Which Expression is Equivalent to 144 Superscript Three-halves

This section presents some sample exam questions for you to review as office of your preparation for the exam. To demonstrate how each competency may exist assessed, sample questions are accompanied by the competency that they measure. While studying, y’all may wish to read the competency before and subsequently yous consider each sample question. Please annotation that the competency statements do not appear on the actual exam.

The correct answer is provided for each sample exam question. The sample questions are included to illustrate the formats and types of questions you will encounter on the exam; however, your performance on the sample questions should not exist viewed as a predictor of your performance on the actual examination.

Domain I—Mathematical Processes and Number Sense

Competency 001—Understand mathematical trouble solving.

ane. In four one-half-cup samples of a cereal containing dried cranberries, the numbers of cranberries were 17, 22, 22, and 18. Diet information on a box of this cereal defines the serving size as i cup or 53 grams. If a box contains 405 grams, which of the following is the all-time estimate of the number of cranberries in one box of this cereal?

  1. less than 300
  2. between 300 and 325
  3. between 326 and 350
  4. more than 350

Enter to expand or plummet answer 1.
Answer expanded
Selection

B is correct.
This question requires the examinee to analyze the use of interpretation in a diverseness of situations (eastward.g., rounding, expanse, plausibility). There are approximately 20 cranberries per
1 half
cup, or forty cranberries per cup. The number of cups in a box is
405 divided by 53, which is approximately equal to
400 divided past l equals 8
(rounding both the numerator and denominator down minimizes the fault). Thus the guess number of cranberries in a box of this cereal is
8 times 40 equals 320, which is within the interval of response B.


Competency 002—Sympathise mathematical advice, connections, and reasoning.

2. Given statements
p
and
q, which of the following is the truth table for the compound argument

p
if and only if left parenthesisq
or not
pright parenthesis
?

  1. p q
    p
    if and only if left parenthesis
    q
    or not
    pright parenthesis
    T T T
    T F F
    F T F
    F F F
  2. p q
    p
    if and only if left parenthesis
    q
    or non
    p
    right parenthesis
    T T T
    T F F
    F T T
    F F F
  3. p q
    p
    if and only if left parenthesis
    q
    or non
    p
    right parenthesis
    T T T
    T F F
    F T T
    F F T
  4. p q
    p
    if and simply if left parenthesis
    q
    or not
    p
    right parenthesis
    T T T
    T F F
    F T F
    F F T

Enter to aggrandize or collapse reply.
Answer expanded
Pick

A is right.
This question requires the examinee to apply principles of logic to solve problems. Beginning a truth value column for
not
p

is computed as F, F, T, T (in vertical order). Then this column is used to compute truth values for the statement

q
or not
p
: T, F, T, T. Finally, this consequence is used to compute truth values for the full expression

p
if and only if left parenthesis
q
or not
p
right parenthesis
: T, F, F, F.


Competency 003—Understand number theory.

3. If
p
and
q
are prime numbers and
4 over
q
cubed equals
p
squared over 50

what is the value of
left parenthesis
p
plus
q
right parenthesis
?

  1. five
  2. seven
  3. 8
  4. 9

Enter to aggrandize or collapse reply.
Answer expanded
Option

B is correct.
This question requires the examinee to use the principles of basic number theory (e.thousand., prime factorization, greatest mutual factor, to the lowest degree common multiple). The variables can exist isolated past multiplying both sides of the equation by
50q
cubed
, which yields
200 =
p
squared
q
cubed
. If
p
and
q
are both prime, then

p
squared
q
cubed

is the prime factorization of 200. Since
200 equals 25 times 8 equals 5 squared times ii cubed, and 5 and 2 are both primes,
p
must be 5 and
q
must exist 2, so

p
plus
q
equals v plus ii equals 7
.

Domain 2—Patterns, Algebra, and Functions

Competency 004—Understand relations and functions.

4. Which of the post-obit equations represents the changed of

y
equals half-dozenx
minus four over ane plus threeten
?


  1. y
    equals
    10
    minus 4 over iiix
    plus vi

  2. y
    equals
    ten
    pluss 4 over 6 minus 310

  3. y
    equals i plus threex
    over 6x
    minus iv

  4. y
    equals ane minus threexover vix
    plus 4

Enter to expand or collapse reply.
Answer expanded
Option

B is correct.
This question requires the examinee to perform operations with functions, including compositions and inverses. To find the inverse of a role of the form

y
=
f
of
x
, the original equation is rearranged by solving information technology for
x

equally a function of
y:

y
equals 6x
minus 4 over 1 plus 3x
which becomes
y
left parenthesis 1 plus 3x
right parenthesis equals 6x
minus 4 which becomes
y
plus 3xy
equals half dozenx
minus iv which becomes
y
plus 4 equals half-dozenx
minus 3xy
which becomes
y
plus 4 equals
x
left parenthesis 6 minus 3y
correct parenthesis which becomes
x
equals
y
plus iv over half dozen minus 3y
. Exchanging the variables
10
and
y
results in the inverse part
f superscript negative 1,
y
equals
10
+ 4 over half-dozen minus iii10
.


Competency 005—Understand linear, quadratic, and college-order polynomial functions.

blank Guild 1 Order two Order 3
soft potable 4 half dozen three
large pizza 1 two ane
garlic bread 1 one 0
Total Cost 19 dollars and 62 cents 34 dollars and 95 cents 16 dollars and 50 cents
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v. Given the table of orders and total costs above, and that at that place is a solution to the trouble, which of the following matrix equations could be used to find
d,
p, and
chiliad, the individual prices for a soft drink, a large pizza, and garlic bread respectively?



  1. 3 by 3 matrix row one 4 6 3 row two 1 2 1 row three 1 1 0, multiplied 3 by 1 matrix row one d row two p row three g equals 3 by 1 matrix row one 19 point 62 row two 34 point 95 row three 16 point 50

  2. 3 by 3 matrix row one 4 1 1 row two 6 2 1 row three 3 1 0, multiplied 3 by 1 matrix row one 19 point 62 row two 34 point 95 row three 16 point 50 equals 3 by 1 matrix row one d row two p row three g

  3. 3 by 3 matrix row one 4 1 1 row two 6 2 1 row three 3 1 0, multiplied 3 by 1 matrix row one d row two p row three g equals 3 by 1 matrix row one 19 point 62 row two 34 point 95 row three 16 point 50

Enter to expand or plummet reply.
Answer expanded
Selection

D is correct.
This question requires the examinee to solve systems of linear equations or inequalities using a variety of methods. The system of linear equations can be solved using matrices. Each order can be expressed as an equation, with all three equations written with the variables in the same sequence. The first society is represented by the equation
fourd
plus
p
plus
thou
equals 19.62
, the 2d order by
sixd
plus 2p
plus
yard
equals 34.95
, and the third guild past
3d
plus
p
equals xvi.50
. The rows of the left-hand matrix contain the coefficients of
d,
p, and
g
for each equation:
left parenthesis 4 1 1 right parenthesis, left parenthesis 6 2 one right parenthesis, and left parenthesis 3 one 0 correct parenthesis. The eye matrix contains the variables,
d,
p,
one thousand. The right-manus matrix vertically arranges the constants of the equations.


Competency 006—Empathise exponential and logarithmic functions.

half-dozen. Which of the post-obit is equivalent to the equation
3 log base of operations 10
x
minus 2 log base ten
y
equals 17
?

  1. 3x
    minus iiy
    equals 10 to the power of 17

  2. x
    cubed minus
    y
    squared equals 10 to the power of 17

  3. x
    cubed over
    y
    squared equals 10 to the power of 17
  4. 310
    over 2y
    equals 10 to the power of 17

Enter to expand or collapse answer.
Answer expanded
Pick

C is correct.
This question requires the examinee to utilize the laws of exponents and logarithms.


Nlog base
a
Chiliad
equals log base
a
M
base
Northward
which becomes 3log base 10
x
equals log base ten
x
cubed and 2log base 10
y
equals log base of operations ten
y
squared
.

Log base
a
M
minus log base
a
N
which becomes log base a
M
over
N
which becomes log base ten
x
cubed minus log base 10
y
squared equals log base 10
x
cubed over
y
squared
.
Since
log base of operations
a
M
equals
North
is equivalent to
a
base
Due north
equals
One thousand, then log base x
x
cubed over
y
squared equals 17 is equivalent to 10 to the power of 17 equals
x
cubed over
y
squared
.


Competency 007—Sympathise rational, radical, absolute value, and piece-wise defined functions.

7. Which of the following represents the domain of the function
f(10) =
The square root of 2x plus 3 over 3x plus 1?

  1. left parenthesis negative 3 halves comma negative 1 third right parenthesis union left parenthesis negative 1 third comma infinity right parenthesis.
  2. left bracket negative 3 halves comma negative 1 third right parenthesis union left parenthesis negative 1 third comma infinity right parenthesis.
  3. left parenthesis negative infinity comma negative 3 halves right parenthesis union left parenthesis negative 3 halves comma negative 1 third right parenthesis union left parenthesis negative 1 third comma infinity right parenthesis.
  4. left bracket negative 3 halves comma negative 1 third right parenthesis union left parenthesis negative 1 third comma 0 right parenthesis union left parenthesis 0 comma infinity right parenthesis.

Enter to expand or collapse answer.
Reply expanded
Selection

B is correct.
This question requires the examinee to analyze rational, radical, absolute value, and piece-wise defined functions in terms of domain, range, and asymptotes. Unless otherwise specified, the domain of a role is the range of values for which the function has a real number value. A rational function must have a nonzero denominator, and solving the equation
3x
plus one equals 0 yields
x
equals negative 1 third
. Thus this value must be excluded from the domain. The radical expression in the numerator must have a non-negative argument and solving the inequality
210
plus 3 is greater than or equal to 0 yields
x
is greater than or equal to negative 3 over 2
. The “or” represents the union of the two sets divers by the inequalities, or the matrimony of the 2 intervals.

Domain Iii—Measurement and Geometry

Competency 008—Understand measurement principles and procedures.

The shape of the letter B is designed as shown below, consisting of rectangles and semicircles.

There is a diagram of a block letter B.

The overall elevation is dimensioned as h. The tiptop of each outer semicircle is dimensioned as h over 2. The height of each semicircular void (one in the upper one-half of the alphabetic character and ane in the lower half of the letter) is dimensioned as h over 6. The thickness of the spine of the letter is dimensioned as h over 6.

8. Which of the post-obit formulas gives the area
A
of the shaded region as a function of its meridian
h?


  1. A
    equals
    h
    squared times left parenthesis one over half dozen plus pi over 18 right parenthesis.”

  2. A
    equals
    h
    squared times left parenthesis 1 over 6 plus pi over ix right parenthesis.”

  3. A
    equals
    h
    squared times left parenthesis 1 over 6 plus 2pi over 9 correct parenthesis.”

  4. A
    equals
    h
    squared times left parenthesis one over six plus 5pi over xviii right parenthesis.”

Enter to aggrandize or plummet answer.
Answer expanded
Choice

A is correct.
This question requires the examinee to employ the concepts of perimeter, circumference, surface area, surface area, and volume to solve real-world problems. The total area of the letter B can be viewed as the area of an

h
times
h
over half-dozen

rectangle plus the expanse of a circle with radius

h over 4

minus the area of a circle with radius


h over 12
, or

h
squared over half-dozen plus pi times the square of left parenthesis
h
over four right parenthesis squared minus pi times the foursquare of left parenthesis
h
over 12 correct parenthesis.

This simplifies to

h
squared over half-dozen plus pi
h
squared over 16 minus pi
h
squared over 144.

and further to

h
squared times left parenthesis i over six plus pi over sixteen minus pi over 144 right parenthesis.

and

h
squared times left parenthesis 1 over 6 plus 9pi over 144 minus pi over 144 right parenthesis and
h
squared times left parenthesis 1 over 6 plus pi over 18 correct parenthesis.

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Competency 009—Understand Euclidean geometry in two and three dimensions.

There is a diagram of a geometric proof.

The effigy is a parallelogram, ABCD. The angles at A and C are astute, and a diagonal is drawn from A to C. The proof reads as follows.
Given: Line segment AB is parallel to line segment DC, line segment A B is congruent to line segment DC.
Prove: Triangle ABC is congruent to triangle CDA.
Argument 1. Line segment AB is congruent to line segment DC. Reason, Given.
Statement 2. Blank. Reason, blank.
Statement 3. Line segment AB is parallel to line segment DC. Reason, Given.
Statement 4. Blank. Reason, bare.
Argument 5. Triangle ABC is coinciding to triangle CDA. Reason, blank.

9. In the proof above, steps ii and 4 are missing. Which of the post-obit reasons justifies stride 5?

  1. AAS
  2. ASA
  3. SAS
  4. SSS

Enter to expand or collapse answer.
Reply expanded
Pick

C is right.
This question requires the examinee to clarify formal and breezy geometric proofs, including the utilise of similarity and congruence. The side-angle-side (SAS) theorem can exist used to testify that
Triangle
ABC

and
triangle
CDA

are congruent if each has two sides and an included angle that are congruent with two sides and an included bending of the other. In the diagram
line segment AB and line segment DC
are given equally congruent, and the missing statement 2 is that
the line segment A C
is congruent to itself past the reflexive property of equality. The included angles
bending
BAC
and angle
DCA

are congruent because they are alternate interior angles synthetic past the transversal
line segment
AC

that crosses the parallel line segments
line segment
AB
and line segment
DC
. Thus
Triangle
ABC
and triangle
CDA

encounter the requirements for using SAS to prove congruence.


Competency 010—Understand coordinate and transformational geometry.

ten. The vertices of triangle
ABC
are

A
left parenthesis negative 5, 3 right parenthesis,
B
left parenthesis ii, 2 right parenthesis, and
C
left parenthesis negative one, negative 5 right parenthesis
. Which of the following is the length of the median from vertex
B
to side
AC?

  1. iv
  2. 2 times the square root of 5
  3. the square root of 34
  4. 4 times the foursquare root of 5

Enter to expand or plummet respond.
Respond expanded
Choice

C is correct.
This question requires the examinee to apply concepts of distance, midpoint, and gradient to classify figures and solve bug in the coordinate plane. The midpoint of side
Air-conditioning
where its median intersects is computed as
left parenthesis negative five plus negative 1 over 2, 3 plus negative 5 over 2 right parenthesis equals left parenthesis negative 3, negative 1 right parenthesis. The distance from

B
left parenthesis two, ii correct parenthesis to
left parenthesis negative 3, negative 1 correct parenthesis

is computed as

d
equals the square root of the quantity open paren negative three negative 2 close paren squared plus open paren negative 1 negative 2 close paren squared equals the foursquare root of 25 plus 9 equals the square root of 34.
.

Domain Iv—Trigonometry and Calculus

Competency 011—Understand trigonometric functions.

11. Which of the post-obit are the solutions to
two sine squared theta equals cosine theta plus 1 for 0 is less than theta is less than or equal to 2pi?

  1. pi over 6 comma pi comma 11 pi over 6
  2. pi over 3 comma pi comma 5 pi over 3
  3. 2 pi over 3 comma 4 pi over 3 comma 2 pi
  4. 5 pi over 6 comma 7 pi over 6 comma 2 pi

Enter to expand or plummet answer.
Answer expanded
Option

B is right.
This question requires the examinee to manipulate trigonometric expressions and equations using techniques such as trigonometric identities. Since
sine squared theta equals 1 minus cosine squared theta, 2 sine squared theta equals cosine theta plus 1 which becomes 2 times 1 minus cosine squared theta equals cosine theta plus 1 which becomes 2 cosine squared theta plus cosine theta negative 1 equals 0 which becomes left parenthesis ii cosine theta minus 1 correct parenthesis times cosine theta plus i equals 0 which becomes cos theta equals 1 half or cosine theta equals negative 1. Thus for null is less than theta is less than or equal to 2pi, theta equals pi over 3 comma 5pi over three comma or pi.


Competency 012—Understand differential calculus.

12. If

f
prime of
x
equals iiix
to the fourth minus eightx
squared plus vi, what is the value of the limit every bit
h
approaches 0 of
f
prime of left parenthesis ane plus
h
right parenthesis minus
f
of left parenthesis one right parenthesis all over
h
?

  1. negative 4
  2. negative 1
  3. 1
  4. 4

Enter to expand or collapse reply.
Respond expanded
Choice

A is correct.
This question requires the examinee to clarify the derivative as the slope of a tangent line and as the limit of the difference caliber. The limit expression is equivalent to the derivative

f
prime of left parenthesis 1 right parenthesis
. Since it is much easier to evaluate the derivative of a polynomial, this is preferred over evaluating the limit expression.

f
prime number of
x
equals 12x
cubed minus 16x, so
f
prime of ane equals 12 minus 16 equals negative 4
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Competency 013—Empathize integral calculus.

13. A sum of
2000 dollars
is invested in a savings account. The amount of coin in the account in dollars later on
t
years is given past the equation

A
equals 2000e to the
0.05t
. What is the estimate average value of the account over the first ii years?

  1. 2103 dollars
  2. 2105 dollars
  3. 2206 dollars
  4. 2210 dollars

Enter to expand or plummet answer.
Answer expanded
Option

A is correct.
This question requires the examinee to apply integration to solve real-world issues. The average value of a continuous part

f
of
ten
over an interval from a to b is 1 over b minus a times the integral from a to b of
f
of
x
d x
. Since the independent variable
t
represents the number of years, the average daily balance over 2 years will be
one half
of the integral of the function evaluated from 0 to two:
1 half the integral from 0 to 2 of 2000 due east to the 0.05t
dt
equals g over 0.05 times e to the 0.1 minus ane, which is approximately equal to 2103
.

Domain V—Statistics, Probability, and Discrete Mathematics

Competency 014—Sympathize principles and techniques of statistics.

A frequency bar graph is shown.

The vertical axis is labeled frequency. The horizontal centrality is labeled consequence. The values of the bars are: Upshot, 1, frequency 10
Result, 2, frequency 30
Outcome, three, frequency 50
Issue, iv, frequency 30
Outcome, five, frequency xx
Outcome, half dozen, frequency 10
Outcome, vii, frequency 10

fourteen. Which of the following statements describes the set up of data represented by the histogram above?

  1. The manner is equal to the mean.
  2. The mean is greater than the median.
  3. The median is greater than the range.
  4. The range is equal to the mode.

Enter to expand or collapse answer.
Answer expanded
Option

B is correct.
This question requires the examinee to clarify information in a variety of representations. The mean tin can be calculated as
left bracket 10 left parenthesis one right parenthesis plus 30 left parenthesis 2 right parenthesis plus fifty left parenthesis 3 right parenthesis plus 30 left parenthesis 4 right parenthesis plus 20 left parenthesis 5 right parenthesis plus 10 left parenthesis six right parenthesis plus 10 left parenthesis 7 right parenthesis correct bracket divided by 160 equals 3.5625. The median is the 50th percentile, which is 3. The mode is the most frequent value, which is three. The range is
seven minus 1 equals 6. Thus “the mean is greater than the median” is the correct response.

Competency 015—Understand principles and techniques of probability.

15. The heights of adults in a large group are approximately normally distributed with a hateful of 65 inches. If 20% of the adult heights are less than 62.5 inches, what is the probability that a randomly chosen adult from this group volition be between 62.5 inches and 67.5 inches tall?

  1. 0.3
  2. 0.iv
  3. 0.5
  4. 0.half-dozen

Enter to expand or collapse answer.
Reply expanded
Pick

D is correct.
This question requires the examinee to analyze compatible, binomial, and normal probability distributions. A normal distribution is symmetric nearly the hateful. Thus if 20% of the heights are less than 62.5 inches (ii.v inches from the hateful), then 20% of the heights will be greater than 67.5 inches (also 2.5 inches from the mean). Thus
100 percent minus left parenthesis 20 percent plus twenty percent right parenthesis equals 60 percent
and the probability is 0.6 that the developed will be between 62.5 and 67.5 inches tall.


Competency 016—Empathise principles of detached mathematics.

sixteen. Five different algebra textbooks, two different calculus textbooks, and iv different geometry textbooks are to be arranged on a shelf. How many different arrangements are possible if the books of each subject must be kept together?

  1. left parenthesis five times ii times four right parenthesis cubed
  2. 11 factorial over 5 factorial 2 factorial four factorial
  3. five factorial 2 factorial iv factorial 3 factorial
  4. 11 cubed over left parenthesis 5 factorial 2 factorial four factorial correct parenthesis cubed

Enter to aggrandize or collapse answer.
Answer expanded
Option

C is correct.
This question requires the examinee to apply concepts of permutations and combinations to solve problems. If the books in each of the 3 subjects must be kept together, and then the number of means the groups of books can be bundled by subject field is represented by
iii factorial. If there are
n
books within a field of study, the number of means the books can be bundled is
n!. Thus the algebra books tin can exist bundled in
5 factorial
unlike ways, the calculus books can exist arranged in
2 factorial
dissimilar means, and the geometry books can be arranged in
four factorial
different ways. Since there is independence between the unlike arrangements computed, the total number of ways the books tin can exist arranged is the production of all the factorials
iii factorial 5 factorial 2 factorial 4 factorial
which is equivalent to
5 factorial 2 factorial iv factorial iii factorial.

Which Expression is Equivalent to 144 Superscript Three-halves

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