Math 2 Review
Answer Section
SHORT ANSWER
1. ANS:
a. F =
or L =
b. This is an example of inverse variation since the rule can be rewritten as F =
y=
which is of the form
.
c. 1,800
d. As the length of the wrench increases, the amount of force needed decreases at a decreasing rate.
PTS: 1
REF: Lesson 1-2
OBJ: 1-2.2 Solve for one variable in terms of the others in situations where the variables are related by
direct and inverse variation.
NAT: 1 | 2 | 6 | 9
2. ANS:
a. This is an example of direct variation since the formula is of the form y =
.
b. S = 6
c. 105.84 =
= 54 cm
, so
=
= 4.2 cm.
PTS: 1
REF: Lesson 1-1
OBJ: 1-1.4 Solve problems involving direct and inverse variation.
NAT: 1 | 2 | 6 | 9
3. ANS:
y=
y=
y=
y=
y=
y=
Explanations should take into consideration increasing/decreasing values, positive/negative values, and
general shape expected.
PTS: 1
REF: Lesson 1-1
OBJ: 1-1.1 Recognize numeric and graphic patterns of change in direct and inverse variation
relationships.
NAT: 1 | 2 | 9
4. ANS:
a. A person’s weight in space will decrease as the person moves away from Earth.
b. Evelyn will weigh more than Marissa in space. If r and d are the same and w increases, then S will
increase also.
c. A person’s weight in space varies directly with a person’s weight on Earth and inversely with a
person’s distance from the center of Earth.
d. d =
e. w =
PTS: 1
REF: Lesson 1-1 | Lesson 1-2
OBJ: 1-1.1 Recognize numeric and graphic patterns of change in direct and inverse variation
relationships. | 1-1.4 Solve problems involving direct and inverse variation. | 1-2.1 Write rules to define
functions of two variables that combine direct and inverse variation.
NAT: 1 | 2 | 6 | 9
5. ANS:
PTS: 1
NAT: 1 | 2 | 3
6. ANS:
a. M = 120n + 30r
REF: Lesson 1-2
OBJ: 1-2.5 Graph linear equations in the form ax + by = c.
b. 1,200 = 120n + 30r
c. 5 newspaper ads
d. 40 radio ads
e. Rule: r =
= 40
4n
Meaning of slope: An increase of 1 newspaper ad will require a decrease of 4 radio ads.
Meaning of y-intercept: If Eli buys no newspaper ads, he can buy 40 radio ads.
PTS: 1
REF: Lesson 1-2
OBJ: 1-2.1 Write rules to define functions of two variables that combine direct and inverse variation. | 12.3 Write equations in the general form ax + by = c to express conditions relating two variables. | 1-2.2
Solve for one variable in terms of the others in situations where the variables are related by direct and
inverse variation.
NAT: 1 | 2 | 6 | 9
7. ANS:
a. 3x + 4y = 58
b. x + 3y = 36
c. 6 small baskets and 10 large baskets
PTS: 1
REF: Lesson 1-1 | Lesson 1-3
OBJ: 1-1.2 Express direct and inverse variation relationships in symbolic forms. | 1-1.4 Solve problems
involving direct and inverse variation. | 1-3.2 Solve linear systems by graphing, substitution, and
elimination methods. NAT: 1 | 2 | 3 | 6 | 9
8. ANS:
Madison is not correct. If (1.5, 7) is a solution to the system, it must satisfy both equations. While it does
satisfy the first equation, it does not satisfy the second since 3(1.5) + 2 is 6.5 not 7.
PTS: 1
REF: Lesson 1-3
OBJ: 1-3.2 Solve linear systems by graphing, substitution, and elimination methods.
NAT: 1 | 2 | 3 | 6
9. ANS:
a. This system has an infinite number of solutions since both equations are equivalent to
y=
x + 2.
b. This system has no solutions since the equations represent lines that have the same slope and different
y-intercepts. The lines are parallel to each other.
c. This system has exactly one solution since the equations represent two lines with different
slopes, so they must intersect in only one point.
PTS: 1
REF: Lesson 1-3
OBJ: 1-3.3 Recognize linear systems with zero or infinitely many solutions by inspecting graphs,
equation forms, and results of reasoning by substitution and elimination. NAT: 1 | 2 | 3
10. ANS:
a.
b.This matrix is the sum of the three given matrices.
c. This matrix is the hits matrix minus the extra-base hits matrix.
d. There is some room for disagreement, but the following seems like a good choice:
Best: Mays
Second: Cepeda
Third: McCovey
Mays had the most hits in two of the three years and the most extra-base hits in all three years. Cepeda
was consistent in his hitting, but his extra-base hits were considerably below Mays’ extra-base hits.
McCovey had a relatively poor second season and was not as consistent in number of hits as the other
two. Mays had a three-year total of 551 hits and 240 extra-base hits compared to 549 hits and 196 extrabase hits for Cepeda. McCovey was well below this in hits with 387 and extra-base hits with 161. On the
other hand, McCovey had a total of 101 home runs in his first three years compared to only 76 for
Cepeda. Mays again was the leader with a total of 128 home runs. If a student argues for another ordering,
the case would need to focus on individual statistics for individual years, such as McCovey’s home runs
in his first and third years.
PTS: 1
REF: Lesson 2-1
OBJ: 2-1.1 Construct matrices to organize, display, and analyze information. | 2-1.2 Interpret given
matrices. | 2-1.3 Understand, carry out, and interpret matrix operations--row and column sums, matrix
addition and subtraction, and scalar multiplication.
NAT: 1 | 2 | 5
11. ANS:
a.
b.
The above matrix is the matrix product:
Explanation: The wages-per-hour matrix must be first and the hours-worked matrix must be
second in order to do the matrix multiplication. Each entry in the wages-per-hour matrix is
multiplied by a corresponding column entry, and then the four results are added together. This is done for
each of the columns separately.
PTS: 1
REF: Lesson 2-1 | Lesson 2-2
OBJ: 2-1.1 Construct matrices to organize, display, and analyze information. | 2-2.1 Understand, carry
out, and interpret matrix multiplication.
NAT: 1 | 2 | 5
12. ANS:
a. A B is a 2 3 matrix. The number of rows in the product will always be the same as the
number of rows in the first matrix, and the number of columns will always be the same as the
number of columns in the second matrix.
b. To find the entry in the second row and the second column of matrix A
following: (1)(3) + (–1)(–1) = 4.
B, you compute the
PTS: 1
REF: Lesson 2-1 | Lesson 2-2
OBJ: 2-1.3 Understand, carry out, and interpret matrix operations--row and column sums, matrix
addition and subtraction, and scalar multiplication. | 2-2.1 Understand, carry out, and interpret matrix
multiplication.
NAT: 1 | 2
13. ANS:
a. s + g = 300
3s + 5g = 1,336
=
=
=
There were 82 student tickets sold and 218 general public tickets sold.
b. Students could use graphing, tables of values, substitution, or linear combination to solve the
system of equations.
PTS: 1
REF: Lesson 1-3 | Lesson 2-3
OBJ: 1-3.1 Write systems of linear equations to match given problem conditions. | 1-3.2 Solve linear
systems by graphing, substitution, and elimination methods. | 2-3.3 Use matrices and their properties to
solve systems of linear equations. | 2-3.4 Review, analyze, and compare various methods for solving
systems of linear equations. NAT:
1|2|3|6
14. ANS:
Any two non-square matrices that are the same size (say m n) will satisfy this condition. It is possible to
add them together because they are the same size. Since they are not square matrices, the number of
columns in A, m, does not equal the number of rows in B, n. Thus, they cannot be multiplied together.
PTS: 1
REF: Lesson 2-3
OBJ: 2-3.1 Examine properties of operations with matrices.
NAT: 1 | 2
15. ANS:
Three properties that students are familiar with and that hold for real numbers but not for matrices are: the
Commutative Property of Multiplication, the property that a multiplicative inverse always exists (except
for 0); and the zero property of multiplication. After identifying the property, students should provide
matrices that illustrate the property not holding.
PTS: 1
REF: Lesson 2-3
OBJ: 2-3.2 Compare properties of matrices with those of real numbers.
NAT: 1 | 2
16. ANS:
a.
b.
The triangle is isosceles since BC = CA.
c.
The coordinates of D are (0, 3). The slope of
slopes of
and
is
and the slope of
is
. Since the
are negative reciprocals, the segments are perpendicular to each other.
PTS: 1
REF: Lesson 3-1
OBJ: 3-1.1 Use coordinates to represent points, lines, and geometric figures in a plane.
NAT: 1 | 2 | 3
17. ANS:
The quadrilateral is a parallelogram. Students could show that opposite sides have the same slope or that
opposite sides have equal lengths.
Slope of
= slope of
=
Slope of
= slope of
=
AB = CD =
AD = BC =
PTS: 1
REF: Lesson 3-1
OBJ: 3-1.1 Use coordinates to represent points, lines, and geometric figures in a plane.
NAT: 1 | 2 | 3
18. ANS:
a. 5
b. (0, -4)
c.
PTS: 1
REF: Lesson 3-1
OBJ: 3-1.1 Use coordinates to represent points, lines, and geometric figures in a plane.
NAT: 1 | 2 | 3
19. ANS:
a.
PTS: 1
REF: Lesson 3-1
OBJ: 3-1.1 Use coordinates to represent points, lines, and geometric figures in a plane. | 3-1.4 Review
and synthesize the major objectives of the unit.
NAT: 1 | 2 | 3
20. ANS:
a.
b. (x, y)
(-3x, 3y)
Reflecting the figure across the y-axis takes (x, y)
(-x, y). Then a size transformation of
magnitude 3 centered at the origin takes (-x, y)
(-3x, 3y).
c. After the reflection, the lengths of the image are the same as the lengths of the original figure. Then
since size transformations change lengths by a factor equal to the magnitude of the size transformation,
the length of
will be 3 times the length of
.
d. The reflection leaves the area unchanged. The size transformation of magnitude 3 will change the area
by a factor of 3 , or 9. So, the area of WXYZ is 9 times the area of ABCD.
PTS: 1
REF: Lesson 3-1
OBJ: 3-1.1 Use coordinates to represent points, lines, and geometric figures in a plane. | 3-1.4 Review
and synthesize the major objectives of the unit.
NAT: 1 | 2 | 3
21. ANS:
A’B’C’ is not a size transformation image of ABC. Students could justify this in several ways:
• Since point A(0, 2) has image A´(0, 5), if the transformation were a size transformation, it
would have to have magnitude 2.5. But then the image of B(2, 3) would have to be (5, 7.5),
but it is not. So, the transformation cannot be a size transformation centered at the origin.
• Lines AA' and BB' should intersect at the origin if the transformation is a size transformation centered
at the origin. It is clear that they do not, so the transformation is not a size transformation centered at the
origin.
•
Students could show that the lengths of corresponding sides are not proportional.
PTS: 1
REF: Lesson 3-1
OBJ: 3-1.2 Use coordinate rules for size transformations centered at the origin to develop corresponding
matrix representations.
NAT: 1 | 2 | 3
22. ANS: