 to 2, three times. 2⟶+1315⟶+1328⟶+1341 Therefore, the next three terms are: 15, 28, and 41. | |

Exercises 4 In an arithmetic sequence, we find each term by adding the common difference to the previous term. For the given sequence, we need to subtract 6 from the first term to find the second term. 18⟶−612… We can continue in the same way to find any number of terms in the sequence. Here we've been asked to find the three terms that come after the first term. This means that we will subtract 6 from 18, three times. 18⟶−612⟶−66⟶−60 Therefore, the next three terms are: 12, 6, and 0. | |

Exercises 5 By observing the change that occurs between each consecutive term, we can determine the common difference of arithmetic sequence. Here, we can see that the change from one term to the next is adding 5. 13→+518→+523→+528… | |

Exercises 6 By observing the change that occurs between each consecutive term, we can determine the common difference of arithmetic sequence. Here, we can see that the change from one term to the next is subtracting 25. 175→−25150→−25125→−25100… | |

Exercises 7 By observing the change that occurs between each consecutive term, we can determine the common difference of arithmetic sequence. Here, we can see that the change from one term to the next is adding 4. -16→+4-12→+4-8→+4-4… | |

Exercises 8 By observing the change that occurs between each consecutive term, we can determine the common difference of arithmetic sequence. Here, we can see that the change from one term to the next is subtracting 31. 4⟶−31332⟶−31331⟶−313… | |

Exercises 9 By observing the change that occurs between each consecutive term, we can determine the common difference of arithmetic sequence. Here, we can see that the change from one term to the next is subtracting 1.5. 6.5⟶−1.55⟶−1.53.5⟶−1.52… | |

Exercises 10 By observing the change that occurs between each consecutive term, we can determine the common difference of arithmetic sequence. Here, we can see that the change from one term to the next is adding 9. -16⟶+9-7⟶+92⟶+911… | |

Exercises 11 By observing the change that occurs between each consecutive term, we can describe the pattern of the sequence. Here, we can see that the common difference from one term to the next is adding 3. 19→+322→+325→+328 To find the next three terms in the sequence, we will extend this pattern three times. 19→+322→+325→+328→+331→+334→+337… | |

Exercises 12 By observing the change that occurs between each consecutive term, we can describe the pattern of the sequence. Here, we can see that the common difference from one term to the next is adding 11. 1→+1112→+1123→+1134 To find the next three in the sequence, we will extend this pattern three times. 1→+1112→+1123→+1134→+1145→+1156→+1167… | |

Exercises 13 By observing the change that occurs between each consecutive term, we can describe the pattern of the sequence. Here, we can see that the common difference from one term to the next is adding 5. 16→+521→+526→+531 To find the next three in the sequence, we will extend this pattern three times. 16→+521→+526→+531→+536→+541→+546… | |

Exercises 14 By observing the change that occurs between each consecutive term, we can describe the pattern of the sequence. Here, we can see that the common difference from one term to the next is subtracting 30. 60⟶−3030⟶−300⟶−30-30 To find the next three in the sequence, we will extend this pattern three times. 60⟶−3030⟶−300⟶−30-30⟶−30-60⟶−30-90⟶−30-120… | |

Exercises 15 By observing the change that occurs between each consecutive term, we can describe the pattern of the sequence. Here, we can see that the common difference from one term to the next is subtracting 0.3. 1.3⟶−0.31⟶−0.30.7⟶−0.30.4 To find the next three in the sequence, we will extend this pattern three times. 1.3⟶−0.31⟶−0.30.7⟶−0.30.4⟶−0.30.1⟶−0.3-0.2⟶−0.3-0.5… | |

Exercises 16 By observing the change that occurs between each consecutive term, we can describe the pattern of the sequence. Before we do that, let's expand the fractions to have a common factor of 6.FractionExpansionResult 323⋅22⋅264 212⋅31⋅363 313⋅21⋅262 Now, we can see that the common difference from one term to the next is subtracting 61. 65⟶−6164⟶−6163⟶−6162… To find the next three in the sequence, we will extend this pattern three times. 65⟶−6164⟶−6163⟶−6162⟶−6161⟶−610⟶−61-61… | |

Exercises 17 Let's make a table to organize the given sequence terms as ordered pairs.Position (n)Term (an) 14 212 320 428 Now, let's plot these ordered pairs using the horizontal axis for position n and the vertical axis for the term an. | |

Exercises 18 Let's make a table to organize the given sequence terms as ordered pairs.Position (n)Term (an) 1-15 20 315 430 Now, let's plot these ordered pairs using the horizontal axis for position n and the vertical axis for the term an. | |

Exercises 19 Let's make a table to organize the given sequence terms as ordered pairs.Position (n)Term (an) 1-1 2-3 3-5 4-7 Now, let's plot these ordered pairs using the horizontal axis for position n and the vertical axis for the term an. | |

Exercises 20 Let's make a table to organize the given sequence terms as ordered pairs.Position (n)Term (an) 12 219 336 453 Now, let's plot these ordered pairs using the horizontal axis for position n and the vertical axis for the term an. | |

Exercises 21 Let's make a table to organize the given sequence terms as ordered pairs.Position (n)Term (an) 10 2421 39 41321 Now, let's plot these ordered pairs using the horizontal axis for position n and the vertical axis for the term an. | |

Exercises 22 Let's make a table to organize the given sequence terms as ordered pairs.Position (n)Term (an) 16 25.25 34.5 43.75 Now, let's plot these ordered pairs using the horizontal axis for position n and the vertical axis for the term an. | |

Exercises 23 There are two ways for us to determine if the given graph represents an arithmetic sequence.Graphically: Check to see if all of the points lie on the same straight line. Numerically: Observe if the difference between each term in the sequence is constant.Graphically Using a straight edge, try to draw a line that passes through all of the points.There is no one straight line on which all of the points lie. Therefore, the graph does not represent an arithmetic sequence.Numerically To observe the difference between each term in the sequence, let's first organize the given ordered pairs in a table.nan 11 24 31 44 To determine if the difference between one term and the next is constant, let's compare the change between an terms. 1→+34→−31→+34 There is no common difference between these terms. We can therefore conclude that the sequence is not an arithmetic sequence. | |

Exercises 24 There are two ways for us to determine if the given graph represents an arithmetic sequence.Graphically: Check to see if all of the points lie on the same straight line. Numerically: Observe if the difference between each term in the sequence is constant.Graphically Using a straight edge, try to draw a line that passes through all of the points.We see that all of the points on the given graph lie on the same invisible line. Therefore, the graph represents an arithmetic sequence.Numerically To observe the difference between each term in the sequence, let's first organize the given ordered pairs in a table.nan 15 212 319 426 To determine if the difference between one term and the next is constant, let's compare the change between an terms. 5→+712→+719→+726 Here we see that there is a common difference between terms in the sequence. We can therefore conclude that the sequence is an arithmetic sequence. | |

Exercises 25 There are two ways for us to determine if the given graph represents an arithmetic sequence.Graphically: Check to see if all of the points lie on the same straight line. Numerically: Observe if the difference between each term in the sequence is constant.Graphically Using a straight edge, try to draw a line that passes through all of the points.We see that all of the points on the given graph lie on the same invisible line. Therefore, the graph represents an arithmetic sequence.Numerically To observe the difference between each term in the sequence, let's first organize the given ordered pairs in a table.nan 170 255 340 425 To determine if the difference between one term and the next is constant, let's compare the change between an terms. 70→−1555→−1540→−1525 Here we see that there is a common difference between terms in the sequence. We can therefore conclude that the sequence is an arithmetic sequence. | |

Exercises 26 There are two ways for us to determine if the given graph represents an arithmetic sequence.Graphically: Check to see if all of the points lie on the same straight line. Numerically: Observe if the difference between each term in the sequence is constant.Graphically Using a straight edge, try to draw a line that passes through all of the points.There is no one straight line on which all of the points lie. Therefore, the graph does not represent an arithmetic sequence.Numerically To observe the difference between each term in the sequence, let's first organize the given ordered pairs in a table.nan 12 210 316 420 To determine if the difference between one term and the next is constant, let's compare the change between an terms. 1→+810→+616→+420 There is no common difference between these terms. We can therefore conclude that the sequence is not an arithmetic sequence. | |

Exercises 27 When a sequence is arithmetic, the difference between two consecutive terms is constant. Examining our sequence, we can see that this is the case. 13 →+13 26 →+13 39 →+13 52 Therefore, the sequence is arithmetic and the common difference is 13. | |

Exercises 28 When a sequence is arithmetic, the difference between two consecutive terms is constant. Examining our sequence, we can see that this is not the case. 5 →+4 9 →+5 14 →+6 20 Therefore, the sequence is not arithmetic. | |

Exercises 29 When a sequence is arithmetic, the difference between two consecutive terms is constant. Examining our sequence, we can see that this is not the case. 48 →−24 24 →−12 12 →−6 6 Therefore, the sequence is not arithmetic. | |

Exercises 30 When a sequence is arithmetic, the difference between two consecutive terms is constant. Examining our sequence, we can see that this is the case. 87 →−6 81 →−6 75 →−6 69 Therefore, the sequence is arithmetic and the common difference is -6. | |

Exercises 31 Counting the number of smileys in each group, we get the following sequence: 4,6,8,10. To determine if the sequence is arithmetic, we can calculate the difference between consecutive terms and check whether they are the same: a4−a3:a3−a2:a2−a1:10−8=28−6 =26−4 =2 Since each term is 2 greater than the previous one, the sequence is arithmetic. | |

Exercises 32 We are given the sequence of numbers rolled on two dices. 2,5,8,11 In order to determine if it is an arithmetic sequence, we have to inspect the differences between consecutive rolls.Since the differences are all the same and equal to 3, the given sequence is arithmetic. | |

Exercises 33 Explicit equations for arithmetic sequences follow a specific format. an=a1+(n−1)d In this form, a1 is the first term in a given sequence, d is the common difference from one term to the next, and an is the nth term in the sequence. For this exercise, the first term is a1=-5. Let's observe the other terms to determine the common difference d. -5→+1-4→+1-3→+1-2… By substituting these two values into the explicit equation and simplifying, we can find the formula for this sequence. an=a1+(n−1)dd=1, a1=-5an=(-5)+(n−1)(1)Distribute 1an=-5+n−1Subtract termsan=n−6 This equation can be used to find any term in the given sequence. To find a10, the 10th term in the sequence, we substitute 10 for n. an=n−6n=10a10=(10)−6Subtract terma10=4 The 10th term in the sequence is 4. | |

Exercises 34 Explicit equations for arithmetic sequences follow a specific format. an=a1+(n−1)d In this form, a1 is the first term in a given sequence, d is the common difference from one term to the next, and an is the nth term in the sequence. For this exercise, the first term is a1=-6. Let's observe the other terms to determine the common difference d. -6→−3-9→−3-12→−3-15… By substituting these two values into the explicit equation and simplifying, we can find the formula for this sequence. an=a1+(n−1)dd=-3, a1=-6an=(-6)+(n−1)(-3)Distribute -3an=-6−3n+3Add termsan=-3n−3 This equation can be used to find any term in the given sequence. To find a10, the 10th term in the sequence, we substitute 10 for n. an=-3n−3n=10a10=-3(10)−3Multiplya10=-30−3Subtract terma10=-33 The 10th term in the sequence is -33. | |

Exercises 35 Explicit equations for arithmetic sequences follow a specific format. an=a1+(n−1)d In this form, a1 is the first term in a given sequence, d is the common difference from one term to the next, and an is the nth term in the sequence. For this exercise, the first term is a1=21. Let's observe the other terms to determine the common difference d. 21⟶+211⟶+21121⟶+212… By substituting these two values into the explicit equation and simplifying, we can find the formula for this sequence. an=a1+(n−1)dd=21, a1=21an=(21)+(n−1)(21)Distribute 21an=21+21n−21Subtract termsan=21n This equation can be used to find any term in the given sequence. To find a10, the 10th term in the sequence, we substitute 10 for n. an=21nn=10a10=21(10)Multiplya10=5 The 10th term in the sequence is 5. | |

Exercises 36 Explicit equations for arithmetic sequences follow a specific format. an=a1+(n−1)d In this form, a1 is the first term in a given sequence, d is the common difference from one term to the next, and an is the nth term in the sequence. For this exercise, the first term is a1=100. Let's observe the other terms to determine the common difference d. 100→+10110→+10120→+10130… By substituting these two values into the explicit equation and simplifying, we can find the formula for this sequence. an=a1+(n−1)dd=10, a1=100an=100+(n−1)(10)Distribute 10an=100+10n−10Subtract terman=10n+90 This equation can be used to find any term in the given sequence. To find a10, the 10th term in the sequence, we substitute 10 for n. an=10n+90n=10a10=10(10)+90Multiplya10=100+90Add termsa10=190 The 10th term in the sequence is 190. | |

Exercises 37 Explicit equations for arithmetic sequences follow a specific format. an=a1+(n−1)d In this form, a1 is the first term in a given sequence, d is the common difference from one term to the next, and an is the nth term in the sequence. For this exercise, the first term is a1=10. Let's observe the other terms to determine the common difference d. 10→−100→−10-10→−10-20… By substituting these two values into the explicit equation and simplifying, we can find the formula for this sequence. an=a1+(n−1)dd=-10, a1=10an=(10)+(n−1)(-10)Distribute -10an=10−10n+10Add termsan=-10n+20 This equation can be used to find any term in the given sequence. To find a10, the 10th term in the sequence, we substitute 10 for n. an=-10n+20n=10a10=-10(10)+20Multiplya10=-100+20Add termsa10=-80 The 10th term in the sequence is -80. | |

Exercises 38 Since we are already told the sequence is arithmetic, we know that the difference is the same for each pair of consecutive terms. Let's use the first and second terms: d=74−73=71. Next, we can substitute the first term and the common difference into the equation for finding the nth term of an arithmetic sequence. an=a1+(n−1)da1=73, d=71an=73+(n−1)71Distribute 71an=73+71n−71Subtract fractionsan=72+71n To find the 10th term a10, we can substitute n=10 into the equation. an=72+71nn=10a10=72+71⋅10b1⋅a=baa10=72+710Add fractionsa10=712 The tenth term is 712. | |

Exercises 39 The common difference is the difference between each pair of consecutive numbers of the arithmetic sequence, which in our case is -1. This has been correctly identified underneath the sequence. Adding -1 to the previous term, we get the next term of the sequence. Thus, the common difference is -1, not 1. | |

Exercises 40 The formula for the nth term of an arithmetic sequence is: an=a1+(n−1)d. Right off the bat, there was a mistake when writing this formula on the note. To correct it, let's first find the common difference by subtracting the first term from the second one to be sure that it's been calculated correctly: d=22−14=8. Now we can substitute the first term and the common difference into the correct formula and simplify. an=a1+(n−1)da1=14, d=8an=14+(n−1)8Distribute 8an=14+8n−8Subtract terman=6+8n The correct equation for the nth term of the arithmetic sequence is: an=6+8n. | |

Exercises 41 Let's first find the common difference of the arithmetic sequence. It is given that it's is 1.5 times the first term, so we have to multiply 3 by 1.5: d=3⋅1.5=4.5. By adding the common difference to the first term, we get the second term. We can then use this information to find each consecutive term: a1=3a2=7.3+4.5=7.5a3=7.5+4.5=12a4=.12+4.5=16.5. To graph the sequence, we have to make ordered pairs out of the information above. We can do this by writing the position of the term as x and the value of the term as y. Essentially we have (n,an).nan(n,an) a13(1,3) a27.5(2,7.5) a312(3,12) a416.5(4,16.5) If we mark these points on a coordinate plane, we get the graph of the sequence. | |

Exercises 42 We are told that the first row has 10 dominoes. This means that a1=10. We are also told that each row has 2 more than the previous row. This means that the arithmetic sequence is found by adding two each time: a1=10a2=10+2=12a3=12+2=14a4=14+2=16a5=16+2=18 We can write this sequence as: {10,12,14,16,18}. Now we can graph the sequence using the number of the terms as the x-axis and the value of the terms as the y-axis. | |

Exercises 43 aThe figures given are a triangle (3 sides), a square (4 sides), and a pentagon (5 sides). To continue this pattern, we need to draw a hexagon (6 sides), a heptagon (7 sides), and an octagon (8 sides).bTo describe the 20th figure in the sequence, we can write an equation for our arithmetic sequence. We know that with each new figure 1 side is being added, so our common difference d is 1. We also know that the first figure has 3 sides, so our equation becomes: an=a1+(n−1)d⇒an=3+(n−1)1. By substituting 20 for n in the equation, we can learn about our 20th figure. an=3+(n−1)1n=20a20=3+(20−1)1 Simplify RHS Subtract terma20=3+19⋅1a⋅1=aa20=3+19Add terms a20=22 The 20th figure will have 22 sides. This shape is called a doicosagon, which is a very little known fact. | |

Exercises 44 aThe given figures are all circles but with, depending on how you look at it, an increasing number of wedges or an increasing number of dividing lines. For the sake of uniformity, let's consider the wedges rather than the lines. The first three figures have 2, 4, and 6 wedges. To continue this pattern, we will draw circles with 8, 10, and 12 wedges.bTo describe the 20th figure in the sequence, we can write an equation for our arithmetic sequence. We know that with each new figure 2 wedges are being added, so our common difference d is 2. We also know that the first figure has 2 wedges, so our equation becomes: an=a1+(n−1)d⇒an=2+(n−1)2. By substituting 20 for n in the equation, we can learn about our 20th figure. an=2+(n−1)2n=20a20=2+(20−1)2 Simplify RHS Subtract terma20=2+19⋅2Multiplya20=2+38Add terms a20=40 The 20th figure will have 40 wedges. | |

Exercises 45 aTo write a function representing an arithmetic sequence, we need both the first term and the common difference. From the table, we can see that the first term is 5. We can also see that the difference between each pair of consecutive terms, the common difference, is 5. Let's substitute these values in the function for an arithmetic sequence. f(n)=a1+(n−1)da1=5, d=5f(n)=5+(n−1)5Distribute 5f(n)=5+5n−5Subtract termf(n)=5n The function f(n)=5n represents the arithmetic sequence.bFrom the table, we can identify the ordered pairs (n,an) as (1,5), (2,10), (3,15), (4,20). Let's plot them on the coordinate plane.cWe will use the function from the previous part to find the value of n when f(n)=100. f(n)=5nf(n)=100100=5nLHS/5=RHS/520=nRearrange equationn=20 20 minutes after midnight on January 1st, 100 babies will be born. | |

Exercises 46 aLet's make a table of the ordered pairs from the graph.n1234 an56484032As we can see, the first term is 56 and by calculating the difference between each pair of consecutive terms, we can tell that the common difference is -8. Let's substitute these values in the function for an arithmetic sequence. f(n)=a1+(n−1)da1=56, d=-8f(n)=56+(n−1)(-8)Distribute -8f(n)=56−8n+8Subtract termf(n)=64−8n The function f(n)=64−8n represents the arithmetic sequence.bWe will use the function from the previous part to find the value of n, when f(n)=16. f(n)=64−8nf(n)=1616=64−8n Solve for n LHS−64=RHS−64-48=-8nLHS/(-8)=RHS/(-8)6=nRearrange equation n=6 In week 6 the movie earns $16 million.cWe know that the common difference is -8 so let's just continue the arithmetic sequence from the last recorded ordered pair until we hit 0: …32→−824→−816→−88→−80, We have the following arithmetic sequence: 56,48,40,32,24,16,8,0 Let's add all of these using a calculator.The sum is 224 which means the movie earns a total of $224 million. | |

Exercises 47 We are told that the area of each small square is 1 square inch. We can add the areas of the small squares in each figure to find the total areas. First figure: Second figure: Third figure: Total Area=1 in2Total Area=5 in2Total Area=9 in2 Four squares have been added each time, creating a common difference d of 4. The first term in this case is 1, the area of the first figure. We can substitute this value into the general form for an arithmetic sequence: an=a1+(n−1)d⇒an=1+(n−1)4. Let's simplify this function a little bit and rewrite it into a function notation, as requested. an=1+(n−1)4Rewrite an as f(n)f(n)=1+(n−1)4Distribute 4f(n)=1+4n−4Subtract termf(n)=4n−3 Now, we can find the 30th term in the sequence by substituting n=30 into the formula. f(n)=4n−3n=30f(30)=4⋅30−3Multiplyf(30)=120−7Subtract termf(30)=117 The 30th figure will have an area of 117 in2. | |

Exercises 48 We are told that the area of each small square is 1 square inch. We can add the areas of the small squares in each figure to find the total areas. First figure: Second figure: Third figure: Total Area=4 in2Total Area=9 in2Total Area=25 in2 The difference between the areas of the first and second figures is 9−4=5 in2 while the difference between the areas of the second and third figures is 25−9=16 in2. This means that there isn't a common difference between the terms and the sequence cannot be arithmetic. | |

Exercises 49 The general form for the equation of an arithmetic sequence is: an=a1+(n−1)d, where n represents the x-coordinates on a graph and an represents the y-coordinates. Let's think about whether the domain and range are continuous or discrete separately.Domain The domain corresponds to the possible inputs of a function. In the case of arithmetic functions, the inputs n are the numbers of the terms (first term is n=1, second term is n=2, third term is n=3, and so on). Therefore, the inputs can only be positive, integer values and the domain is discrete.Range Because the domain is discrete, we can only substitute positive integers for n, the function will only give outputs corresponding to those specific inputs. This means that the range is also discrete. | |

Exercises 50 First of all, there are a few things that your friend understood correctly.The domain is only positive integers. This is because the domain represents the position of the term within the sequence. The outputs will either constantly increase or decrease. This is because of the common difference.However, these things don't imply that the range will be exclusively positive or exclusively negative. What if the sequence is increasing and begins on a negative value? Or what if the sequence is decreasing and begins on a positive value? Let's look at an example: an=-5+(n−1)2. In the equation above, the first term is -5 and the common difference is 2. If we were to list out the first five terms of the sequence, we would have: -5→+2-3→+2-1→+21→+23. As you can see, the sequence begins on a negative value but ends up with positive values. Therefore, your friend is incorrect. | |

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##### Other subchapters in Writing Linear Functions

- Maintaining Mathematical Proficiency
- Mathematical Practices
- Writing Equations in Slope-Intercept Form
- Writing Equations in Point-Slope Form
- Writing Equations of Parallel and Perpendicular Lines
- Quiz
- Scatter Plots and Lines of Fit
- Analyzing Lines of Fit
- Piecewise Functions
- Chapter Review
- Chapter Test
- Cumulative Assessment