ACT Math Test A2

7/26/2019 ACT Math Test A2

1/181

Name: Date:

Teacher: Class/Period:

1)

2)


2/181

3)

4)


3/181

5)

6)

7)


4/181

8)

9)


5/181

10)

11)

12)


6/181

13)

14)

15)


7/181

16)

17)

18)


8/181

19)

20)

21)


9/181

22)

23)

24)


10/181

25)

26)

27)


11/181

28)

29)

30)


12/181

31)

32)

33)

34)


13/181

35)

36)

37)

38)

39)


14/181

40)

41)

42)


15/181

43)

44)

45)


16/181

46)

47)


17/181

48)

49)

50)


18/181

51)

52)

53)


19/181

54)

55)


20/181

56)

57)

58)


21/181

59)

60)

61)

62)


22/181

63)

64)

65)


23/181

66)

67)

68)

69)


24/181

70)

71)

72)


25/181

73)

74)

75)


26/181

76)

77)


27/181

78)

79)

80)


28/181

81)

82)

83)

84)


29/181

85)

86)

87)

88)

89)


30/181


31/181

93)

94)

95)

96)


32/181

97)

98)

99)


33/181

100)

101)

102)

103)


34/181

104)

105)

106)

107)


35/181

108)

109)

110)


36/181

111)

112)

113)


37/181

114)

115)

116)


38/181

117)

118)

119)


39/181

120)

121)

122)


40/181

123)

124)


41/181

125)

126)

127)


42/181

128)

129)

130)

131)

132)


43/181

133)

134)

135)


44/181

136)

137)

138)

139)


45/181

140)

141)

142)

143)


46/181

144)

145)

146)


47/181

147)

148)


48/181

149)


49/181

150)


50/181

151)

152)


51/181

153)

154)


52/181

155)

156)


53/181

157)

158)

159)


54/181

160)


55/181

161)

162)

163)

164)


56/181

165)

166)

167)


57/181

168)

169)


58/181

170)

171)

172)


59/181

173)

174)


60/181

175)

176)

177)

178)


61/181

179)

180)

181)


62/181

182)

183)

184)


63/181

185)

186)

187)


64/181

188)

189)

190)


65/181

191)

192)


66/181

193)


67/181

194)


68/181

195)


69/181

196)

197)

198)


70/181

199)

200)

201)


71/181

202)

203)

204)


72/181

205)

206)

207)

208)


73/181

209)

210)

211)

212)


74/181

213)

214)

215)

216)


75/181

217)

218)


76/181

219)

220)

221)


77/181

222)

223)

224)


78/181

225)


79/181

226)


80/181

227)


81/181

228)

229)


82/181

230)

231)


83/181

232)

233)

234)


84/181

235)

236)


85/181

237)

238)


86/181

239)

240)

241)

242)


87/181

243)

244)

245)


88/181

246)


89/181

247)

248)

249)


90/181

250)

251)

252)


91/181

253)

254)

255)


92/181

256)

257)

258)


93/181

259)

260)

261)


94/181

262)

263)

264)


95/181

265)

266)

267)

268)


96/181

269)

270)

271)


97/181

272)

273)


98/181

274)

275)

276)


99/181

277)

278)


100/181

279)

280)

281)


101/181

282)

283)

284)


102/181

285)

286)

287)

288)

289)

290)


103/181

Please use the space below to write your response(s) to the writing assignment provided by your

teacher. If there are multiple tasks to the question, please clearly label the number or letter of each

task in the column to the left of your answers. If you need additional pages for your response, yourteacher can provide them.

Please write the name of the writing assignment here: _____________________________________

Task


104/181

!"#"$"%&" '("") #*$ )(" +,-./)01*$"!"

!"#$%&' (( )*+,-.,/-0&1$ !11$112$*3

)40'35-*1 -. ' 65*$

')-%2-$2 3*$4 !" ! #$ "

'.*5"67%)"$&"5) 3*$4 $ " &" ! '

8*/%)6'.*5" 3*$4 $ # $9 " &:" # "9;

70'+&'3581

')-%2-$2 3*$4 *# - ("< ! '" ! ) " = (> '> -%2 )-$" &*%?)-%)?> @("$" ( =A

+,-2$-)/& BC,-)/*%

+,-2$-)/& 3*$4,.- " "

-*58 $835-*1

1/$&." :" # * ;< ! : $ # +;< " , +; @("$" )< " ( < # ' &"%)"$ :*>+;

B../5?" ! " 9 #*&/ :*> + $ ); @("$" )< " ( < # ' &"%)"$ :*>+;

E05"$D*.- # " 9 #*&/ :* $ )> +; @("$" )< " ( < ! ' &"%)"$ :*>+;

E05"$D*.- # " 9 #*&/ :*> + $ ); @("$" )< " ( < ! ' &"%)"$ :*>+;

65*$1 '*+ :-5*31

'.*5" & "

F/25*/%) - " >

G/?)-%&" . " :"< # "9;+;

2/$"&)$/J " " * # #*&,? :* ! > +;

:"9>$9; -%2 :"$ > -%2 -$" &*%?)-%)? @/)(! -%2 %*)

D*)( "C,-. )* L"$*A

:"9>$9; /? - 5*/%)A

& " ?.*5"

' " $6/%)"$&"5)


105/181

!"#$%&&'(%)*#

!"#$%&'() *%$() +",( ! ! " #

-",./( 0&$(1(#$ $ ! %" #

23,.34&5 0&$(1(#$ & ! % 6 "'#

78$9%:31(%& +9(31(, ( ; " ) ; ! *;

+',# ) "(%# '(/ )#"(%#

-"&(# ! !

23#"&(# ( ; ! ) ; " *; #;)*' 3# &

.%1*%($%#2 .%3"%#2 '(/ 0)*(4"(5

?1" $9, ($" ' -(@4(&'( (' ! (6 " A' # 6 B+

?1" $9, ($" ' -(1"(# ,' ! A(6 " ('B

C(3,($1"' -(@4(&'( (' ! (6A"' #6B

C(3,($1"' -(1"(# ,' ! =9(1( " 6

23,D"&%$"3 -./ ! .A-)/B !

7(1,4$%$"3 -0/ ! 0A-)/B !

"3$*6 %3%($%2 73%'2 '(/ 8)&*6%

+1"%&:/( & ! )1

7%1%//(/3:1%, & ! )1

+1%.(E3"5 & ! A)6 " );B1

2"1'/( & ! ";

. ! +

C(&(1%/ 71"#, 2 ! 31

*":9$ 2"1'4/%1 28/"&5(1 2 ! ";1

781%,"5 2 ! 31

*":9$ 2"1'4/%1 23&( 2 ! ";1

-.9(1( 2 ! "FGHHF

6HHF

6HHF

6HH;

6HH;

-IHHHHHHHA- #/BI

-IHHHHHHHHH

A- #

/BI/I

(6 #(6"'

HHHHHHHH6 #"

'HH;

#"& .HHHHH*

#"&3HHHHH)

#"&&HHHHH(

"HH'

J;KKL D8 ?2+) 0&'M ?// 1":9$# 1(#(1N(5M !" #$%&'(#)#

! ! 5"#$%&'(

" ! 1%$(

# ! $",(

$ ! "&$(1(#$

% !.1"&'".%/

& ! %,34&$ 3> ,3&(8 %>$(1#8(%1#

' ! &4,D(1 3> $",(# "&$(1(#$ "#'3,.34&5(5 %&&4%//8

( % &5 ) ! /(:# 3> 1":9$ $1"%&:/(

* ! 98.3$(&4#(

(' ! '$9 $(1,

' !&4,D(1 3> $9( $(1,

+ ! '3,,3& 5">>(1(&'(

,' !#4, 3> $9( >"1#$ ' $(1,#

" ! '3,,3& 1%$"3

- ! &4,D(1 3> 3DO('$# "& $9( #($

/ ! &4,D(1 3> 3DO('$# #(/('$(5

& ! %1(%

) !D%#(

1 !9(":9$

" ! 1%5"4#

! '"1'4,>(1(&'(

+ ! 5"%,($(1

2 ! N3/4,(3 ! %1(% 3> D%#(

FM6G

)

(.

&

*

3


106/181

Answer Key

1) A

2) C

3) C

4) A

5) B6) B

7) A

8) C

9) B

10) D

11) A

12) A

13) D

14) C

15) A

16) C

17) B

18) D

19) C

20) C

21) B

22) D23) A

24) A

25) C

26) B

27) B

28) B

29) B

30) D

31) D

32) A

33) C

34) D

35) A

36) D

37) D


107/181

38) A

39) C

40) B

41) C

42) B43) C

44) A

45) D

46) A

47) C

48) C

49) B

50) B

51) B

52) A

53) C

54) B

55) D

56) A

57) B

58) C

59) D60) A

61) C

62) D

63) A

64) D

65) B

66) C

67) B

68) A

69) B

70) B

71) C

72) D

73) B

74) C


108/181

75) A

76) B

77) D

78) C

79) A80) D

81) D

82) B

83) D

84) A

85) D

86) B

87) D

88) C

89) C

90) D

91) A

92) D

93) A

94) A

95) D

96) C97) B

98) B

99) C

100) A

101) C

102) A

103) B

104) C

105) A

106) C

107) D

108) C

109) B

110) D

111) A


109/181

112) A

113) C

114) B

115) D

116) B117) C

118) C

119) B

120) A

121) B

122) D

123) D

124) C

125) A

126) C

127) D

128) B

129) C

130) A

131) D

132) A

133) A134) B

135) C

136) A

137) B

138) C

139) B

140) C

141) A

142) C

143) D

144) C

145) B

146) A

147) A

148) D


110/181

149) B

150) B

151) C

152) C

153) D154) A

155) A

156) B

157) C

158) B

159) A

160) D

161) C

162) B

163) C

164) D

165) D

166) A

167) B

168) B

169) A

170) C171) C

172) C

173) C

174) B

175) D

176) D

177) A

178) D

179) B

180) D

181) A

182) B

183) A

184) A

185) D


111/181

186) C

187) C

188) A

189) B

190) C191) A

192) A

193) D

194) B

195) B

196) A

197) B

198) D

199) B

200) D

201) C

202) D

203) B

204) A

205) D

206) D

207) D208) C

209) C

210) A

211) B

212) B

213) B

214) A

215) C

216) C

217) B

218) D

219) D

220) C

221) A

222) A


112/181

223) A

224) B

225) D

226) B

227) C


113/181

Scoring Criteria:

A 4-point response will include, but is not limited to, the following points:

A. Domain of the graph of the parabola:All real numbers or ( , )

Explanation of how the domain was found:This function is defined for all realnumbers. A quadratic function does not involve roots or rational expressions with

variables in the denominator; therefore, there are no restrictions on the value forx. Thismakes the domain of f(x) all real numbers.

OR

Looking at the graph, one can see that the parabola continues to expand horizontally,suggesting that any value ofxwill have a corresponding y-value. Therefore,xcan beany real number.

Range of the graph of the parabola:All real numbers less than or equal to 1, y 1, or( , 1]

Explanation of how the range was found:Since this parabola opens downward, thevertex, (4,1), is a maximum for the graph. Therefore, the range is all real numbers equalto or less than 1.

OR

Looking at the graph, one can see that all y-coordinates for points on the graph are lessthan or equal to 1.

B. Correct equation in vertex form:y= 8(x4)2+ 1

Appropriate algebraic work:y= a(xh)2+ ky= a(x4)2+ 17 = a(3 4)2+ 17 = a(1)

2+ 1

7 = a+ 18 = a

Using substitution, the equation becomes: y= 8(x4)2+ 1

Explanation of how the points were used to determine the equation:The vertex(4,1) and a point on the parabola (3,7) are given. Unknowns hand kin the vertex formfor the equation of the parabola correspond to the coordinates of the vertex of theparabola. Since the vertex is identified on the graph, I know h= 4 and k=1. The graphitself represents solutions to the equation, which means that the additional pointidentified on the graph is a solution. Therefore, I can substitute 3 forxand 7 for y.Given the values for h, k,x, and y, I can substitute them into the equation for theparabola and solve for a. Finally, I use the value a, h, k, to write the equation for theparabola in vertex form.

228)


114/181

Note:A student could also use symmetry of parabola to locate the point (5,7). Usingthe 3 known points (4,1), (3,7), and (5,7), the students could write 3 equations with 3unknowns (a, h, and k). This system could then be solved for the unknowns, allowingthe students to write the equation of the parabola.

C. Additional transformation:Shift up 1 unit

Explanation of how the answer was found:After reflecting f(x) across thex-axis, the

graph would be a parabola that opens up with a vertex at (4,1). This function has2 zeros (or roots) because the graph intersects thex-axis at 2 different points. Sinceg(x) is supposed to be a quadratic function with 1 real root, the graph can only intersectthex-axis at 1 point. This means that the vertex must be on thex-axis. If you shift theentire function up 1 unit, the vertexwill be (4,0), which is on thex-axis. Therefore, toobtain g(x) after the reflection, you have to shift the function up 1 unit.

Note:Although not required, a student may also include sketches of the transformationsto aid in the explanation.

Note:Since g(x) is identified as a function in the task, transformations that lead to a

horizontally oriented parabola are not appropriate for this portion of the response.


115/181

Rubric:

4 A response at this level provides evidence of thorough knowledge andunderstanding of the subject matter.

The response addresses all parts of the question or problem correctly.

The response demonstrates efficient and accurate use of appropriate procedures.

The explanation of strategies used in the response shows evidence of a goodunderstanding of mathematical concepts and principles, and it does not contain any

misconceptions. The explanation in the response is clear and coherent.

3 A response at this level provides evidence of competent knowledge andunderstanding of the subject matter.

The response addresses most parts of the question or problem correctly.

The response includes some minor errors but generally uses appropriate proceduresaccurately.

The explanation of strategies used in the response shows some evidence of a goodunderstanding of mathematical concepts and principles, and it contains few, if any,misconceptions.

The explanation in the response is mostly clear and coherent.

2 A response at this level provides evidence of a basic knowledge andunderstanding of the subject matter.

The response addresses some parts of the question or problem correctly.

The response includes a number of errors but demonstrates some use ofappropriate procedures.

The explanation of strategies used in the response shows a little evidence ofunderstanding of mathematical concepts and principles, but it may contain someevidence of misconceptions.

The explanation in the response is partially clear, but some parts may be difficult tounderstand.

1 A response at this level provides evidence of minimal knowledge andunderstanding of the subject matter.

The response addresses a few parts of the problem correctly, but the response ismostly incorrect.

The response includes inappropriate procedures or simple manipulations that showlittle or no understanding of correct procedures.

The explanation of strategies used in the response shows little or no evidence ofunderstanding of mathematical concepts and principles, and it may contain evidence

of significant misconceptions. Many parts of the explanation are difficult to understand.

0 A response at this level is not scorable.The response is off-topic, blank, hostile, orotherwise not scorable.


116/181

229) C

230) A

231) B

232) A

233) D234) D


117/181

Scoring Criteria:

A 4-point response may include, but is not limited to, the following points:

A. Correct year: Year 3

Appropriate computations used to make the decision:

Job A:Year 2: $27,000 + $3,000 = $30,000Year 3: $30,000 + $3,000 = $33,000

Job B:Year 2: $30,000(1.04) = $31,200Year 3: $31,200(1.04) = $32,448

Explanation of the approach used to find the answer: For Job A, I started with$27,000 and added $3,000 each year.For Job B, I started with $30,000, and multipliedby 1.04 to get the salary for each successive year. Year 3 is the first year for which thesalary for Job A exceeds the salary for Job B.

NOTE: Students could also use the formulas for arithmetic and geometric series.

B. Correct year: Year 5

Appropriate computations used to make the decision:

Job A:Year 4: $33,000 + $3,000 = $36,000Year 5: $36,000 + $3,000 = $39,000

Totals:Year 2: $27,000 + $30,000 = $57,000Year 3: $57,000 + $33,000 = $90,000

235)


118/181

Year 4: $90,000 + $36,000 = $126,000Year 5: $126,000 + $39,000 = $165,000

Job B:Year 4: $30,000(1.04) = $31,200Year 5: $31,200(1.04) = $32,448

Totals:

Year 2: $30,000 + $31,200 = $61,200Year 3: $61,200 + $32,488 = $93,648Year 4: $93,648 + $33,745.92 = $127,393.92Year 5: $127,393.92 + $35,095 = $162,489.68

Explanation of the approach used to find the answer: For Job A salaries, Icontinued to add $3,000 each year.For Job B, I continued to multiply by 1.04 to get thesalary for each successive year.For both job totals, I added the Year 1 and Year 2salaries together. For Year 3 and beyond, I added that years salary to the sum of theprevious years salaries.Year 5 is the first year for which the total amount of moneyearned for Job A exceeds the total amount of money earned for Job B.

NOTE: Students could also use the formulas for arithmetic and geometric series.

C. Job A and B salaries and totals: Job A yearly salary = $84,000; Job A total =$1,110,000; Job B yearly salary $63,205.48 ; Job B total $893,342.36

Appropriate work needed to find the answer:Job A salary: 27,000 3,000(20 1)+

Job A total: ( )+20

227,000 84,000

Job B salary: 20 130,000(1.04)

Job B total:201 1.04

30,0001 1.04

Explanation of how the formulas were used to find the answer: For the Job A yearlysalary, I substituted 27,000 for the initial term, 3,000 for the difference, and 20 for the numberof terms.For the Job A total, I substituted 20 for the number of terms, 27,000 for the initialterm, and 84,000 for the final term.For the Job B yearly salary, I substituted 30,000 for theinitial term, 1.04 for the ratio, and 20 for the number of terms. For the Job B total, I substituted30,000 for the initial term, 1.04 for the ratio, and 20 for the number of terms.


119/181

Rubric:























120/181

Scoring Criteria:


A. Correct value for the 500th term:1495

Appropriate work needed to find the answer:

500 3(500) 5 1500 5 1495a = = =

B. Correct expression:500

1

(3 5)n

n

=

C. Correct sum: 373,250


1 3(1) 5 2a = =

( )500500

2

2 1495 250(1493) 373,250S = + = =

Explanation of the approach used to find the answer: I found 1a by substituting 1 for

nin the given expression. Then, I substituted 500n = , 1 2a = , 500 1495a = into the

formula for an arithmetic series.

D. Correct sum: 1,123,250


a = =1000 3(1000) 5 2995

( )S = + = =10001000

2 2 2995 500(2993) 1,496,500

1000 5001,496,500 373,250 1,123,250S S = =

Explanation of the approach used to find the answer: I found a1000 by substituting

1000 for nin the given expression. I found the sum of the first 1000 terms using themethod in Part C and then subtracted the sum of the first 500 terms from the sum of thefirst 1000 terms.

OR


= =5013(501) 5 1498a

236)


121/181

Rubric:























122/181

Scoring Criteria:


A. Correct system of inequalities:

0 100

0 75

0 150

x

y

x y

+

Note:Students may include the inequalities 0x and 0y and not include the zeros

in the other inequalities. The examinees may eliminate the zeros in either the first 2inequalities or the last inequality in the list above.

B. Correct graph of the feasible region:

Explanation of how the graph was constructed: The equation 100x = is a verticalline through (100,0).I shaded to the left of 100x = since 0 100 , which means I shouldshade toward the origin. The line 75y = is a horizontal line through (0,75).I shaded

below 75y = since 0 100 , which means I should shade toward the origin.The

equation 150x y+ = goes through (150,0) and (0,150). I shaded toward the origin since0 0 150+ .

C. Correct vertices of the feasible region: (0,0), (100,0), (0,75), (100,50), and (75,75)

D. Correct objective function: 12 8P x y= +

E. Correct maximum profit: $1,600


237)


123/181

( )0,0 12(0) 8(0) 0P = + =

( )100,0 12(100) 8(0) 1,200P = + =

( )0,75 12(0) 8(75) 600P = + =

( )100,50 12(100) 8(50) 1,600P = + =

( )75,75 12(75) 8(75) 1,500P = + =

Explanation of the approach used to find the answer: I evaluated the objectivefunction for each vertex.The maximum profit is the highest value from the evaluation ofthe vertices of the objective functions.

F. Correct numbers of chairs made: 100 basic chairs and 50 deluxe chairs


124/181

Rubric:























125/181

Scoring Criteria:


A. Correct explanation of the inequality:All acceptable widths are within 0.05 cm of3 cm.

B. Explanation of what Mark did incorrectly:In step 1, Mark used 3w + instead of3w and 0.05 and 0.05 are in the wrong inequalities. First, Mark should have written

the original inequality without the absolute value bars. Then, Mark should have rewrittenthe original inequality, changing the inequality sign to greater than and the right-handside of inequality to 0.05. In step 2, Mark added 3 to the right-hand side of bothinequalities but subtracted 3 from the left-hand side of both inequalities. He should havesubtracted 3 from both sides of the inequalities. In step 4, Mark used closed dotsinstead of open dots. Closed dots indicate that 2.95 and 3.05 are part of the solutionset. However, these values should not be included in the solution set.

C. Correct solution to the inequality: 2.95 3.05w< <

Correct graph:

Appropriate work needed to find the answer:0.05 3 0.05w < <

3 0.05 3 0.05w < < +

Explanation of why the solution is correct: I made a compound inequality with 0.05

on the left of the given inequality and then removed the absolute value bars. Then, Iadded 3 to each of the 3 parts of the inequality. I put open dots at 2.95 and 3.05because inequality uses less than signs instead of less than or equal to signs. I shadedbetween 2.95 and 3.05 because there are less than signs instead of greater than signs.

Note: In Part C, the student does not have to explain the things that Mark has alreadydone correctly in Part B.

238)


126/181

Rubric:























127/181

Scoring Criteria:


Correct solution: (2, 5, 1)


3 2 3

2 4 5

5 2 8

x y z

x y z

x z

+ =

+ =

+ =

22 4 5 4 2 8 10

4 2 3 1

8 5 11

x y z x y z

x y z

x z

+ = + =

+ =

+ =

5

( 2)

5 2 8 25 10 40

8 5 11 16 10 22

9 18

x z x z

x z x z

x

+ = + =

+ = =

=

9 18

9 9

x=

2x =

5(2) 2 8z+ =

10 2 8z+ =

2 2z = 1z =

3(2) 2( 1) 3y+ =

6 2 3y+ + =

8 3y+ =

5y =

Explanation of the approach used to find the answer: I picked 2 pairs of equations andeliminated yfrom each. For the second pair, I had to multiply the top equation by 2 in order toeliminate y. I added down the columns to eliminate ybecause ( ) 0y y+ = and 2 2 0y y + = .

Then, I decided to eliminate z. I multiplied the top equation by 5 and the bottom equation by 2so that 10zand 10zwould cancel out. Then, I solved for x, and substituted 2x = back intoone of the equations with no y. Next, I substituted 2x = and 1z = back into one of the

original equations. To verify my answer, I substituted my solutions for x, y, and zback into all 3original equations.

Verification of the answer:3(2) ( 5) 2( 1) 6 5 2 3+ = + = yes

2(2) ( 5) 4( 1) 4 5 4 5 + = + = yes

4(2) 2( 5) 3( 1) 8 10 3 1+ = + = yes

239)


128/181

Rubric:























129/181


130/181

1

12a = and

1

8b =

( )1122 0c+ = and ( )1

82 0d =

1

60c+ = and

1

40d =

1

6c = and

1

4d =

OR

= =

1 12 3

1 12 824 24

24 4 6 1 1

24 24 6 4

2 3

4 6

Explanation of how the inverse was found: I multiplied Kbya b

c d

and set the

product equal to the identity matrix.Then, I formed 2 systems of equations by settingcorresponding elements of the matrices equal. I formed the systems by keeping theequations with the same variables in one system.I divided through by 2 in the equationsthat are equal to 0 to simplify the calculations.Then, I used elimination to solve thesystems of equations. I had to multiply one equation by 3 to cancel 3cand 3cso that Icould solve for a. I had to multiply another equation by 2 to cancel 2dand 2dso that Icould solve for b.I substituted aand bback into the simplified equations that were equalto 0 and solved for cand d.

OR

I used 1det

d bK c a

, substituted my values from matrix K, and simplified.


131/181


132/181

Scoring Criteria:


A. Correct matrix:6 3 5

1 1 5G

=

Appropriate work leading to the answer:

2 0 4 4 3 1

1 3 0 2 4 5

a b c

d e f

+ =

4 3 1 2 0 4

2 4 5 1 3 0

a b c

d e f

=

4 ( 2) 3 0 1 4

2 1 4 3 5 0

a b c

d e f

=

B. Correct product for H K: The matrix product cannot be computed.

Explanation of why the expression cannot be computed: Matrix His a 2 3 matrix

and matrix Kis a 2 2 matrix.The expression H Kcannot be computed because theinner dimensions of the product are not equal: (2 3)(2 2) ; 3 2 .If I try to multiply H

by K, I will run out of elements in the columns of Kbefore I use all the elements in therows of H.

C. Correct product of K J:4 8 10

14 23 24


0 2 4 3 1 0(4) ( 2)(2) 0(3) ( 2)(4) 0( 1) ( 2)(5)

1 5 2 4 5 1(4) 5(2) 1(3) 5(4) 1( 1) 5(5)

+ + + =

+ + +

Explanation of the approach used to find the answer: I multiplied the matricestogether by multiplying across the rows of the first matrix and down the columns of thesecond matrix.

241)


133/181

Rubric:























134/181

Scoring Criteria:


A. Description of the transformations that could occur to the graph of f(x) = x2to getg(x):I shifted the graph left 3 units. I reflected the graph across the x-axis. Then, Ishifted the graph up 1 unit.

B. Correct graph:

2

2

4

6

8

10

Ox

y

4 2 26

(4,0)(3,1)

(2,0)

(1,3)

(0,8)


2 2( 6) ( 6 3) 1 ( 3) 1 9 1 8g = + + = + = + = 2 2( 5) ( 5 3) 1 ( 2) 1 4 1 3g = + + = + = + =

2 2( 4) ( 4 3) 1 ( 1) 1 1 1 0g = + + = + = + = 2 2( 3) ( 3 3) 1 (0) 1 0 1 1g = + + = + = + = 2 2( 2) ( 2 3) 1 (1) 1 1 1 0g = + + = + = + =

2 2( 1) ( 1 3) 1 (2) 1 4 1 3g = + + = + = + = 2 2(0) (0 3) 1 (3) 1 9 1 8g = + + = + = + =

242)


135/181

C. Correct domain of g(x):All real numbersCorrect range of g(x): 1y

Explanation of how the answers were determined: The domain is all real numbersbecause any number can be substituted in for xin ( )g x . The range is 1y because

( )g x is a parabola with the vertex at (3,1). Since the parabola opens downward, the

range will be all numbers less than or equal to the y-value of the vertex.


136/181

Rubric:























137/181

Scoring Criteria:


A. Correct domain for f(x):All real numbers

Correct range for f(x): 6y

Correct domain forg(x):All real numbers

Correct range for g(x):All real numbers

Explanation of how the answer was determined:The domain is all real numbersbecause any number can be substituted for xin f(x). The range is 6y because f(x) is

a parabola with the vertex at (0,6). Since the coefficient on the x2term is positive, theparabola opens upward, and the range will be all numbers greater than or equal to they-value of the vertex, 6. The domain is all real numbers because any number can besubstituted for xin g(x). The range is all real numbers because g(x) is a linear functionand each y-value is used in a linear function.

B. Correct expression for f(g(x)): 24 12 3x x +


( ) = = = + = +2 2 2( ) (2 3) (2 3) 6 4 12 9 6 4 12 3f g x f x x x x x x

Explanation of the approach used to find the answer: I substituted in 2 3x for g(x).Then, I substituted 2 3x for each xin g(x). I squared 2 3x using FOIL, thensimplified and subtracted 6.

C. Correct domain for f(g(x)):All real numbers

Correct range for f(g(x)): 6y

Appropriate work leading to the answer:( 12) 12 3

2(4) 8 2x

= = =

( ) ( )( ) ( ) ( ) ( )2

3 3 3 9

2 2 2 4( ) 4 12 3 4 18 3 9 15 6f g x f g = = + = + = =

Explanation of the approach used to find the answer:The domain is all real

numbers because any number can be substituted for xin f(g(x)). Also, the domain ofg(x), the inner function, is all real numbers. The range is 6y because f(g(x)) is a

parabola with its vertex at ( )32 , 6 . Since the coefficient on the x2term is positive, the

parabola opens upward, and the range will be all numbers greater than or equal to they-value of the vertex, 6.

243)


138/181

Rubric:























139/181

Scoring Criteria:


A. Correct center: (2,3)

Correct radius: 8

Explanation of how the answer was determined: To find the center, I took theexpressions inside each set of parentheses and solved for the variable( 2 0 2x x = = and 3 0 3y y+ = = ). To find the radius, I took the square root of

64.

OR

The standard form of an equation of a circle is 2 2 2( ) ( )x h y k r + = , where (h,k) is the

center and ris the radius.

B. Determination if Sandy was correct: No, the point is not on the circle.

Appropriate work needed to make the decision:2 2 2 2( 2 2) ( 10 3) ( 4) ( 7) 16 49 65 + + = + = + = ; 65 64

Explanation of how your decision was made: I substituted 2 for xand 10 for yinthe left-hand side of the equation for the circle. I evaluated the expression and obtained65. Since 65 is not equal to 64, then the point (2,10) is not on the circle.

C. Correct graph:

Explanation of the procedure used to graph the circle:I graphed the center of thecircle at (2,3). I then used the radius to find 4 points on the circle by moving up 8 unitsto (2,5), down 8 units to (2,11), right 8 units to (10,3), and left 8 units to (6,3).Then, I connected the points to form a circle.

244)


140/181

Rubric:























141/181

Scoring Criteria:


A. Correct difference: 4 5i +


+ = + + = +5 2 3(3 ) 5 2 9 3 4 5i i i i i

B. Correct product: 17 i+


+ = + = + = + + = +2(3 )(5 2 ) 15 6 5 2 15 2( 1) 15 2 17i i i i i i i i

C. Correct quotient:13 11

29 29i


( )

( )

( )

( )

i i i i i i i

i i i i i

+ + = = =

+ +

2

2

3 5 2 15 6 5 2 15 11 2( 1) 13 11

5 2 5 2 25 10 10 4 25 4( 1) 29

Explanation of the approach used to find the answer: First, I identified the complexconjugate of the denominator. I changed the imaginary part of the denominator frompositive to negative: 5 2 5 2i i+ . Then, I multiplied the numerator and denominator

by the complex conjugate using FOIL. Then, I substituted 1 for 2i . Finally, I added liketerms and put my answer in a bi+ form.

245)


142/181

Rubric:























143/181

Scoring Criteria:


A. Explanation of what Ethan did correctly:In step 2, Ethan said that there are 2nonreal solutions because the value he computed for the discriminant is less than 0.

Explanation of what Ethan did incorrectly:In step 1, Ethan only squared the 5 andleft the negative sign in front of the 5. He should have calculated

2( 5) 4(1)(10) 25 40 15 = = .

B. Identification of what Ethan did incorrectly, and explanation of what he should

have done:In step 2, Ethan attempted to square 5. He should have squared5

2

instead of 5. His result should have been ( )2

5 25

2 4 = . Also in step 2, Ethan only

squared 5. He forgot to square 5, which would have given him 25. In step 4, Ethanfactored x2 5x 25 incorrectly. The expression x2 5x 25 is not factorable. In step 5,

Ethan added 35 to the left-hand side of the equation but subtracted 35 from the right-hand side of the equation. He should have added 35 to both sides of the equation. Instep 8, Ethan divided 35 by 2. He should have found the square root of 35.

C. Determination of whether Ethans solutions match his answer to part A:No

Explanation of why the determination was made: Ethans solutions do not match his

answer to part A because 452

and25

2 are not nonreal solutions.

246)


144/181

Rubric:























145/181

Scoring Criteria:


A. Correct graph of the system of inequalities:


Points for the boundary equation of 2( 2)y x :

( )y = = = 2 2( 1 2) ( 3) 9 1,9

( )y = = = 2 2(0 2) ( 2) 4 0,4

( )y = = = 2 2(1 2) ( 1) 1 1,1

( )y = = = 2 2(2 2) (0) 0 2,0

( )y = = = 2 2(3 2) (1) 1 3,1

( )y = = = 2 2(4 2) (2) 4 4,4

Note:Only 3 points are necessary and other points are possible.

Points for the boundary equation of 2 5y x< + :

( )y = + = + = 2( 2) 5 4 5 1 2,1

( )y = + = + = 2( 1) 5 1 5 4 1,4 ( )y = + = + = 2(0) 5 0 5 5 0,5

( )y = + = + = 2(1) 5 1 5 4 1,4

( )y = + = + = 2(2) 5 4 5 1 2,1

( )y = + = + = 2(3) 5 9 5 4 3, 4

Note:Only 3 points are necessary and other points are possible.

247)


146/181

Appropriate work needed to determine the direction of shading for the boundaryequations:

Is 20 (0 2) ?

Is 20 ( 2) ?

0 is not greater than or equal to 4

Is 20 (0) 5?< +

0 is less than 5

B. Correct points of intersection:6 5

1 , 62 2

+

and

6 51 , 6

2 2

+

Appropriate work needed to find the answer:2 2( 2) 5x x = +

2 24 4 5x x x + = + 22 4 1 0x x =

2( 4) ( 4) 4(2)( 1) 4 16 8 4 24 61

2(2) 4 4 2x

+ = = = =

2 26 6 6 6 5

1 2 1 1 2 62 2 2 4 2

y

= = = + + = +

2 2

6 6 6 6 51 2 1 1 2 6

2 2 2 4 2y

= + = + = + =

Explanation of the approach used to find the answer:I set the boundary equationsequal to each other. Then, I simplified the equation so that it was equal to 0. I used thequadratic formula to solve for x. Once I found the 2 values of x, I substituted them both

back into 2( 2)x and simplified. This gave me the 2 points of intersection.


147/181

Rubric:























148/181


149/181

Rubric:























150/181

Scoring Criteria:


Correct additional zeros: 0,2 13

Appropriate work needed to find the answer:4 2 3( ) 25 36 ( 25 36)f x x x x x x x = =

4 1 0 25 36

4 16 36

1 4 9 0

2 4 9 0x x = 2( 4) ( 4) 4(1)( 9) 4 16 36 4 52 4 2 13

2 132(1) 2 2 2

x +

= = = = =

Note:An examinee could also use polynomial long division instead of synthetic division to findthe answer.

Explanation of the approach used to find the answer:First, I factored out an x. That gave

me 0x = as a zero of the function. Since ( )4,0 is an x-intercept of the graph of the function,

4x = is a zero of the function. Since 4x = is a zero of the function, I used synthetic division

to find3 25 36

4

x x

x

+. I put in 0 for the coefficient of the x2term in the synthetic division

process. Performing synthetic division told me that3

225 36 4 94

x xx x

x

=

+. Then, I used

the quadratic formula to find the remaining zeros.

249)


151/181

Rubric:























152/181

Scoring Criteria:


A. Correct number of distinct real roots for each graph:Graph 1: 1 distinct real rootGraph 2:4 distinct real rootsGraph 3:3 distinct real roots

Explanation of how the information shown in the graph proves this:I examined each graph and looked at how many times it intersected with the x-axis. Thenumber of times that each graph touches the x-axis gives the number of distinct realroots. Graph 1 intersected the x-axis 1 time. Graph 2 intersected the x-axis 4 times.Graph 3 intersected the x-axis 3 times.

B. Correct numbers of real and non-real roots:Graph 1: 1 real root and 4 non-real rootsGraph 2: 5 real rootsGraph 3:3 real roots and 2 non-real roots

Explanation of how the information shown in the graph proves this:Graph 1 has 1 distinct real root. There can only be an even number of non-real roots, soGraph 1 must have 4 non-real roots.

Graph 2 has 4 distinct real roots. The graph intersects the x-axis 4 times, but one of thepositive x-intercepts intersects but does not cross the x-axis, which means it is a doubleroot. That means there are 5 real roots and 0 non-real roots.

Graph 3 crosses the x-axis 3 times and has no double roots, so there must be 3 realroots and 2 non-real roots.

250)


153/181

Rubric:























154/181

Scoring Criteria:


A. Correct amount of time when working together: 10.3 hr


18 6 3= 24 6 4= LCM of 18 and 24 6 3 4 72= =

( ) ( )1 1 118 2472 72 x+ = 72

4 3x

+ =

7 72x = 72

7x =

10.3=x

Explanation of the approach used to find the answer: First, I found the leastcommon multiple of the numerical denominators. I multiplied both sides of the equationsby 72. Then, I simplified and solved for x, rounding my answer to the nearest tenth.

Note:There are other ways to solve the equation.

B. Explanation of why the students answer is not reasonable: It cannot take Annabeland Makenna more hours to paint the inside walls of a house together than it takeseither one of them by themselves.

251)


155/181

Rubric:























156/181

Scoring Criteria:


A. Correct value: 1 3 3+


( 10) 1 7 2( 10) 1 7 20 1 27 1 9 3 1 3 3 = + = + + = + = + = +

f

Explanation of the approach used to find the answer:I substituted 10 for xin ( )f x .

I simplified the expression under the radical to 27. Then, I broke 27 into 9, the highest

perfect square that is a factor of the expression, and 3. Then, I simplified 9 as 3.

B. Correct domain: { }: 5, except 4x x x =

Explanation of how the answer was determined: To find the domain, I needed todetermine when the denominator equaled zero and when the expression under the

radical sign was negative. First, I set the denominator equal to zero. I added 5x+ toboth sides of the equation, and squared both sides. I subtracted 5 from both sides of theequation. The denominator equaled zero when xwas 4, so I eliminated 4 from thedomain of the function. Then, I set up an inequality to show when the expression underthe radical was less than zero (negative). I subtracted 5 from both sides of theinequality. The expression under the radical was negative when 5x < , so I eliminated

5x < from the domain. Using both sets of exclusions, I found that the domain had tobe greater than or equal to 5 except for the case when 4=x .

C. Correct expression:23 5 3 7 2 2 3 35

4

+ + + + +

+

x x x x

x


( ) x

f x g x x x x

+ = + =

+ +

1 1 7 2( ) ( ) 1 7 2

3 5 3 5

( )( )

( )( )

( )( )

( )

x x x x x x

x x x x x

+ + + + + + + += =

+ + + + + + +

2

1 7 2 3 5 3 5 3 7 2 7 2 5

3 5 3 5 9 3 5 3 5 5

( )

23 5 3 7 2 2 3 35

9 5

+ + + + +

+

x x x x

x

23 5 3 7 2 2 3 35

4

+ + + + +

+

x x x x

x

252)


157/181

Rubric:




The explanation o

Documents

ACT Math Test A2