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Algebra I EOC Review (Part 3)brombergwestglades.pbworks.com/w/file/fetch/117634536/EOC...

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Algebra I EOC Review (Part 3) 1. Statement Reason 1. 2.5(6.25 + 0.5) = 11 1. Given 2. 15.625 + 1.25 = 11 2. Distribution Property 3. 15.625 = 9.75 3. Subtraction Property of Equality 4. = 0.624 4. Division Property of Equality Answer: D 2. Total cost = (total of scooter rental) + (total cost of helmet rental) = (50 + 2β„Ž) + (12 + 1.50β„Ž) = 62 +3 .50β„Ž 3. Set expressions to indicate which integer is which. Remember that consecutive means one after the other. 1 st integer: 2 nd integer: +1 3 rd integer: +2 β€œ3 consecutive integers that have a sum of 87” (1 st integer) + (2 nd integer) + (3 rd integer) = 87 () + ( + 1) + ( + 2) = 87 3 + 3 = 87 3 = 84 = 28 1 st integer: = 28 2 nd integer: + 1 = 29 3 rd integer: + 2 = 30 Check: (1 st integer) + (2 nd integer) + (3 rd integer) = 87 28 + 29 + 30 = 87 87 = 87 Answers: 28 , 29 , 30 4. Set expressions to indicate which integer is which. Remember that consecutive means one after the other. Note that β€œconsecutive even integers” are supposed 2 apart from each other. 1 st integer: 2 nd integer: +2 3 rd integer: +4 β€œTwice the largest number is six less than three times the smallest number” 2(largest integer) = 3(smallest integer) – 6 2( + 4) = 3() βˆ’ 6 2 + 8 = 3 βˆ’ 6 8=βˆ’6 14 = 1 st integer: = 14 2 nd integer: + 2 = 16 3 rd integer: + 4 = 18 Check: 2(largest integer) = 3(smallest integer) – 6 2(18) = 3(14) βˆ’ 6 36 = 42 βˆ’ 6 36 = 36 Answers: 14 , 16 , 18
Transcript

Algebra I EOC Review (Part 3)

1.

Statement Reason 1. 2.5(6.25π‘₯ + 0.5) = 11 1. Given

2. 15.625π‘₯ + 1.25 = 11 2. Distribution Property

3. 15.625π‘₯ = 9.75 3. Subtraction Property of Equality

4. π‘₯ = 0.624 4. Division Property of Equality

Answer: D

2. Total cost = (total of scooter rental) + (total cost of helmet rental) 𝑐 = (50 + 2β„Ž) + (12 + 1.50β„Ž) 𝑐 = 62 + 3.50β„Ž

3. Set expressions to indicate which integer is which. Remember that consecutive means one after the other. 1

st integer: π‘₯ 2

nd integer: π‘₯ + 1 3

rd integer: π‘₯ + 2

β€œ3 consecutive integers that have a sum of 87” (1

st integer) + (2

nd integer) + (3

rd integer) = 87

(π‘₯) + (π‘₯ + 1) + (π‘₯ + 2) = 87 3π‘₯ + 3 = 87 3π‘₯ = 84 π‘₯ = 28 1

st integer: π‘₯ = 28 2

nd integer: π‘₯ + 1 = 29 3

rd integer: π‘₯ + 2 = 30

Check: (1

st integer) + (2

nd integer) + (3

rd integer) = 87

28 + 29 + 30 = 87 87 = 87 Answers: 28, 29, 30

4. Set expressions to indicate which integer is which. Remember that consecutive means one after the other. Note that β€œconsecutive even integers” are supposed 2 apart from each other. 1

st integer: π‘₯ 2

nd integer: π‘₯ + 2 3

rd integer: π‘₯ + 4

β€œTwice the largest number is six less than three times the smallest number”

2(largest integer) = 3(smallest integer) – 6 2(π‘₯ + 4) = 3(π‘₯) βˆ’ 6 2π‘₯ + 8 = 3π‘₯ βˆ’ 6 8 = π‘₯ βˆ’ 6 14 = π‘₯ 1

st integer: π‘₯ = 14 2

nd integer: π‘₯ + 2 = 16 3

rd integer: π‘₯ + 4 = 18

Check:

2(largest integer) = 3(smallest integer) – 6 2(18) = 3(14) βˆ’ 6 36 = 42 βˆ’ 6 36 = 36 Answers: 14, 16, 18

Algebra I EOC Review (Part 3)

5. Matthew = 20

Susie =3

4(Matthew) βˆ’ 12

Susie =3

4(20) βˆ’ 12

Susie = 15 βˆ’ 12 Susie = 3 Answer: 3

6. length = 2(width) βˆ’ 3 𝑙 = 2𝑀 βˆ’ 3 Perimeter of rectangle = (width) + (width) + (length) + (length) 36 = (𝑀) + (𝑀) + (2𝑀 βˆ’ 3) + (2𝑀 βˆ’ 3) 36 = 6𝑀 βˆ’ 6 42 = 6𝑀 7 = 𝑀 Area of rectangle = (length)(width) = (2𝑀 βˆ’ 3)(𝑀) = (2(7) βˆ’ 3)(7) = (14 βˆ’ 3)(7) = (11)(7) = 77 Answer: 77 in2

7. β€œShe has read 9 pages so far, which accounts for three-eighths of the assignment” 9 =3

8(assignment)

Let π‘₯ = number of pages need to be read

9 =3

8π‘₯

(8

3) 9 = (

8

3) (

3

8π‘₯)

24 = π‘₯

Answer: 9 =3

8π‘₯ and 24 pages

8. XYZ Company salary > ABC Company salary $38,000 + 2% annual commission on sales > $45,000 38,000 + 0.02𝑠 > 45,000 0.02𝑠 > 7,000 𝑠 > 350,000 Note that it requires MORE THAN $350,000 in sales in XYZ Company to have a salary greater than ABC Company Answer: D

Algebra I EOC Review (Part 3)

9. Note how her first 3 test scores are averaged:

56+75+91

3= 74

If her next test/final counts double, then you need to account for 5 tests. Note that to have a minimum mean score of 80 can be equal or greater (β‰₯).

56+75+91+π‘₯+π‘₯

5β‰₯ 80

222+2π‘₯

5β‰₯ 80

222 + 2π‘₯ β‰₯ 80(5) 222 + 2π‘₯ β‰₯ 400 2π‘₯ β‰₯ 178 π‘₯ β‰₯ 89 Note that Jennifer would need to score AT LEAST an 89 on her next test/final.

Answer: 222+2π‘₯

5β‰₯ 80 and π‘₯ β‰₯ 89

10. Note that β€œmiles per minute” is written as miles

minute .

Also, note that the hours need to be converted to minutes. Remember that a unit rate is a rate with a denimonator of 1.

15 miles

2 hoursβˆ™

1 hour

60 minutes=

15 miles

120 minutes=

0.125 miles

1 minute= 0.125 miles/minute

Answer: 0.125 miles per minute

11. Note that the however many blocks there are, the numerator is still represented in meters. 𝑛𝑑

𝑑=

(number of blocks)(meters)

minutes=

π‘šπ‘’π‘‘π‘’π‘Ÿπ‘ 

π‘šπ‘–π‘›π‘’π‘‘π‘’π‘ = π‘šπ‘’π‘‘π‘’π‘Ÿπ‘ /π‘šπ‘–π‘›π‘’π‘‘π‘’ = π‘šπ‘’π‘‘π‘’π‘Ÿπ‘  π‘π‘’π‘Ÿ π‘šπ‘–π‘›π‘’π‘‘π‘’

Answer: D

12. When using dimensional analysis (converting units), check to make sure units cancel each other.

15 L βˆ™1000 mL

1 Lβˆ™

0.91 g

1 mLβˆ™

1 kg

1000 g

Answer: D

Algebra I EOC Review (Part 3)

13. Note that there are two sets of triangles, and by the way they are drawn, they are considered SIMILAR (proportional). Note that the BASE of the LARGER triangle includes both the 40.5 AND 54 feet.

Writing these similar triangles as a proportion: Larger triangle ~ smaller triangle

35

40.5+54=

π‘₯

54 must add 40.5 and 54 together for base of the larger triangle

35

94.5=

π‘₯

54

(35)(54) = π‘₯(94.5) cross multiply (Multiplication Property of Equality) 1890 = 94.5π‘₯ π‘₯ = 20 Answer: 20 feet

14. Note that there are two sets of triangles, and by the way they are drawn, they are considered SIMILAR (proportional). Note that the BASE of the LARGER triangle includes both the π‘₯ AND 3.

Writing these similar triangles as a proportion: Larger triangle ~ smaller triangle π‘₯+6

π‘₯+3=

4

3 must add π‘₯ and 3 together for base of the larger triangle

(3)(π‘₯ + 6) = (4)(π‘₯ + 3) cross multiply (Multiplication Property of Equality) 3π‘₯ + 18 = 4π‘₯ + 12 6 = π‘₯ Answer: π‘₯ = 6

Algebra I EOC Review (Part 3)

15. I recommend drawing two rectangles to organize your work.

Note that since the scale factor is 1

4, the new rectangle should be SMALLER.

a. Multiply each dimension by 1

4 to get the new rectangle.

120 cm βˆ™1

4= 30 cm 100 cm βˆ™

1

4= 25 cm

Answer: 30 cm by 25 cm

b. Perimeter (larger poster) = 100 + 100 + 120 + 120 = 440 cm

Perimeter (smaller poster) = 25 + 25 + 30 + 30

= 110 cm

Ratio of the perimeters of the smaller poster to larger poster =smaller poster

larger poster=

110

440=

1

4

Remember that the ratio of the PERIMETERS is the SAME as the scale factor (scale factor)1

Answer: 1

4

c. Area (larger poster) = (120)(100) = 12000 cm2

Area (smaller poster) = (30)(25)

= 750 cm2

Ratio of the areas of the smaller poster to larger poster =smaller poster

larger poster=

750

12000=

1

16

Remember that the ratio of the AREAS is the scale factor SQUARED (scale factor)2

Answer: 1

16

Algebra I EOC Review (Part 3)

16. Note that β€œPremium” is the cost of the insurance that the employee would need to pay. Note that β€œDeductible” is the cost that the employee would need to pay, regardless of the cost of the medical bill, only if he/she has any medical work done. If no work is needed, then they just pay the premium to have the insurance. Note that since π‘₯ represents the total medical expenses (or medical bill). Note that β€œCo-payment” is another payment the employee needs to pay, EXCEPT only a partial amount of the bill.

For this insurance, the employee would need to pay 20% of the medical bill AFTER subtracting $500 as the deductible. What may make this more confusing is that since the answer choices already show $500 being added into the equation, they again subtract it from the medical bill before applying the 20%. Also, note that if the bill is less than $500, then the insurance will not cover the amount because the employee would already be paying the entire bill from the deductible. That is why it says π‘₯ β‰₯ 500.

As an example, if the medical bill is $20,000, the employee would pay $500 (regardless of the bill amount). So, $20,000 βˆ’ $500 = $19,500 shows the amount before applying the 20%.

Then 0.20($19,500) = $3,900 shows the β€œCo-payment” amount that the employee would pay for this type of insurance.

Note that subtracting $500 from $20,000 does not show $500 being added yet. Therefore, $3,626 + $500 + $3,900 = $8,026 shows how much the employee would need to pay if the medical bill is $20,000.

Answer: C

Algebra I EOC Review (Part 3)

17. Remember that direct variation refers to proportional relationships, and they use 𝑦

π‘₯ to find the constant of variation

(or constant of proportionality).

Remember that βˆ†π‘¦

βˆ†π‘₯ is used to find the rate of change (

βˆ†π‘¦

βˆ†π‘₯ tests if the relation is LINEAR)

a. To test if the relation is a direct variation, use 𝑦

π‘₯.

𝑦

π‘₯=

2

6=

4

12=

5

15=

8

24=

1

3

Since every ordered pair can written as 1

3, then there is a constant of variation.

Answer: Yes; π‘˜ =1

3

b. To test if the relation is linear, use βˆ†π‘¦

βˆ†π‘₯.

βˆ†π‘¦

βˆ†π‘₯=

4βˆ’2

12βˆ’6=

5βˆ’4

15βˆ’12=

8βˆ’5

24βˆ’15=

1

3

Note that if the graph is direct (or proportional), it already is considered linear.

Answer: Yes (π‘š =1

3)

Algebra I EOC Review (Part 3)

18. Remember that direct variation refers to proportional relationships, and they use 𝑦

π‘₯ to find the constant of variation

(or constant of proportionality).

Remember that βˆ†π‘¦

βˆ†π‘₯ is used to find the rate of change (

βˆ†π‘¦

βˆ†π‘₯ tests if the relation is LINEAR)

a. To test if the relation is a direct variation, use 𝑦

π‘₯.

𝑦

π‘₯=

5

βˆ’3β‰ 

βˆ’1

1β‰ 

βˆ’4

3β‰ 

βˆ’13

9

Since not every ordered pair is equal to each other, then there is no constant of variation. Answer: None

b. To test if the relation is linear, use βˆ†π‘¦

βˆ†π‘₯.

Note that if the relation is not direct (or nonproportional), it STILL can be LINEAR. βˆ†π‘¦

βˆ†π‘₯=

βˆ’1βˆ’5

1βˆ’(βˆ’3)=

βˆ’4βˆ’(βˆ’1)

3βˆ’1=

βˆ’13βˆ’(βˆ’4)

9βˆ’3=

3

2

Answer: Yes (π‘š =3

2)

Algebra I EOC Review (Part 3)

19. Remember to use 𝑦 βˆ’ 𝑦1 = π‘š(π‘₯ βˆ’ π‘₯1) and 𝑦 = π‘šπ‘₯ + 𝑏 when dealing with linear relationships.

Slope of 𝑃𝑄̅̅ Μ…Μ… = π‘š =βˆ†π‘¦

βˆ†π‘₯=

12βˆ’10

6βˆ’11= βˆ’

2

5

Using (6, 12): 𝑦 βˆ’ 𝑦1 = π‘š(π‘₯ βˆ’ π‘₯1)

𝑦 βˆ’ 12 = βˆ’2

5(π‘₯ βˆ’ 6)

𝑦 βˆ’ 12 = βˆ’2

5π‘₯ +

12

5

𝑦 = βˆ’5

2π‘₯ +

72

5

𝑦 = βˆ’5

2π‘₯ + 14.4

Answer: C

20. a. Note that speed, in this case, refers to slope.

Aidan’s speed (or slope) is π‘š = βˆ’0.07 To find Brett’s speed (or slope):

Speed = slope =βˆ†π‘¦

βˆ†π‘₯=

0.78βˆ’1.35

9.5βˆ’0= βˆ’

0.57

9.5= βˆ’0.06

Answer: Aidan walks faster (his slope is steeper on the graph, so moves quicker)

b. Note that to determine who gets to the park first will be based on the 𝒙-intercept. To find Aidan’s time (π‘₯-intercept): 𝑦 = βˆ’0.07π‘₯ + 1.47 0 = βˆ’0.07π‘₯ + 1.47 0.07π‘₯ = 1.47 π‘₯ = 21 minutes To find Brett’s time (π‘₯-intercept):

Time Since Leaving Home (in minutes), 𝒙

βˆ’0 9.5 22.5

Distance from the Park (in miles), π’š

1.35 0.78 0

Since to find the π‘₯-intercept is setting 𝑦 = 0, Brett’s time would be 22.5 minutes. Answer: Aidan

Algebra I EOC Review (Part 3)

21. Let π‘š = cost of medium sandwiches Let 𝑙 = cost of larges sandwiches

{ 3π‘š + 2𝑙 = 29.95

4π‘š + 𝑙 = 28.45

Solving by elimination

3π‘š + 2𝑙 = 29.95βˆ’2(4π‘š + 𝑙 = 28.45)

3π‘š + 2𝑙 = 29.95

βˆ’8π‘š βˆ’ 2𝑙 = βˆ’56.90

---------------------------- βˆ’5π‘š = βˆ’26.95 π‘š = $5.39 for a medium sandwich

Substituting π‘š = 5.39 into first equation: 3π‘š + 2𝑙 = 29.95 3(5.39) + 2𝑙 = 29.95 16.17 + 2𝑙 = 29.95 2𝑙 = 13.78 𝑙 = $6.89 for a large sandwich Answer: C

22. Note that the variables must be set as: - π‘₯ = two-point field goals - 𝑦 = three-point field goals

a. Answer: {π‘₯ + 𝑦 = 18

2π‘₯ + 3𝑦 = 39

<βˆ’ number of attempted field goals<βˆ’ number of points

b. Note that since the question asks to find ONLY the number of three-point field goals, solve for 𝑦. Solving by elimination βˆ’2(π‘₯ + 𝑦 = 18)

2π‘₯ + 3𝑦 = 39

βˆ’2π‘₯ βˆ’ 2𝑦 = βˆ’362π‘₯ + 3𝑦 = 39

------------------------ 𝑦 = 3 three-point field goals Substituting 𝑦 = 3 in the 1

st equation (although not needed for this question)

π‘₯ + 𝑦 = 18 π‘₯ + 3 = 18 π‘₯ = 15 two-point field goals

Answer: 3 three-point field goals

Algebra I EOC Review (Part 3)

23. First identify which variables to be 1st

and 2nd

number.: - Let π‘₯ = first number - Let 𝑦 = second number

a. Answer: {π‘₯𝑦 = 323π‘₯ βˆ’ 𝑦 = 2

<βˆ’ product of two numbers <βˆ’ difference between the two numbers

b. Choose any method. I will choose substitution. Solving 2

nd equation for π‘₯ Substituting π‘₯ = 𝑦 + 2 into the 1

st equation

π‘₯ βˆ’ 𝑦 = 2 π‘₯𝑦 = 323 π‘₯ = 𝑦 + 2 (𝑦 + 2)(𝑦) = 323 𝑦2 + 2𝑦 = 323 𝑦2 + 2𝑦 βˆ’ 323 = 0 (𝑦 + 19)(𝑦 βˆ’ 17) = 0 𝑦 + 19 = 0 𝑦 βˆ’ 17 = 0 𝑦 = βˆ’19 𝑦 = 17 Note that you can choose either 𝑦 = βˆ’19 or 𝑦 = 17, BUT remember that 𝑦 is the SECOND number! Substituting 𝑦 = βˆ’19 into 2

nd equation or Substituting 𝑦 = 17 into 2

nd equation

π‘₯ βˆ’ 𝑦 = 2 π‘₯ βˆ’ 𝑦 = 2 π‘₯ βˆ’ (βˆ’19) = 2 π‘₯ βˆ’ 17 = 2 π‘₯ + 19 = 2 π‘₯ = 19 π‘₯ = βˆ’17 If 𝑦 = βˆ’19, then the FIRST number is π‘₯ = βˆ’17 If 𝑦 = 17, then the FIRST number is π‘₯ = 19 Answer: First number: βˆ’17 or First number: 19 Second number βˆ’19 Second number: 17

24. Let π‘š = number of months Let 𝑦 = total cost

{𝑦 = 35π‘š + 50𝑦 = 40π‘š + 35

Since both equations are solve for 𝑦, use substitution

𝑦 = 𝑦 35π‘š + 50 = 40π‘š + 35 50 = 5π‘š + 35 15 = 5π‘š 3 = π‘š Substituting π‘š = 3 into the 1

st equation or Substituting π‘š = 3 into the 2

nd equation

𝑦 = 35π‘š + 50 𝑦 = 40π‘š + 35 𝑦 = 35(3) + 50 𝑦 = 40(3) + 35 𝑦 = 105 + 50 𝑦 = 120 + 35 𝑦 = 155 𝑦 = 155 Answer: 3 months; $155

Algebra I EOC Review (Part 3)

25. Let π‘š = number of multiple choice questions Let 𝑒 = number of open-ended questions

{π‘š + 𝑒 = 85 4π‘š + 12𝑒 = 500

<βˆ’ the number of questions<βˆ’ the number of points

Solving by elimination βˆ’4(π‘š + 𝑒 = 85) 4π‘š + 12𝑒 = 500

βˆ’4π‘š βˆ’ 4𝑒 = βˆ’3404π‘š + 12𝑒 = 500

--------------------------- 8𝑒 = 160 𝑒 = 20 open-ended questions

Substituting 𝑒 = 20 into the first equation π‘š + 𝑒 = 85 π‘š + 20 = 85 π‘š = 65 multiple choice questions Answer: 65 multiple choice questions and 20 open-ended questions

26. Let π‘ž = the number of quarters Let 𝑛 = the number of nickels

{π‘ž + 𝑛 = 20 0.25π‘ž + 0.05 = 2.80

<βˆ’ the number of quarters and nickels<βˆ’ the amount of money

Solving by elimination βˆ’0.25(π‘ž + 𝑛 = 20)

0.25π‘ž + 0.05𝑛 = 2.80

βˆ’0.25π‘š βˆ’ 0.25𝑛 = βˆ’5 0.25π‘ž + 0.05𝑛 = 2.80

---------------------------------- βˆ’0.20𝑛 = βˆ’2.20 𝑛 = 11 nickels

Substituting 𝑛 = 11 into the first equation π‘ž + 𝑛 = 20 π‘ž + 11 = 20 π‘ž = 9 quarters Answer: 9 quarters and 11 nickels

Algebra I EOC Review (Part 3)

27. Let 𝑠 = the number of band shirts Let 𝑝 = the number of pairs of pants

{4𝑠 + 3𝑝 = 1593𝑠 + 2𝑝 = 113

<βˆ’ the total cost for Bianca<βˆ’ the total cost for Gianna

Solving by elimination 3(4𝑠 + 3𝑝 = 159)βˆ’4(3𝑠 + 2𝑝 = 113)

12𝑠 + 9𝑝 = 477

βˆ’12𝑠 βˆ’ 8𝑝 = βˆ’452

--------------------------- 𝑝 = $25 for a pair of pants

Substituting 𝑝 = 25 into the 1

st equation

4𝑠 + 3𝑝 = 159 4𝑠 + 3(25) = 159 4𝑠 + 75 = 159 4𝑠 = 84 𝑠 = $21 for a band shirt Note that you need to find the total of 5 band shirts and 5 pairs of pants Total cost = (5 band shirts) + (5 pairs of pants)

= [5($21)] + [5($25)] = $105 + $125 = $230

Answer: $230

28. Consider setting the system of equations as {π‘₯ + 𝑦 = 1

π‘₯ βˆ’ 𝑦 = βˆ’7 to help.

Since (βˆ’3,4) is the solution, you can substitute into each answer choice to check. Note that you should simplify each equation before substituting to make it easier. π‘₯ + 𝑦 βˆ’ 2(π‘₯ + 𝑦) = 1 βˆ’ 2(βˆ’7) 3(π‘₯ + 𝑦) + π‘₯ βˆ’ 𝑦 = 3(1) βˆ’ 7 π‘₯ + 𝑦 βˆ’ 2π‘₯ βˆ’ 2𝑦 = 1 + 14 3π‘₯ + 3𝑦 + π‘₯ βˆ’ 𝑦 = 3 βˆ’ 7 βˆ’π‘₯ βˆ’ 𝑦 = 15 4π‘₯ + 2𝑦 = βˆ’4 βˆ’(βˆ’3) βˆ’ (4) = 15 4(βˆ’3) + 2(4) = βˆ’4 3 βˆ’ 4 = 15 βˆ’12 + 8 = βˆ’4 βˆ’1 β‰  15 βˆ’4 = βˆ’4 π‘₯ + 𝑦 + 5(π‘₯ βˆ’ 𝑦) = 5(1) βˆ’ 7 βˆ’4(π‘₯ + 𝑦) + π‘₯ βˆ’ 𝑦 = 1 βˆ’ 4(βˆ’7) π‘₯ + 𝑦 + 5π‘₯ βˆ’ 5𝑦 = 5 βˆ’ 7 βˆ’4π‘₯ βˆ’ 4𝑦 + π‘₯ βˆ’ 𝑦 = 1 + 28 6π‘₯ βˆ’ 4𝑦 = βˆ’2 βˆ’3π‘₯ βˆ’ 5𝑦 = 29 6(βˆ’3) βˆ’ 4(4) = βˆ’2 βˆ’3(βˆ’3) βˆ’ 4(4) = βˆ’2 βˆ’18 βˆ’ 16 = βˆ’2 9 βˆ’ 16 = βˆ’2 βˆ’34 β‰  βˆ’2 βˆ’7 β‰  βˆ’2 π‘₯ + 𝑦 βˆ’ π‘₯ + 𝑦 = 1 + 7 2𝑦 = 8 2(4) = 8 8 = 8 Answers: [ ] π‘₯ + 𝑦 βˆ’ 2(π‘₯ + 𝑦) = 1 βˆ’ 2(βˆ’7) [ x ] 3(π‘₯ + 𝑦) + π‘₯ βˆ’ 𝑦 = 3(1) βˆ’ 7 [ ] π‘₯ + 𝑦 + 5(π‘₯ βˆ’ 𝑦) = 5(1) βˆ’ 7 [ ] βˆ’4(π‘₯ + 𝑦) + π‘₯ βˆ’ 𝑦 = 1 βˆ’ 4(βˆ’7) [ x ] π‘₯ + 𝑦 βˆ’ π‘₯ + 𝑦 = 1 + 7

Algebra I EOC Review (Part 3)

29. Remember that average rate of change refers to slope. To find the average rate of change (or slope), first identify the (π‘₯, 𝑦) coordinates. You can choose points such as (1975, 1060) and (1980, 1040)

Then average rate of change = slope =βˆ†π‘¦

βˆ†π‘₯=

acres

years=

1060βˆ’1040

1975βˆ’1980= βˆ’

20

5= βˆ’4 acres/years

Answer: B

30. Note that to find the maximum height involves finding the VERTEX. Note that to the find how long it takes to reach the maximum height uses the AXIS OF SYMMETRY.

Remember that to find the vertex is (βˆ’π‘

2π‘Ž, 𝑓 (βˆ’

𝑏

2π‘Ž)) find the axis of symmetry and then plug into the function

Remember that to find the axis of symmetry is π‘₯ = βˆ’π‘

2π‘Ž

It would be helpful to identify π‘Ž, 𝑏, and 𝑐 of the given function β„Ž = βˆ’5𝑑2 + 30𝑑 π‘Ž = βˆ’5 𝑏 = 30 𝑐 = 0 note that this is the 𝑦-intercept (so the rocket starts from the ground) Axis of symmetry

π‘₯ = βˆ’π‘

2π‘Ž

π‘₯ = βˆ’30

2(βˆ’5)

π‘₯ = 3 seconds to reach the maximum height Vertex (3, 𝑓(3)) β„Ž = βˆ’5𝑑2 + 30𝑑 β„Ž(3) = βˆ’5(3)2 + 30(3) β„Ž(3) = βˆ’5(9) + 90 β„Ž(3) = βˆ’45 + 90 note that the 2

nd term is doubled the 1

st term

β„Ž(3) = 45 meters Answer: it takes the rocket 3 seconds to reach a maximum height of 45 meters

Algebra I EOC Review (Part 3)

31. Note that to find the maximum height involves finding the VERTEX. Note that to the find how long it takes to reach the maximum height uses the AXIS OF SYMMETRY.

Remember that to find the vertex is (βˆ’π‘

2π‘Ž, 𝑓 (βˆ’

𝑏

2π‘Ž)) find the axis of symmetry and then plug into the function

Remember that to find the axis of symmetry is π‘₯ = βˆ’π‘

2π‘Ž

It would be helpful to identify π‘Ž, 𝑏, and 𝑐 of the given function 0 = βˆ’16𝑑2 + 32𝑑 + 48 π‘Ž = βˆ’16 𝑏 = 32 𝑐 = 48 note that this is the 𝑦-intercept (the golf ball starts at an initial height of 48 feet) Axis of symmetry

π‘₯ = βˆ’π‘

2π‘Ž

π‘₯ = βˆ’32

2(βˆ’16)

π‘₯ = 1 second to reach the maximum height Vertex (1, 𝑓(1)) 𝑓(𝑑) = βˆ’16𝑑2 + 32𝑑 + 48 𝑓(1) = βˆ’16(1)2 + 32(1) + 48 𝑓(1) = βˆ’16 + 32 + 48 note that the 2

nd term is doubled the 1

st term

𝑓(1) = 64 feet Answer: it takes the golf ball 1 second to reach a maximum height of 64 feet

32. Note that to find the time for the rock to hit the water is finding the π‘₯-intercept because the height would be 𝑦 = 0. Remember that finding the π‘₯-intercept is setting 𝑦 = 0. Note that since the question says to approximate the answer, there will be decimals involved, which means that factoring will most likely NOT work. Therefore, solve by completing the square or using the quadratic formula.

Remember that the quadratic formula is =βˆ’π‘Β±βˆšπ‘2βˆ’4π‘Žπ‘

2π‘Ž , and it is used to find the zeros, roots, or π‘₯-intercepts.

It would be helpful to identify π‘Ž, 𝑏, and 𝑐 of the given function β„Ž(𝑑) = βˆ’π‘‘2 βˆ’ 2𝑑 + 30 π‘Ž = βˆ’1 𝑏 = βˆ’2 𝑐 = 30 note that this is the 𝑦-intercept (the rock starts at an initial height of 30 meters)

π‘₯ =βˆ’π‘Β±βˆšπ‘2βˆ’4π‘Žπ‘

2π‘Ž

π‘₯ =2±√(βˆ’2)2βˆ’4(βˆ’1)(30)

2(βˆ’1)

π‘₯ =2±√4+120

βˆ’2

π‘₯ =2±√124

βˆ’2

π‘₯ β‰ˆ βˆ’6.57 or 4.57 seconds Note that negative time should not be considered (extraneous solution) Answer: B

Algebra I EOC Review (Part 3)

33. Consider the function 𝑓(π‘₯) = 2π‘₯2 + 4π‘₯ βˆ’ 30. Classify each statement. Axis of symmetry Vertex (βˆ’1, 𝑓(βˆ’1))

π‘₯ = βˆ’π‘

2π‘Ž 𝑓(π‘₯) = 2π‘₯2 + 4π‘₯ βˆ’ 30

π‘₯ = βˆ’4

2(2) 𝑓(βˆ’1) = 2(βˆ’1)2 + 4(βˆ’1) βˆ’ 30

π‘₯ = βˆ’1 𝑓(βˆ’1) = 2 βˆ’ 4 βˆ’ 30 𝑓(βˆ’1) = βˆ’32 So, the vertex is (βˆ’1, βˆ’32) Zero (π‘₯-intercepts) 𝑦-intercepts 𝑓(π‘₯) = 2π‘₯2 + 4π‘₯ βˆ’ 30 𝑓(π‘₯) = 2π‘₯2 + 4π‘₯ βˆ’ 30 0 = 2π‘₯2 + 4π‘₯ βˆ’ 30 𝑓(π‘₯) = 2(0)2 + 4(0) βˆ’ 30 0 = 2(π‘₯2 + 2π‘₯ βˆ’ 15) 𝑓(π‘₯) = βˆ’30 0 = 2(π‘₯ βˆ’ 3)(π‘₯ + 5) 2 β‰  0 π‘₯ βˆ’ 3 = 0 π‘₯ + 5 = 0 π‘₯ = 3 π‘₯ = βˆ’5 Answers:

a. The vertex of the graph of 𝑓(π‘₯) is (1, βˆ’32). [ ] True [ x ] False b. The zeros of 𝑓(π‘₯) are π‘₯ = 3 and π‘₯ = βˆ’5. [ x ] True [ ] False c. The graph of 𝑓(π‘₯) opens down. [ ] True [ x ] False d. The axis of symmetry is π‘₯ = βˆ’1. [ x ] True [ ] False e. The 𝑦-interept of 𝑓(π‘₯) is βˆ’30. [ x ] True [ ] False

34. A pine cone drops from the top of a slash pine tree. The height of the pine cone after π‘₯ seconds is given by 𝑓(π‘₯) = 125 βˆ’ 15π‘₯2.

Answers: [ x ] the time when the pine cone is at the highest point [ x ] the height of the pine cone when it drops off the slash pine [ ] the distance of the pine cone from the slash pine when it hits the ground [ ] the time it takes the pine cone to hit the ground [ x ] the time when the pine cone drops off the slash pine

Algebra I EOC Review (Part 3)

35. Substitute π‘₯-values into 𝑓(π‘₯) and 𝑔(π‘₯) until the numbers are close enough to be considered equaled to each other. For π‘₯ = 3.4 𝑓(π‘₯) = π‘₯2 + 13 𝑔(π‘₯) = 3π‘₯ + 14 𝑓(3.4) = (3.4)2 + 13 𝑔(3.4) = 3(3.4) + 14 𝑓(3.4) = 11.56 + 13 𝑔(3.4) = 10.2 + 14 𝑓(3.4) = 24.56 𝑔(3.4) = 24.2 For π‘₯ = 3.3 𝑓(π‘₯) = π‘₯2 + 13 𝑔(π‘₯) = 3π‘₯ + 14 𝑓(3.3) = (3.3)2 + 13 𝑔(3.3) = 3(3.3) + 14 𝑓(3.3) = 10.89 + 13 𝑔(3.3) = 9.9 + 14 𝑓(3.3) = 23.89 𝑔(3.3) = 23.9 these are the closet to each other For π‘₯ = 3.2 𝑓(π‘₯) = π‘₯2 + 13 𝑔(π‘₯) = 3π‘₯ + 14 𝑓(3.2) = (3.2)2 + 13 𝑔(3.2) = 3(3.2) + 14 𝑓(3.2) = 10.24 + 13 𝑔(3.2) = 9.6 + 14 𝑓(3.2) = 23.24 𝑔(3.2) = 23.6 ALTERNATIVE: Note that 𝑓(π‘₯) = 𝑔(π‘₯) and since the questions asks for an answer to the nearest tenth, solve for π‘₯ by completing the square or using the quadratic formula. 𝑓(π‘₯) = 𝑔(π‘₯) π‘₯2 + 13 = 3π‘₯ + 14 π‘₯2 βˆ’ 3π‘₯ + 13 = 14 π‘₯2 βˆ’ 3π‘₯ βˆ’ 1 = 0 It would be helpful to identify π‘Ž, 𝑏, and 𝑐 of the equation. π‘Ž = 1 𝑏 = βˆ’3 𝑐 = βˆ’1

π‘₯ =βˆ’π‘Β±βˆšπ‘2βˆ’4π‘Žπ‘

2π‘Ž

π‘₯ =3±√(βˆ’3)2βˆ’4(1)(βˆ’1)

2(1)

π‘₯ =2±√9+4

2

π‘₯ =2±√13

2

π‘₯ β‰ˆ 3.3 or βˆ’0.3 Note that the question requires the POSITIVE solution. Answer: 3.3

Algebra I EOC Review (Part 3)

36. Remember that the exponential growth function is 𝑦 = π‘Ž(1 + π‘Ÿ)𝑑

a. 𝑦 = π‘Ž(1 + π‘Ÿ)𝑑 𝑦 = 15,000(1 + 0.025)𝑑 𝑦 = 15,000(1.025)𝑑 Answer: 𝐴(𝑑) = 15,000(1.025)𝑑

b. 𝑦 = π‘Ž(1 + π‘Ÿ)𝑑 𝑦 = 15,000(1 + 0.035)𝑑 𝑦 = 15,000(1.035)𝑑 Answer: 𝐡(𝑑) = 15,000(1.035)𝑑

c. 𝐴(𝑑) = 15,000(1.025)𝑑 𝐴(5) = 15,000(1.025)5 𝐴(5) β‰ˆ 16,971 people Answer: about 16,971 people

d. 𝐡(𝑑) = 15,000(1.035)𝑑 𝐡(5) = 15,000(1.035)5 𝐡(5) β‰ˆ 17,815 people Answer: about 17,815 people

37. Remember that the geometric sequence is π‘Žπ‘› = π‘Ž1 βˆ™ π‘Ÿπ‘›βˆ’1 Note that π‘Ž1 = 1218 Note that to find the ratio, π‘Ÿ, take a 𝑦-value and divide it by the previous 𝑦-value.

So, 1236.27

1218= 1.015

π‘Žπ‘› = π‘Ž1 βˆ™ π‘Ÿπ‘›βˆ’1 π‘Žπ‘› = 1218 βˆ™ (1.015)π‘›βˆ’1 𝑏(𝑑) = 1218 βˆ™ (1.015)π‘›βˆ’1 Answer: D

Algebra I EOC Review (Part 3)

38. Substitute 𝑑-values to compare which is greater.

For 𝑑 = 5 𝑉𝐴(𝑑) = 𝑑2 + 1.50 𝑉𝐡(𝑑) = 10(1.25)𝑑 𝑉𝐴(5) = (5)2 + 1.50 𝑉𝐡(5) = 10(1.25)5 𝑉𝐴(5) = 25 + 1.50 𝑉𝐡(5) β‰ˆ 30.52 𝑉𝐴(5) = 26.5

For 𝑑 = 6 𝑉𝐴(𝑑) = 𝑑2 + 1.50 𝑉𝐡(𝑑) = 10(1.25)𝑑 𝑉𝐴(6) = (6)2 + 1.50 𝑉𝐡(6) = 10(1.25)6 𝑉𝐴(6) = 36 + 1.50 𝑉𝐡(6) β‰ˆ 38.15 𝑉𝐴(6) = 37.5

For 𝑑 = 7 𝑉𝐴(𝑑) = 𝑑2 + 1.50 𝑉𝐡(𝑑) = 10(1.25)𝑑 𝑉𝐴(7) = (7)2 + 1.50 𝑉𝐡(7) = 10(1.25)7 𝑉𝐴(7) = 49 + 1.50 𝑉𝐡(7) β‰ˆ 47.68 𝑉𝐴(7) = 50.5

For 𝑑 = 11 𝑉𝐴(𝑑) = 𝑑2 + 1.50 𝑉𝐡(𝑑) = 10(1.25)𝑑 𝑉𝐴(11) = (11)2 + 1.50 𝑉𝐡(11) = 10(1.25)11 𝑉𝐴(11) = 121 + 1.50 𝑉𝐡(11) β‰ˆ 116.42 𝑉𝐴(11) = 122.5

For 𝑑 = 12 𝑉𝐴(𝑑) = 𝑑2 + 1.50 𝑉𝐡(𝑑) = 10(1.25)𝑑 𝑉𝐴(12) = (12)2 + 1.50 𝑉𝐡(12) = 10(1.25)12 𝑉𝐴(12) = 144 + 1.50 𝑉𝐡(12) β‰ˆ 145.52 𝑉𝐴(12) = 145.5 Answers: [ x ] 𝑑 = 5 [ x ] 𝑑 = 6 [ ] 𝑑 = 7 [ ] 𝑑 = 11 [ x ] 𝑑 = 12

39. Consider creating a table.

𝒙 π’š 0 12 1 18 2 27 3 40.5

Note that the table represents an EXPONENTIAL function, and you can find the common ratio by taking a 𝑦-value and dividing it by the previous 𝑦-value.

So, 18

12=

3

2= π‘Ÿ

Note that β€œincreasing the π‘₯ by 1” simply means the β€œnext π‘₯-value”.

𝒙 π’š 0 + 1 18 1 + 1 27 2 + 1 40.5 3 + 1 60.75

Regardless of the increase, the common ratio is STILL 3

2 .

Answer: B

Algebra I EOC Review (Part 3)

40. Remember that the exponential function is 𝑦 = π‘Ž βˆ™ 𝑏π‘₯ Remember that π‘Ž is the 𝑦-intercept. Remember that to find the common ratio, 𝑏, take a 𝑦-value and divide it by the previous 𝑦-value.

So, 16.20

18=

9

10

𝑦 = π‘Ž βˆ™ 𝑏π‘₯

𝑦 = 18 (9

10)

π‘₯

Answer: 𝑦 = 18 (9

10)

π‘₯

41. Note that a β€œconstant percent” refers to a common ratio. Remember that to find the common ratio take a 𝑦-value and divide it by the previous 𝑦-value. Car A Car B Car C 18,000

21,000β‰ 

15,000

18,000

15,625

18,000β‰ 

13,250

15,625

22,500

25,000=

20,250

22,500= 0.9 = 90%

Answer: C

42. Note that a β€œconstant rate” refers to rate of change (or slope).

Remember that to find the rate of change (or slope) use βˆ†π‘¦

βˆ†π‘₯.

Note that since each year has a βˆ†π‘₯ = 1, then to find the rate of change, simply take each 𝑦-value and subtract it by the previous 𝑦-value. Answer: B

43. Note that a β€œconstant rate” refers to rate of change (or slope).

Remember that to find the rate of change (or slope) use βˆ†π‘¦

βˆ†π‘₯.

Note that since each year has a βˆ†π‘₯ = 1, then to find the rate of change, simply take each 𝑦-value and subtract it by the previous 𝑦-value. Answer: C

44. Note that a β€œconstant factor” refers to a common ratio. Remember that to find the common ratio take a 𝑦-value and divide it by the previous 𝑦-value. Answer: A

Algebra I EOC Review (Part 3)

45. Note that Town 𝐴 has a constant rate of change. Note that Town 𝐡 has a common ratio.

a. Remember that to find the constant rate of change (or slope), use βˆ†π‘¦

βˆ†π‘₯ .

Note that since each year has a βˆ†π‘₯ = 1, then to find the rate of change, simply take each 𝑦-value and subtract it by the previous 𝑦-value.

Slope = π‘š =βˆ†π‘¦

βˆ†π‘₯= 300

Note that 1500 is the 𝑦-intercept. Therefore, 𝑦 = π‘šπ‘₯ + 𝑏 𝑦 = 300π‘₯ + 1500 Answer: 𝐴(𝑑) = 300𝑑 + 1500

b. Remember that to find the common ratio take a 𝑦-value and divide it by the previous 𝑦-value.

So, 1725

1500= 1.15

Note that 1500 is the 𝑦-intercept. Therefore, 𝑦 = π‘Ž βˆ™ 𝑏π‘₯ 𝑦 = 1500(1.15)π‘₯ Answer: 𝐡(𝑑) = 1500(1.15)𝑑

c. Answers:

Time, 𝒕 (years)

Town A population

𝑨(𝒕)

Town B Population

𝑩(𝒕) 0 1500 1500 1 1800 1725 2 2100 1984 3 2400 2281 4 2700 2624 5 3000 3017 6 3300 3470 7 3600 3390 8 3900 4589

Algebra I EOC Review (Part 3)

46. 𝑦 = 𝑓 (1

5π‘₯) refers to

1

5 of what values would give π‘₯-values of 𝑦 = 𝑓(π‘₯). The 𝑦-values do not get affected.

1

5𝑛 = βˆ’4

1

5𝑛 = βˆ’1

1

5𝑛 = 0

1

5𝑛 = 3

1

5𝑛 = 6

𝑛 = βˆ’20 𝑛 = βˆ’5 𝑛 = 0 𝑛 = 15 𝑛 = 30 The table below shows the values for the function 𝑦 = 𝑓(π‘₯).

𝒙 π’š βˆ’4 7 βˆ’1 βˆ’2 0 3 3 βˆ’4 6 5

Answers:

𝒙 π’š βˆ’20 7 βˆ’5 βˆ’2 0 3

15 βˆ’4 30 5

47. Remember that a residual is the VERTICAL DISTANCE between a data point and a line of fit.

Note that (5,21) is NOT a point on 𝑦 =1

2π‘₯ + 15. You can test this by substituting π‘₯ = 5 into the equation.

Substituting π‘₯ = 5

𝑦 =1

2π‘₯ + 15

𝑦 =1

2(5) + 15

𝑦 = 17.5 So, (5, 17.5) is a point on the line, while (5, 21) is a point that is VERTICALLY DISTANT from it.

Then finding the difference in the 𝑦-values is 21 βˆ’ 17.5 = 3.5 Answer: 3.5

Algebra I EOC Review (Part 3)

48. Note that to verify there are 40% more girls than there are boys, multiply the amount of boys by 1.40.

For example, if there are 50 boys, then 50(1.40) = 70 girls. So, 50 boys + 70 girls = 120 total

Since we do not know how many boys and girls there are, we can replace boys with π‘₯ and girls with 1.40π‘₯, since that is the method used above to obtain 40% more girls.

Math English Science Total

Boys 45 π‘₯

Girls 77 65 1.40π‘₯

Total 90 300

Since there are 300 boys and girls: π‘₯ + 1.40π‘₯ = 300 2.40π‘₯ = 300 π‘₯ = 125 boys So, since there 125 boys, then 125(1.40) = 175 girls. Answer:

Math English Science Total

Boys 13 67 45 125

Girls 77 65 33 175

Total 90 132 78 300

49. Note that how there is a certain amount times as many RIGHT-HANDED boys/girls as there are left-handed boys/girls. So, setting up the table can be done as (letting π‘₯ represent boys and 𝑦 represent girls):

Left Right

Boys π‘₯ 9π‘₯

Girls 𝑦 5𝑦

Total 18 122

Thus, we can create a system of equations for this:

{π‘₯ + 𝑦 = 18 9π‘₯ + 5𝑦 = 122

<βˆ’ the number of leftβˆ’handed students <βˆ’ the number of rightβˆ’handed students

Solving by elimination βˆ’9(π‘₯ + 𝑦 = 18)

9π‘₯ + 5𝑦 = 122

βˆ’9π‘₯ βˆ’ 9𝑦 = βˆ’1629π‘₯ + 5𝑦 = 122

------------------------ βˆ’4𝑦 = βˆ’40 𝑦 = 10 girls

Substituting 𝑦 = 10 into the first equation π‘₯ + 𝑦 = 18 π‘₯ + 10 = 18 π‘₯ = 8 boys Therefore, the completed table would be:

Left Right

Boys 8 72

Girls 10 50

Total 18 122

The probability that a right-handed STUDENT at random is female: 50 female

122 studentsβ‰ˆ 0.410

Answer: A

Algebra I EOC Review (Part 3)

50. Note the differences in taking the TOTAL (uses entire total) versus the CONDITIONAL TOTAL (uses marginal total).

Answer Choice Using the Table to Check True or False

40% of the participants were girls 80 girls

200 participants= 0.40 = 40% true

70% of the participants preferred β€œA” 70 preferred "A"

200 participants= 0.35 = 35% false

20

120 of the boys preferred β€œD”

15 preferred "D"

120 boys=

15

120 false

10

35 of the participants who preferred β€œB” were girls

10 girls

35 preferred "B"=

10

35 true

The proportion of boys who preferred β€œC” is equal to the proportion of girls who preferred β€œC”

30 preferred "C"

120 boys≠

30 preferred "C"

80 girls false

Answers: [ x ] 40% of the participants were girls. [ ] 70% of the participants preferred β€œA”.

[ ] 20

120 of the boys preferred β€œD”.

[ x ] 10

35 of the participants who preferred β€œB” were girls.

[ ] The proportion of boys who preferred β€œC” is equal to the proportion of girls who preferred β€œC”.


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