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Algebra 2 Benchmark Review Ms. Cheung

Algebra 2 Benchmark Review Ms. Cheung. Standard 1.0 1) The absolute value of any number will always be positive, thus the equation |5 – 6m| > - 7 will

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Page 1: Algebra 2 Benchmark Review Ms. Cheung. Standard 1.0 1) The absolute value of any number will always be positive, thus the equation |5 – 6m| > - 7 will

Algebra 2

Benchmark ReviewMs. Cheung

Page 2: Algebra 2 Benchmark Review Ms. Cheung. Standard 1.0 1) The absolute value of any number will always be positive, thus the equation |5 – 6m| > - 7 will

Standard 1.0

1)

The absolute value of any number will always be positive, thus the equation |5 – 6m| > - 7 will always be true!!

Page 3: Algebra 2 Benchmark Review Ms. Cheung. Standard 1.0 1) The absolute value of any number will always be positive, thus the equation |5 – 6m| > - 7 will

Standard 1.0

2)

627510 x 55510 x

55510 x 55510 x

9

455

x

x13

655

x

x

Page 4: Algebra 2 Benchmark Review Ms. Cheung. Standard 1.0 1) The absolute value of any number will always be positive, thus the equation |5 – 6m| > - 7 will

Standard 2.03. Which order pair is the solution to the system of equations below?

207 yx

1642074 yy

1648028 yy168032 y

6432 y2y

Solve for xSubstitute for xDistributeCombine Like Term

Combine Like termSolve for y

6

2014

20)2(7

207

x

x

x

yx

Page 5: Algebra 2 Benchmark Review Ms. Cheung. Standard 1.0 1) The absolute value of any number will always be positive, thus the equation |5 – 6m| > - 7 will

Standard 2.04. Which order pair is the solution to the system of equations below?

242

2028

yx

yx

Multiply 2 to equation 2

Add the two equations1

44

264

24

x

x

x

yx

428

2028

yx

yx

244 y 6y

Page 6: Algebra 2 Benchmark Review Ms. Cheung. Standard 1.0 1) The absolute value of any number will always be positive, thus the equation |5 – 6m| > - 7 will

5) Ms. Cheung’s current phone plan is about to be expired. She has been offered two promotions from two different phone providers. AT&T provides a 500 minute plan at $30 with 40 cents per text message. T-Mobile provides a 500 minute plan at $40 with 30 cents per text message. If x represents the number of text sent and y is the total amount charge for both plans, how do you write a system of equations that model this situation?

Standard 2.0

xy

xy

3.040

4.030

Page 7: Algebra 2 Benchmark Review Ms. Cheung. Standard 1.0 1) The absolute value of any number will always be positive, thus the equation |5 – 6m| > - 7 will

6) Gerardo wrote a matrix equation.

What is the solution to Gerardo’s equation?

Standard 2.0

6

20

21

34

y

x

62

2034

6

20

21

34

yx

yx

y

x

4 ,262

2034

yx

yx

Page 8: Algebra 2 Benchmark Review Ms. Cheung. Standard 1.0 1) The absolute value of any number will always be positive, thus the equation |5 – 6m| > - 7 will

7) Jackie wrote a matrix equation.

What is the solution to Gerardo’s equation?

Standard 2.0

6

3

23

12

y

x

623

32

6

3

23

12

yx

yx

y

x

3 ,0623

32

yx

yx

Page 9: Algebra 2 Benchmark Review Ms. Cheung. Standard 1.0 1) The absolute value of any number will always be positive, thus the equation |5 – 6m| > - 7 will

Standard 3.0

8)

234

234

723

835

nnn

nnn

234 152 nnn

Page 10: Algebra 2 Benchmark Review Ms. Cheung. Standard 1.0 1) The absolute value of any number will always be positive, thus the equation |5 – 6m| > - 7 will

Standard 3.0

1414a

66 2

aa

14206 2 aa

Page 11: Algebra 2 Benchmark Review Ms. Cheung. Standard 1.0 1) The absolute value of any number will always be positive, thus the equation |5 – 6m| > - 7 will

Standard 3.0

105

126 2

k

kk

10176 2 kk

Page 12: Algebra 2 Benchmark Review Ms. Cheung. Standard 1.0 1) The absolute value of any number will always be positive, thus the equation |5 – 6m| > - 7 will

Standard 3.0

342486 3434 xxxx

11)

48x 312x 1

Page 13: Algebra 2 Benchmark Review Ms. Cheung. Standard 1.0 1) The absolute value of any number will always be positive, thus the equation |5 – 6m| > - 7 will

Given a polynomial expression such that it has two factors of

A)B) C)D)

24503510 234 xxxx

).4( and )1( xx

).2( and )1( xx

).3( and )1( xx

).3( and )2( xx

).4( and )3( xx

441

6424 632

Standard 3.012)

Page 14: Algebra 2 Benchmark Review Ms. Cheung. Standard 1.0 1) The absolute value of any number will always be positive, thus the equation |5 – 6m| > - 7 will

13) Which of the following represent the graph of ?23 i

Standard 5.0

A

D

C

B

Page 15: Algebra 2 Benchmark Review Ms. Cheung. Standard 1.0 1) The absolute value of any number will always be positive, thus the equation |5 – 6m| > - 7 will

14. Kevin knows that . What will he get if he simplifies

A)B) C)D)

i 1

4 and 4

Standard 5.0

?16

4

i4

ii 4 and 4

Page 16: Algebra 2 Benchmark Review Ms. Cheung. Standard 1.0 1) The absolute value of any number will always be positive, thus the equation |5 – 6m| > - 7 will

15. Jocelyn knows that . What will he get if he simplifies

A)B) C)D)

i 1

8 and 8

Standard 5.0

?64

8

i8

ii 8 and 8

Page 17: Algebra 2 Benchmark Review Ms. Cheung. Standard 1.0 1) The absolute value of any number will always be positive, thus the equation |5 – 6m| > - 7 will

16. Jocelyn knows that . What will she get if he simplifies

A)B) C)D)

i 1

Standard 6.0

?252

i10

i5

i10

i5

Page 18: Algebra 2 Benchmark Review Ms. Cheung. Standard 1.0 1) The absolute value of any number will always be positive, thus the equation |5 – 6m| > - 7 will

Standard 6.017)

2339

)3)(3(

iii

ii

)1(9

10

Page 19: Algebra 2 Benchmark Review Ms. Cheung. Standard 1.0 1) The absolute value of any number will always be positive, thus the equation |5 – 6m| > - 7 will

Standard 6.018)

i

ii

32

523

Page 20: Algebra 2 Benchmark Review Ms. Cheung. Standard 1.0 1) The absolute value of any number will always be positive, thus the equation |5 – 6m| > - 7 will

Standard 6.019)

ii

ii

43512

)43(512

i9

Page 21: Algebra 2 Benchmark Review Ms. Cheung. Standard 1.0 1) The absolute value of any number will always be positive, thus the equation |5 – 6m| > - 7 will

Standard 8.0

20)

042 p 4 ,0 ,1 cba

)1(2

)4)(1(400 2 x 216 2

4 2or 2 xx

Page 22: Algebra 2 Benchmark Review Ms. Cheung. Standard 1.0 1) The absolute value of any number will always be positive, thus the equation |5 – 6m| > - 7 will

Standard 8.0

21)

02556 2 xx 25 ,5 ,6 cba

)6(2

)25)(6(455 2 x 12600255 12

6255

12255

12255 or xx

12255

35

35 or xx

Page 23: Algebra 2 Benchmark Review Ms. Cheung. Standard 1.0 1) The absolute value of any number will always be positive, thus the equation |5 – 6m| > - 7 will

22. Given the graph f(x)=ax2+bx+c. If a is negative, what would my graph look like?

A) A linear graph going upB) A linear graph going upC) A quadratic graph going upD) A quadratic graph going down

Standard 9.0

f(x)=ax2+bx+c is a Quadratic Equation since it has a power of 2.

Since a is negative, the graph will be going downward.

Page 24: Algebra 2 Benchmark Review Ms. Cheung. Standard 1.0 1) The absolute value of any number will always be positive, thus the equation |5 – 6m| > - 7 will

23)

Standard 9.0

Vertex(-3, -2)

Vertex(2, 2)

From (-3, -2) to (2, 2)We get

)2 ,3(

)2 ,2(

2)2 3,(2

4) (5,

Positive 5, so it’s 5 unit to the right.

Positive 4, 4 units UP!!

Page 25: Algebra 2 Benchmark Review Ms. Cheung. Standard 1.0 1) The absolute value of any number will always be positive, thus the equation |5 – 6m| > - 7 will

24)

Standard 9.0

Vertex(5, 1)

Vertex(-5, 1)

Both functions have the same coefficient: 3, both have the same shape.

Both functions are positive, so both vertices are minimum.

Both functions are positive, so their vertices are minimums.

Page 26: Algebra 2 Benchmark Review Ms. Cheung. Standard 1.0 1) The absolute value of any number will always be positive, thus the equation |5 – 6m| > - 7 will

25) What is the minimum value of the function f(x)=x2 – 9?

A) – 9 B) – 3 C) 0D) 9

Standard 10.0

Minimum The lowest Point!Find the vertex!

X= - b/2aX = 0/2 = 0

y = 0 – 9= -9

The lowest Point!

Page 27: Algebra 2 Benchmark Review Ms. Cheung. Standard 1.0 1) The absolute value of any number will always be positive, thus the equation |5 – 6m| > - 7 will

Standard 10.0

26)

X= - b/2aX = -6/2 = - 3

y=(-3)2+6(-3)+5= 9 – 18 + 5 = - 4

Vertex: ( - 3, - 4)

Page 28: Algebra 2 Benchmark Review Ms. Cheung. Standard 1.0 1) The absolute value of any number will always be positive, thus the equation |5 – 6m| > - 7 will

Standard 24.0

27)𝐅𝐢𝐧𝐝 𝐟 (𝟐 )𝐛𝐲 𝐬𝐮𝐛𝐬𝐭𝐢𝐭𝐮𝐭𝐢𝐧𝐠𝐱=𝟐𝐢𝐧𝐭𝐨𝐟 (𝐱)

𝑔 ( 𝑓 (2 ))¿𝑔 (3 (2 )2−4)

¿𝑔 (12−4)

¿𝑔 (8)

-6

¿10

Page 29: Algebra 2 Benchmark Review Ms. Cheung. Standard 1.0 1) The absolute value of any number will always be positive, thus the equation |5 – 6m| > - 7 will

Standard 24.0

28)

𝐒𝐮𝐛𝐬𝐭𝐢𝐭𝐮𝐭𝐢𝐧𝐠𝐠 (𝐱)𝐢𝐧𝐭𝐨𝐟 (𝐱)

𝑓 (𝑔 (𝑥 ))¿ 𝑓 (𝑥+3) S

¿ (𝑥+3)2−1

¿ 𝑥2+6 𝑥+9−1

¿ 𝑥2+6 𝑥+8

Page 30: Algebra 2 Benchmark Review Ms. Cheung. Standard 1.0 1) The absolute value of any number will always be positive, thus the equation |5 – 6m| > - 7 will

Standard 24.029) Which of the following functions is the

inverse of f(x) = 3x – 2?

321

321

231

231

)(

)(

)(

)(

x

x

x

x

xf

xf

xf

xfA)

B)

C)

D)

𝐒𝐭𝐞𝐩𝟐 :𝐒𝐰𝐢𝐭𝐜𝐡 𝐱𝐚𝐧𝐝 𝐲 .𝐒𝐭𝐞𝐩𝟑 :𝐒𝐨𝐥𝐯𝐞𝐟𝐨𝐫 𝐲 .

𝐒𝐭𝐞𝐩𝟏 :𝐒𝐮𝐛𝐬𝐭𝐢𝐭𝐮𝐭𝐞𝐲 𝐟𝐨𝐫 𝐟 (𝐱 ) .

𝑓 (𝑥 )=3 x−2𝐒𝐭𝐞𝐩𝟏 :𝐒𝐮𝐛𝐬𝐭𝐢𝐭𝐮𝐭𝐞𝐲 𝐟𝐨𝐫 𝐟 (𝐱 ) .y=3 x−2𝐒𝐭𝐞𝐩𝟐 :𝐒𝐰𝐢𝐭𝐜𝐡 𝐱𝐚𝐧𝐝 𝐲 .x=3 y−2𝐒𝐭𝐞𝐩𝟑 :𝐒𝐨𝐥𝐯𝐞𝐟𝐨𝐫 𝐲 .+𝟐 +𝟐

x+2=3 y𝟑 𝟑𝑥+2

3=𝑦

Page 31: Algebra 2 Benchmark Review Ms. Cheung. Standard 1.0 1) The absolute value of any number will always be positive, thus the equation |5 – 6m| > - 7 will

Standard 7.030) Rewrite the following exponents to positive

exponents.

24

52

3

2

)(

)(

yx

yxxf

y

xxfA)

B)

𝑓 (𝑥 )= 1

𝑥2 𝑦3

𝑓 (𝑥 )=𝑥2 𝑦2

𝑥4 𝑦5𝑥2 𝑦 3

𝑓 (𝑥 )= 1

𝑥2 𝑦3

Page 32: Algebra 2 Benchmark Review Ms. Cheung. Standard 1.0 1) The absolute value of any number will always be positive, thus the equation |5 – 6m| > - 7 will

Standard 7.031) 𝐒𝐭𝐞𝐩𝟐 :𝐍𝐞𝐠𝐚𝐭𝐢𝐯𝐞𝐄𝐱𝐩𝐨𝐧𝐞𝐧𝐭

𝐒𝐭𝐞𝐩𝟑 :𝐒𝐢𝐦𝐩𝐥𝐢𝐟𝐲

𝐒𝐭𝐞𝐩𝟏 :𝐃𝐢𝐬𝐭𝐫𝐢𝐛𝐮𝐭𝐞𝐞𝐱𝐩𝐨𝐧𝐞𝐧𝐭

¿ 3𝑎2𝑏3𝑐− 2

𝑎− 3𝑏6𝑐3

¿ 3𝑎2𝑏3𝑎3

𝑏6𝑐3𝑐2

¿ 3𝑎5𝑏3

𝑏6𝑐5 𝐒𝐭𝐞𝐩𝟑 :𝐒𝐢𝐦𝐩𝐥𝐢𝐟𝐲

¿ 3 𝑎5

𝑏3𝑐5

Page 33: Algebra 2 Benchmark Review Ms. Cheung. Standard 1.0 1) The absolute value of any number will always be positive, thus the equation |5 – 6m| > - 7 will

Standard 7.032) 𝐒𝐭𝐞𝐩𝟐 :𝐒𝐢𝐦𝐩𝐥𝐢𝐟𝐲

𝐒𝐭𝐞𝐩𝟏 :𝐅𝐚𝐜𝐭𝐨𝐫𝐢𝐳𝐞

¿𝑥(𝑥+4)𝑥+3

∙(𝑥+3)(𝑥−3)(𝑥+4)(𝑥−3)

¿ 𝑥

Page 34: Algebra 2 Benchmark Review Ms. Cheung. Standard 1.0 1) The absolute value of any number will always be positive, thus the equation |5 – 6m| > - 7 will

Standard 7.033) 𝐒𝐭𝐞𝐩𝟐 :𝐍𝐞𝐠𝐚𝐭𝐢𝐯𝐞𝐄𝐱𝐩𝐨𝐧𝐞𝐧𝐭

𝐒𝐭𝐞𝐩𝟑 :𝐒𝐢𝐦𝐩𝐥𝐢𝐟𝐲

𝐒𝐭𝐞𝐩𝟏 :𝐃𝐢𝐯𝐢𝐝𝐞𝐬𝐰𝐢𝐭𝐜𝐡𝐭𝐨𝐌𝐮𝐥𝐭𝐢𝐩𝐥𝐲

¿ 20 𝑥− 4

27 𝑦2 ×15 𝑦−5

8 𝑥− 3𝐒𝐭𝐞𝐩𝟏 :𝐃𝐢𝐯𝐢𝐝𝐞𝐬𝐰𝐢𝐭𝐜𝐡𝐭𝐨𝐌𝐮𝐥𝐭𝐢𝐩𝐥𝐲

¿ 2027 𝑦 2𝑥4×

15 𝑥3

8 𝑦5𝐒𝐭𝐞𝐩𝟐 :𝐍𝐞𝐠𝐚𝐭𝐢𝐯𝐞𝐄𝐱𝐩𝐨𝐧𝐞𝐧𝐭

¿5

9 𝑦 2𝑥×

5

2 𝑦5

𝑥

5

2

59

𝐒𝐭𝐞𝐩𝟑 :𝐒𝐢𝐦𝐩𝐥𝐢𝐟𝐲

¿25

18𝑥 𝑦7

Page 35: Algebra 2 Benchmark Review Ms. Cheung. Standard 1.0 1) The absolute value of any number will always be positive, thus the equation |5 – 6m| > - 7 will

Standard 7.034)

𝐒𝐭𝐞𝐩𝟐 :𝐒𝐢𝐦𝐩𝐥𝐢𝐟𝐲𝐒𝐭𝐞𝐩𝟏 :𝐅𝐚𝐜𝐭𝐨𝐫𝐢𝐳𝐞

¿4 (𝑥2−4)

2−𝑥

¿4 (𝑥+2)(𝑥−2)

2−𝑥-1

¿−4 (𝑥+2)

Page 36: Algebra 2 Benchmark Review Ms. Cheung. Standard 1.0 1) The absolute value of any number will always be positive, thus the equation |5 – 6m| > - 7 will

Standard 22.035)

Strategy: Convert to decimal

=0.5 + 0.25+ 0.125 + 0.625 + …

=0.5 + 0.25+ 0.125 + 0.0625 + …

=0.9375

Closest to 1

How about

=

= =

Page 37: Algebra 2 Benchmark Review Ms. Cheung. Standard 1.0 1) The absolute value of any number will always be positive, thus the equation |5 – 6m| > - 7 will

Standard 22.036)

Strategy:Substitute n=1, 2, 3, 4, 5 into the answer choice to see if the term generated is the same as the question= 4(1) = 4 ≠3 (first term)

1st term 2nd term3rd term 4th term5th term

n=1 n=2 n=3 n=5n=4

= 3+4(1) = 7 ≠3 (first term)

= 2(1)+1 = 3 ¿3 (first term)= 2(2)+1 = 5 ≠7 (second term)

= 4(1)-1 = 3 (first term) = 4(2)-1 = 7 (second term)= 4(3)-1 = 11 (third term) = 4(4)-1 = 15 (fourth term)= 4(5)-1 = 19 (fifth term)

Page 38: Algebra 2 Benchmark Review Ms. Cheung. Standard 1.0 1) The absolute value of any number will always be positive, thus the equation |5 – 6m| > - 7 will

¿1525

Standard PS 1.0

37)Strategy: Identify Keywords

P(cm label & no labels)=P(cm label) P(no labels)

¿28

×1024

Multiply

No. of rulers with cm labels

Total No. of Rulers

No. of rulers without labels

One less ruler

5

3

1

21

8

14¿

14

Page 39: Algebra 2 Benchmark Review Ms. Cheung. Standard 1.0 1) The absolute value of any number will always be positive, thus the equation |5 – 6m| > - 7 will

38)

Standard PS 1.0

Strategy: Identify Keywords

Not raining in San Francisco and not raining in Sydney

Multiply

P(not raining in SF & not raining in Sydney)=P(not raining in SF) P(not raining in Sydney)

¿ (1−80 % )×(1−30 % )¿20 %×70 %¿14 %

Page 40: Algebra 2 Benchmark Review Ms. Cheung. Standard 1.0 1) The absolute value of any number will always be positive, thus the equation |5 – 6m| > - 7 will

39)

Standard PS 1.0Strategy: Identify Keywords

Green and green Multiply

P(green & green)=P(green) P(green)

¿26×

34

No. of green marble in 1st

bag

Total No. of Marbles in

1st bag

No. of green marbles in 2nd bag

Total No. of marbles in 2nd

bag

12

1

2

¿14

Page 41: Algebra 2 Benchmark Review Ms. Cheung. Standard 1.0 1) The absolute value of any number will always be positive, thus the equation |5 – 6m| > - 7 will

40)

Standard PS 2.0

Page 42: Algebra 2 Benchmark Review Ms. Cheung. Standard 1.0 1) The absolute value of any number will always be positive, thus the equation |5 – 6m| > - 7 will

41) Standard PS 2.0