Download ACE Answers Investigation 5

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Investigation 5
Applications
1. Students may explain the result using an
7. a. Possible answers: The product of three
equation such as (2x + 6) ÷ 2 – x, where x
is the number picked. In order to simplify
this equation, students will need to make
sense of (2x + 6) ÷ 2. They need to rewrite
2x + 6 in two parts as x + x + 3 + 3, with
the result that (2x + 6) ÷ 2 = x + 3. Thus
(2x + 6) ÷ 2 – x = x + 3 – x = 3 for any
number x a student picks.
consecutive whole numbers will always
yield an even number. The product of
three consecutive whole numbers is
always a multiple of 6.
b. The product of three consecutive
whole numbers is even. There are
two cases for a set of 3 consecutive
numbers. Case 1: Two evens and an
odd. Using the facts from Exercise 5
and Problem 2.2, (even × odd) ×
even = even × even = even.
Case 2: Two odds and an even, (odd ×
even) × odd = even × odd = even.
The product of three consecutive whole
numbers is a multiple of 6 since there
is at least one even (multiple of 2) and
at least one multiple of 3 (there are
3 numbers so one must be a multiple
of 3), so the product must be a multiple
of 6.
2. Students may explain the result using
an equation such as (2(x + 4) – 6) ÷
2 – x = 1. In order to simplify this equation,
students will need to make sense of
(2(x + 4) – 6) ÷ 2, which is equal to
(2x + 2) ÷ 2. Since 2x + 2 = x + x + 1 + 1,
dividing x + x + 1 + 1 in two parts results
in x + 1, or [2(x + 4) – 6] ÷ 2 = (2x + 2) ÷
2 = x + 1. Thus [(2(x + 4) – 6) ÷ 2] – x =
x + 1 – x = 1 for any number x a student
picks.
3. Answers will vary.
8. a. The product of four consecutive
4. Answers may vary, but they should make
whole numbers will always be even.
It is also divisible by 24. For any four
consecutive numbers, one of them is
divisible by 4, another is divisible by 2
(but not 4), and at least one of the four
is divisible by 3. For example,
(4 × 3 × 2) × 1 = 24 × 1. Therefore,
24 is a factor of the product of any four
consecutive numbers.
logical sense. Possible argument:
Let n and m represent any integers.
Then 2n + 1 and 2m + 1 are two odd
numbers.
But, (2n + 1) + (2m + 1) = 2n + 2m + 2 =
2(n + m + 1), and
2(n + m + 1) is an even number because it
is a multiple of 2.
So, the sum of two odd integers is even.
b. In a set of four consecutive whole
5. Again, answers may vary; one possible
numbers, there will be two odd whole
numbers and two even whole numbers.
This is symbolically shown with E = any
even whole number and O = any odd
whole number:
O × E × O × E; since O × E = E,
O×E×O×E=E×E=E
argument:
Let n and m represent any integers.
Then 2n and 2m are two even numbers.
But (2n)(2m) = 4nm = 2(2nm).
2(2nm) is an even integer.
So, the product of two even numbers must
be even.
9. a. rational
6. Answers may vary; one possible argument:
Take two odd numbers represented by
(2n + 1) and (2m + 1), where m and n are
integers.
Then 2n and 2m are two even numbers.
But, (2n + 1)(2m + 1) = 4nm + 2n + 2m +
1 = 2(2nm + n + m) + 1, and
2(2nm + n + m) + 1 is an odd number,
since it is 1 more than a multiple of 2.
So, the product of two odd integers is odd.
Say It With Symbols
b. irrational
c. irrational
d. rational
e. rational
f. rational
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Investigation 5
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Answers |
Investigation 5
10. a. 10,034 is divisible by 2; 69,883 is not.
Students may reason that they could
tell because 4 is divisible by 2, but 3
is not.
11. Answers will vary, but students should note
that the last two digits must be divisible
by 4. For example, 4,516 is divisible by
4 whereas 4,519 is not, because 16 is
divisible by 4 and 19 is not. This is not easy
to see. Since all powers of 10 greater than
10 (100, 10,000, etc.) are divisible by 4,
you only have to check the last two digits.
b. Answers will vary. Some students may
suggest that as long as the last digit is
even, then the number is divisible by 2,
using the reasoning of long division.
One explanation is that any digit except
the last is a multiple of 10, which is
clearly divisible by 2, and thus the only
important digit is the ones digit.
12. Answers will vary, but students should note
that the last digit must be 0 or 5. This is
true because the other digit is a multiple
of 10, and thus is divisible by 5. The ones
digit is the only important one to check.
Connections
13. a. Answers will vary. Some possibilities are
14. a. Students may list all the combinations
below.
These buildings are composed of a
central cube and 5 arms that contain
0 cubes, then 1 cube, then 2 cubes,
then 3 cubes, and so on. This pattern is
described in the student edition.
The buildings are composed of a
central tower that contains 1 cube, then
2 cubes, and so on. Each new building
is the previous building with 5 cubes
added.
or make a table. For 2 participants,
2 games must be played; for 3
participants, 6 games must be played;
for 4 participants, 12 games must be
played; for 5 participants, 20 games
must be played; and for 6 participants,
30 games must be played.
b. The expression for the number of
games played g in relation to the
number of participants in the league n
is n(n – 1) = g or n2 – n = g.
b. There are many possible equivalent
15. a. n2, where n is the number of participants
expressions for the number of cubes
in the nth building. Some examples:
1 + 5(n – 1), n + 4(n – 1), or 5n – 4.
in the league
b. n; the diagonal is not filled in because
participants do not play themselves.
c. The fifth building contains 21 cubes,
which can be found by substituting
5 into any of the above equations or
other correct equations.
c. Students should combine the two
equations to create a new expression
n2 – n, but some students may
recognize that this is the same as
n(n – 1) and use this as their expression.
d. Answers will vary since they depend
on what the students had for part (b).
However, the expression 5n – 4 is
the simplest form of the expression
possible. This expression can be
thought of as representing an addition
of five blocks for every new tower
starting with n = 1.
d. Answers may vary since they depend
on which expressions students used for
parts (a) and (b). Students should note
that, regardless, the expressions are
equivalent.
16. 4 and –3; (x – 4)(x + 3) = 0, so x = 4 or
x = –3.
e. Students may either find a different
pattern that relates to one of the
previous expressions, or they may
find a new expression altogether.
Other possibilities include rewriting
n + 4(n – 1) as n + 4n – 4.
Say It With Symbols
17. 0 and –4; x2 + 4x = x(x + 4) = 0, so x = 0
or x + 4 = 0, and thus x = 0 or x = –4.
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Investigation 5
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Investigation 5
18. –4 and –5; x2 + 9x + 20 =
(x + 4)(x + 5) = 0, so (x + 4) = 0 or
(x + 5) = 0, and thus x = –4 or x = –5.
make sense since the number of dimes
and quarters each should be greater
than or equal to zero. So, x should be
less than or equal to 1,000 and greater
than or equal to zero, since you want to
account for only a nonnegative number
of coins. Thus, having the value of the
coins add up to anything less than $100
is not possible.
19. –8 and 1; x2 + 7x – 8 = (x + 8)(x – 1) = 0,
so (x + 8) = 0 or (x – 1) = 0, and thus
x = –8 or x = 1.
20. 10 and 1; x2 – 11x + 10 =
(x – 10)(x – 1) = 0, so (x – 10) = 0 or
(x – 1) = 0, and thus x = 10 or x = 1.
To check, substitute the values into the
equation:
0.10x + 0.25(1,000 – x)
0.10(1600) + 0.25(1,000 – 1,600)
160 – 150 = 10, which is the amount of
money in dollars in the bag.
21. 9 and –3; x2 – 6x – 27 =
(x – 9)(x + 3) = 0, so (x – 9) = 0 or
(x + 3) = 0, and thus x = 9 or x = –3.
22. 5 and –5; x2 – 25 = (x + 5)(x – 5) = 0,
so x = –5 or x = 5.
28. a. P = 2(2w + 4) + 2w = 4w + 8 + 2w =
6w + 8 = 2(3w + 4)
23. 10 and –10; x2 – 100 = (x + 10)(x – 10) =
0, so x = –10 or x = 10.
b. Find the dimension of the rectangle if
the perimeter is 80. If the perimeter
of a rectangle is 80, then you could
substitute 80 into your simplest
equation, solve for w, and calculate the
length using the value of w.
80 = 6w + 8
80 – 8 = 6w + 8 – 8
72 = 6w
24. –0.5 and –1; 2x2 + 3x + 1 =
(2x + 1)(x + 1) = 0, so x = –0.5 or x = –1.
25. – 4 and –2; 3x2 + 10x + 8 =
3
(3x + 4)(x + 2) = 0, so x = – 4 or x = –2.
3
26. a. 6r + 2(r – 30) = 8r – 60 = distance
traveled
72
6
b. Possible question: How many miles are
traveled if the rate is 70 miles per hour?
Substitute 70 for r, the rate of speed,
in the simplest expression and get
4(70) – 30 = 280 – 30 = 250 miles
traveled.
29. a. 15x + 9(4,000 – x) = 15x + 36,000 –
9x = 6x + 36,000 = 6(x + 6,000)
b. If the total amount of money collected
at a concert was $60,000, find the
number of reserved and unreserved
seats. Set 60,000 equal to your simplest
expression and solve for the number of
tickets.
60,000 = 6x + 36,000
60,000 – 36,000 = 6x
24,000 = 6x
b. Possible question: If the amount of
money in the bag is $10, how many of
each coin are in the bag? Set $10 equal
to your expression and solve for the
number of coins:
10 = 250 – 0.15x
10 – 250 = 250 – 250 – 0.15x
–240 = –0.15x
24,000
6
= 6x
6
4,000 = x
The number of reserved seats was
4,000 and the number of unreserved
seats was 4,000 – 4,000, or 0.
= –0.15 x
–0.15
1,600 = x
So, in theory, there are 1,600 dimes and
–600 quarters. Note: Students may
recognize that this answer does not
Say It With Symbols
6
12 = w
Therefore, the width is 12 feet and the
length is 2(12) + 4 or 28 feet.
27. a. If x is the number of dimes, then
0.10x + 0.25(1,000 – x) = 0.10x +
250 – 0.25x = 250 – 0.15x = amount
of money in dollars
–240
–0.15
= 6w
3
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Investigation 5
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Answers |
Investigation 5
30. a. 32 feet. Students may substitute the
value 2 into the equation for t and
get 32.
33. The expression is linear.
b. 36 feet. To find the maximum
height, students would first find the
x-intercepts (See part (c).). Then, the
maximum height occurs between the
2
t=
h=
–16  3 
2
2
+
48  3 
2
–7
–5
–3
5
y = 4(x –7) + 6
–50
–42
–34
–2
34. The expression is linear.
two intercepts, so it is at t = 3 . When
3
,
2
x
= 36.
x
–7
–5
–3
5
y = –3– 7(x + 9)
–17
–31
–45
–101
35. The expression is exponential. The values
for y in the table are approximations.
c. 3 seconds; students can use the
equation h = –16t2 + 48t and
the factored form h = –16t(t – 3),
or h = 16t(–t + 3). They find the
x-intercepts, which are 0 and 3. The
intercepts are the points where the
height is 0, so the ball is in the air from
0 to 3 seconds.
–7
x
y = 2(3)x
–5
–3
5
0.001 0.008 0.074
486
36. The expression is quadratic.
d. Answers will vary; students who use a
graph may see the highest point on the
graph as the maximum height and use
the x-intercepts to find the total time
the ball was in the air. Students who
use a table will look for the greatest
y-value in the table for part (b) and the
x-values when y is 0 for part (c).
x
–7
–5
–3
5
y = 3x2 – x – 1
153
79
29
69
37. The expression is quadratic.
x
–7
–5
–3
5
y = 5(x – 2)(x + 3)
180
70
0
180
38. For Exercise 33, 4x – 28 + 6, or 4x – 22;
for Exercise 34, –3 – 7x – 63, or –66 – 7x.
39. Table 1: linear, y = 9 – 3x
Table 2: quadratic, y = x2 – 16
Table 3: none of these
Table 4: exponential, y = 3(4x)
Table 5: none of these. Note: This is
31–32. a. The two expressions in both
Exercises 31 and 32 are not equivalent.
For part (a), students should make a
table or graph and find that the two
equations have either the same table
and graph, or that they are different, in
which case they are not equivalent.
inverse variation, y = 4 .
x
40. a. Yes, the perimeter can be a rational
b. They can simplify the expressions
number if the three numbers form a
using the Distributive Property and/
Pythagorean Triple. For example, a
or Commutative Property to see
3–4–5 right triangle will have perimeter
if the expressions are equivalent.
12 units, which is rational since all
For Exercise 31, the first expression
integers are rational numbers.
simplifies to 4x – 5, and for Exercise 32,
the first simplifies to 6x – 7 and the
b. Yes, the perimeter can be irrational
second to 6x – 3. To show that the
if the three numbers do not form a
expressions are not equivalent,
Pythagorean Triple. For example, 1, 1,
students can show that a value
and 2 .
for x when substituted into both
expressions yields different values.
Say It With Symbols
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Investigation 5
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Investigation 5
41. a. No, the perimeter cannot be a rational
number. If the length of the hypotenuse
is an integer, then the length of the
shortest leg will be rational. However,
the length of the longer leg will be
irrational because it is square root of
three times the length of the shorter leg.
The perimeter is the sum of two rational
numbers plus an irrational number.
For example, a 30–60–90 triangle with
a hypotenuse of 10 inches will have a
shorter leg measuring 5 inches, and a
longer leg measuring 5 3 inches. The
perimeter will be 15 + 5 3 inches.
b. Using similar reasoning as in part (a),
the perimeter will always be irrational.
Extensions
42. a. 52 + 62 = 312 – 302
43. a. Answers may vary depending on the
choice for n.
b. Let a be the first number and a + 1 be
the second number.
Then a2 + (a + 1)2 = (a(a + 1) + 1)2 –
(a(a + 1))2.
b. George is correct. All it takes is
one counterexample to disprove
a conjecture. So, even if there are
many examples that work, that is not
sufficient proof. This conjecture fails
when n is 41, for example.
c. Begin with the right-hand side of the
equation to show that it is equal to
the left-hand side.
a2 + (a + 1)2 = (a(a + 1) + 1)2
– (a(a + 1))2
= (a2 + a + 1)2 – (a2 + a)2
by the Distributive
Property
= (a2 + a + 1)(a2 + a + 1)
– (a2 + a)(a2 + a) by
Definition of Exponents
= (a4 + 2a3 + 3a2 +
2a + 1) – (a4 + 2a3 + a2)
by the Distributive
Property
= 2a2 + 2a + 1 by
combining like terms
= a2 + a2 + 2a + 1 by
rewriting the first term
= a2 + a2 + a + a + 1 by
rewriting the third term
= a2 + a(a + 1) + 1(a + 1)
= a2 + (a + 1)2 by
factoring the last two
terms
Since the left-hand side and the righthand side of the equation are equal,
the conjecture is true.
Say It With Symbols
44. a.
b.

2 +  2
0.5  2  +   2  ; the sum
0.5 +
c.

2 is an irrational number.
2  2  2 2 , which is irrational
because it is the product of a rational
number and an irrational number.
45. If the sum of the digits of a number is
divisible by 3, then the number must
also be divisible by 3. Also, if a number
is divisible by 3, then the sum of the
digits is divisible by 3. The following
proofs are written symbolically. This is not
something that students are expected
to be able to do at this point. Students
may talk informally about parts of the
proof without the symbols. Suppose
n is a three-digit number whose digits
sum to a number that is divisible by 3.
Then n = a(100) + b(10) + c, where a,
b, and c are the digits of n. Rewriting,
n = 99a + a + 9b + b + c. Using the
Commutative Property, n = (99a + 9b) +
(a + b + c). Using the Distributive Property,
(99a + 9b) is divisible by 3, or 3(33a + 3b).
In order for the entire number to be
divisible by 3, a + b + c must also be
divisible by 3.
5
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Investigation 5
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Investigation 5
For example,
n = 651
= 6 × 100 + 5 × 10 + 1
= (6 × 99) + (6 × 1) + (5 × 9) + (5 × 1) + 1
= (6 × 99) + (5 × 9) + (6 × 1) + (5 × 1) + 1
= (99 × 6 + 9 × 5) + (6 × 1 + 5 × 1 + 1)
= 3(33 × 6 + 3 × 5) + (6 + 5 + 1)
= 3(33 × 6 + 15) + (12)
= 3(198 + 15 + 4)
So, given that n is divisible by 3,
6 + 5 + 1 = 12 is divisible by 3.
46. If the sum of the digits of a number is
divisible by 3 and it is even, then the
number is divisible by 6. Suppose n is
an even number whose digits sum to a
number that is divisible by 3. Since it is
even, it must be divisible by 2. It must also
be divisible by 3, by Exercise 45. By the
Fundamental Theorem of Arithmetic (the
prime factorization of), n must be of the
form n = 2 × 3 ×…If we rewrite,
n = 6 ×…Thus, n is divisible by 6.
Now let’s look at another case: If a number
is divisible by 3, then the sum of its digits
is divisible by 3. Say a number n is divisible
by 3, then we can write n = 3k, where k is
an integer. Thus, n = a(100) + b(10) + c =
3k. Again, n = 99a + 9b + a + b + c = 3k,
so dividing each side by 3, k = 33a + 3b +
47. a. Answers will vary.
abc
. But
3
then a  b  c
3
b. Students may find it easiest to explain
why this method works by forming an
equation to represent the value of any
number ending in five, such as 10x + 5,
where x can be any whole number.
Then a student taking the square of
this value will get (10x + 5)(10x + 5) =
100x2 + 100x + 25 = 100(x2 + x) + 25.
This equation represents Judy’s method
of finding the square.
since k is an integer,
must be an integer, so
a + b + c (the sum of the digits of n)
must be divisible by 3.
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Investigation 5