Find each square root.
1.
SOLUTION:
= 4
ANSWER:
4
2.
SOLUTION:
= −10
ANSWER:
−10
3.
SOLUTION:
= ±9
ANSWER:
±9
Estimate each square root to the nearest
integer.
4.
SOLUTION:
The first perfect square less than 27 is 25. =5
The first perfect square greater than 27 is 36. =
6
The square root of 27 is between the integers 5 and
6. Since 27 is closer to 25 than to 36, iscloser
to 5 than to 6.
ANSWER:
5
5.
SOLUTION:
The first perfect square less than 48 is 36. =6
The first perfect square greater than 48 is 49. =
7
The negative square root of 48 is between the
integers −6 and −7. Since 48 is closer to 49 than to
36, − is closer to −7 than to −6.
ANSWER:
−7
6.
SOLUTION:
The first perfect square less than 39 is 36.
=6
The first perfect square greater than 39 is 49. =
7
The positive and negative square roots of 39 are
between the integers ±6 and ±7. Since 39 is closer to
36 than to 49, ± is closer to ±6 than to ±7.
ANSWER:
±6
7.A baseball diamond is actually a square with an area
of 8100 square feet. Most baseball teams cover their
diamond with a tarp to protect it from the rain. The
sides are all the same length. How long is the tarp on
each side?
SOLUTION:
The length of each side of the tarp is equal to the
square root of the area.
The tarp is 90 feet long on each side.
ANSWER:
90 ft
Find each cube root.
8.
SOLUTION:
Since 8 •8 •8 = 512, the cube root of 512 is 8.
ANSWER:
8
9.
SOLUTION:
Since 13 •13 •13 = 2197, the cube root of 2197 is
13.
ANSWER:
13
10.
SOLUTION:
Since −10 • −10 • −10 = −1000, the cube root of
−1000 is −10.
ANSWER:
−10
11.
SOLUTION:
Since −7 • −7 • −7 = −343, the cube root of −343 is
−7.
ANSWER:
−7
Estimate each cube root to the nearest integer.
12.
SOLUTION:
The first perfect cube less than 74 is 64.
The first perfect cube greater than 74 is 125.
Approximate the placement of onanumber
line relative to 4 and 5. Since 74 is closer to 64 than
to 125, willbecloserto4.
.
ANSWER:
4
13.
SOLUTION:
The first perfect cube less than 39 is 27.
The first perfect cube greater than 39 is 64.
Approximate the placement of onanumber
line relative to 3 and 4. Since 39 is closer to 27 than
to 64, willbecloserto3.
ANSWER:
3
14.
SOLUTION:
The first perfect cube less than –636 is –729.
The first perfect cube greater than –636 is –512.
Approximate the placement of onanumber
line relative to –9 and –8. Since –636 is closer to –
729 than to –512, willbecloserto–9.
ANSWER:
−9
15.
SOLUTION:
The first perfect cube less than –879 is –1000.
The first perfect cube greater than –879 is –729.
Approximate the placement of onanumber
line relative to –10 and –9. Since –879 is closer to –
1000 than to –729, willbecloserto–10.
ANSWER:
−10
Find each square root.
16.
SOLUTION:
= 6
ANSWER:
6
17.
SOLUTION:
= 3
ANSWER:
3
18.
SOLUTION:
= −13
ANSWER:
−13
19.
SOLUTION:
= −12
ANSWER:
−12
20.
SOLUTION:
There is no real square root because no number
times itself is equal to −25.
ANSWER:
no real solution
21.
SOLUTION:
= ±1
ANSWER:
±1
Estimate each square root to the nearest
integer.
22.
SOLUTION:
The first perfect square less than 83 is 81. =9
The first perfect square greater than 83 is 100.
=10
The square root of 83 is between the integers 9 and
10. Since 83 is closer to 81 than to 100, iscloser
to 9 than to 10.
ANSWER:
9
23.
SOLUTION:
The first perfect square less than 34 is 25. =5
The first perfect square greater than 34 is 36. =
6
The square root of 34 is between the integers 5 and
6. Since 34 is closer to 36 than to 25, iscloser
to 6 than to 5.
ANSWER:
6
24.
SOLUTION:
The first perfect square less than 102 is 100. =
10
The first perfect square greater than 102 is 121.
=11
The negative square root of 102 is between the
integers –10 and –11. Since 102 is closer to 100 than
to 121, iscloserto–10 than to –11.
ANSWER:
−10
25.
SOLUTION:
The first perfect square less than 14 is 9. =3
The first perfect square greater than 14 is 16. =
4
The negative square root of 14 is between the
integers –3 and –4. Since 14 is closer to 16 than to 9,
iscloserto–4 than to –3.
ANSWER:
−4
26.
SOLUTION:
The first perfect square less than 78 is 64. =8
The first perfect square greater than 78 is 81. =
9
The positive and negative square roots of 78 are
between the integers ±8 and ±9. Since78 is closer to
81 than to 64, ± is closer to ±9 than to ±8.
ANSWER:
±9
27.
SOLUTION:
The first perfect square less than 146 is 144. =
12
The first perfect square greater than 146 is 169.
=13
The positive and negative square roots of 146 are
between the integers ±12 and ±13. Since 146 is
closer to 144 than to 169, ± is closer to ±12
than to ±13.
ANSWER:
±12
28.The table shows the heights of the tallest roller
coasters at Cedar Point. Use the formula from
Example 3 to determine how far a rider can see
from the highest point of each ride. Round to the
nearest tenth.
a.MillenniumForce
b.MeanStreak
c.HowmuchfarthercanariderseeontheTop
Thrill Dragster than on the Magnum XL-200?
SOLUTION:
a.
A rider can see about 21.5 miles from the highest
point on the Millennium Force roller coaster.
b.
A rider can see about 15.5 miles from the highest
point on the Mean Streak roller coaster.
c.
Top Thrill Dragster:
Magnum XL-200:
A rider can see 25.0 −17.5 or 7.5 miles farther on
the Top Thrill Dragster than on the Magnum XL-200.
ANSWER:
a.21.5mi
b.15.5mi
c.7.5mi
Find each cube root.
29.
SOLUTION:
Since −12 • −12 • −12 = −1728, the cube root of
−1728 is −12.
ANSWER:
−12
30.
SOLUTION:
Since −14 • −14 • −14 = −2744, the cube root of
−2744 is −14.
ANSWER:
−14
31.
SOLUTION:
Since 6 •6 •6 = 216, the cube root of 216 is 6.
ANSWER:
6
32.
SOLUTION:
Since 11 •11 •11 = 1331, the cube root of 1331 is
11.
ANSWER:
11
Estimate each cube root to the nearest integer.
Do not use a calculator.
33.
SOLUTION:
The first perfect cube less than 499 is 343.
The first perfect cube greater than 499 is 512.
Approximate the placement of onanumber
line relative to 7 and 8. Since 499 is closer to 512
than to 343, willbecloserto8.
ANSWER:
8
34.
SOLUTION:
The first perfect cube less than 576 is 512.
The first perfect cube greater than 576 is 729.
Approximate the placement of onanumber
line relative to 8 and 9. Since 576 is closer to 512
than to 729, willbecloserto8.
ANSWER:
8
35.
SOLUTION:
The first perfect cube less than –79 is –125.
The first perfect cube greater than –79 is –64.
Approximate the placement of onanumber
line relative to –5 and –4. Since –79 is closer to –64
than to –125, willbecloserto–4.
ANSWER:
–4
36.
SOLUTION:
The first perfect cube less than –1735 is –2197.
The first perfect cube greater than –1735 is –1728.
Approximate the placement of ona
number line relative to –13 and –12. Since –1735 is
closer to –1728 than to –2197, willbe
closer to –12.
ANSWER:
–12
37.The area of a square is 215 square centimeters. Find
the length of a side to the nearest tenth. Then find its
approximate perimeter.
SOLUTION:
The length of each side of the square is equal to the
square root of its area.
A side length of the square is about 14.7 centimeters.
Use the formula to find the perimeter of the
square.
The approximate perimeter of the square is 58.8
centimeters.
ANSWER:
14.7 cm; 58.8 cm
38.Order , −8, , 9, −10, , from
least to greatest.
SOLUTION:
Write the given integers as square roots, and then
order the numbers from least to greatest.
−8 =
9 =
−10 =
< , < < < < <
So, the correct order is −10, , , −8,
, , 9.
ANSWER:
−10, , , −8, , , 9
39.Reason Abstractly Write a number that completes
the analogy. x2 is to 121 as x3 is to ?.
SOLUTION:
121 is a perfect square of 11 since 11 •11 = 121.
1331 is a perfect cube of 11 since 11 •11 •11 =
1331.
So,x2isto121asx3 is to 1331.
ANSWER:
1331
40.Identify Structure Find a square root that lies
between 17 and 18.
SOLUTION:
and . Any number between 289
and 324 will have a square root between 17 and 18.
For example, .
ANSWER:
Sample answer:
41.PerseverewithProblemsUse inverse operations
to evaluate the following.
a.
b.
c.
SOLUTION:
a.
b.
c.
ANSWER:
a. 246
b.811
c.732
42.Building on the Essential Question Describe the
difference between an exact value and an
approximation when finding square roots of numbers
that are not perfect squares. Give an example of
each.
SOLUTION:
Sample answer: The exact value of a square root is
given using the square root symbol, such as . An
approximation is a decimal value, such as ≈6.
ANSWER:
Sample answer: The exact value of a square root is
given using the square root symbol, such as .
An approximation is a decimal value, such as
≈6.
43.Which point on the number line best represents
?
AA
BB
CC
DD
SOLUTION:
Use a calculator to estimate the square root of 210.
Point C best represents .
Choice C is the correct answer.
ANSWER:
C
44.Short Response Estimate the cube root of 65 to the
nearest integer.
SOLUTION:
The first perfect cube less than 65 is 64.
The first perfect cube greater than 65 is 125.
Approximate the placement of onanumber
line relative to 4 and 5. Since 65 is closer to 64 than
to 125, willbecloserto4.
ANSWER:
4
45.The new gymnasium at Oakdale Middle School has a
hardwood floor in the shape of a square. If the area
of the floor is 62,500 square feet, what is the length
of one side of the square floor?
F200ft
G225ft
H250ft
J275ft
SOLUTION:
The length of a side of the gymnasium floor is equal
to the square root of its area.
Each side has length 250 units.
Choice H is the correct answer.
ANSWER:
H
46.A surveyor determined the distance across a field
was feet.Whatistheapproximatedistance?
A25.6ft
B30.6ft
C39.6ft
D42.6ft
SOLUTION:
Use a calculator to estimate the square root of 1568
to the nearest tenth.
Choice C is the correct answer.
ANSWER:
C
Solve.
47.Joseph bought four books at a book sale. Each book
cost $4.50. He paid with $20.00. How much change
did he receive?
SOLUTION:
To find the cost of four books, multiply the number of
books by the cost for each book.
4×4.50=18.00
So, the four books cost $18.00.
To find the amount of change he received, subtract
the cost of the books from the amount that Joseph
paid.
20 – 18 = 2
So, Joseph received $2.00 in change.
ANSWER:
$2.00
48.Mrs. Tanner paid $18.00 for six boxes of pencils.
How much did each box of pencils cost?
SOLUTION:
To find the cost of each box of pencils, divide the
total cost by the number of boxes of pencils.
18÷6=3
So, each box of pencils cost $3.00.
ANSWER:
$3.00
49.Kathryn had $367.50 in her bank account. She wrote
a check for $25.00, and then withdrew $50.00 in
cash. She made a deposit of $100.00. How much
money is in her bank account now?
SOLUTION:
Kathryn has $392.50 in her bank account now.
Kathrynhad$367.50
in her bank account. $367.50
Shewroteacheckfor
$25.00. $367.50– $25.00 = $342.50
Shewithdrew$50.00
in cash. $342.50– $50.00 = $292.50
Shemadeadepositof
$100.00. $292.50+$100.00=$392.50
ANSWER:
$392.50
Name the property shown by each statement.
50.3 + 6 = 6 + 3
SOLUTION:
The order of the numbers changed. This is the
Commutative Property of Addition.
ANSWER:
Commutative Property (+)
51.13 + 0 = 13
SOLUTION:
When zero is added to any number, the sum is the
number. This is the Additive Identity Property.
ANSWER:
Additive Identity Property
52.(2 •5) + 6 = 6 + (2 •5)
SOLUTION:
The order of the numbers changed. This is the
Commutative Property of Addition.
ANSWER:
Commutative Property (+)
53.(x + 3) + 9 = x + (3 + 9)
SOLUTION:
The grouping of the numbers changed. This is the
Associative Property of Addition.
ANSWER:
Associative Property (+)
54.(y •2) •3 = y •(2 •3)
SOLUTION:
The grouping of the numbers changed. This is the
Associative Property of Multiplication.
ANSWER:
AssociativeProperty(×)
55.28 •1 = 28
SOLUTION:
When 1 is multiplied by any number, the product is
the number. This is the Multiplicative Identity
Property.
ANSWER:
Multiplicative Identity Property
56.n + t = t + n
SOLUTION:
The order of the variables has changed. This is the
Commutative Property of Addition.
ANSWER:
Commutative Property (+)
57.1652 •0 = 0
SOLUTION:
When zero is multiplied by any number, the product is
zero. This is the Zero Property of Multiplication.
ANSWER:
Zero Property of Multiplication
58.Suppose that four-tenths of the rectangle below is
shaded.
a. What fraction of the rectangle is not shaded?
b. What fraction of the rectangle would still need to
be shaded for half of the rectangle to be shaded?
c. If an additional of the original rectangle were
to be shaded, what fraction of the rectangle would be
shaded?
SOLUTION:
a. To find the fraction of the rectangle that is not
shaded, subtract the fraction that is shaded from one
whole.
of the rectangle is not shaded.
b. To find the fraction of the rectangle that would still
need to be shaded, subtract the fraction that is
already shaded from one-half.
of the rectangle would still need to be shaded for
half of the rectangle to be shaded.
c.
If an additional of the original rectangle were
shaded, of the rectangle would be shaded.
ANSWER:
a.
b.
c.
Find each product or quotient.
59.8 •(–8)
SOLUTION:
The product of two integers with different signs is
negative. So, 8 •(–8) = –64.
ANSWER:
–64
60.–4 •(–12)
SOLUTION:
The product of two integers with the same sign is
positive. So, –4 •(–12) = 48.
ANSWER:
48
61.40÷(–5)
SOLUTION:
The quotient of two integers with different signs is
negative.So,40÷(–5) = –8.
ANSWER:
–8
62.–150÷(–25)
SOLUTION:
The quotient of two integers with the same sign is
positive. So, –150÷(–25) = 6.
ANSWER:
6
Find each square root.
1.
SOLUTION:
= 4
ANSWER:
4
2.
SOLUTION:
= −10
ANSWER:
−10
3.
SOLUTION:
= ±9
ANSWER:
±9
Estimate each square root to the nearest
integer.
4.
SOLUTION:
The first perfect square less than 27 is 25. =5
The first perfect square greater than 27 is 36. =
6
The square root of 27 is between the integers 5 and
6. Since 27 is closer to 25 than to 36, iscloser
to 5 than to 6.
ANSWER:
5
5.
SOLUTION:
The first perfect square less than 48 is 36. =6
The first perfect square greater than 48 is 49. =
7
The negative square root of 48 is between the
integers −6 and −7. Since 48 is closer to 49 than to
36, − is closer to −7 than to −6.
ANSWER:
−7
6.
SOLUTION:
The first perfect square less than 39 is 36.
=6
The first perfect square greater than 39 is 49. =
7
The positive and negative square roots of 39 are
between the integers ±6 and ±7. Since 39 is closer to
36 than to 49, ± is closer to ±6 than to ±7.
ANSWER:
±6
7.A baseball diamond is actually a square with an area
of 8100 square feet. Most baseball teams cover their
diamond with a tarp to protect it from the rain. The
sides are all the same length. How long is the tarp on
each side?
SOLUTION:
The length of each side of the tarp is equal to the
square root of the area.
The tarp is 90 feet long on each side.
ANSWER:
90 ft
Find each cube root.
8.
SOLUTION:
Since 8 •8 •8 = 512, the cube root of 512 is 8.
ANSWER:
8
9.
SOLUTION:
Since 13 •13 •13 = 2197, the cube root of 2197 is
13.
ANSWER:
13
10.
SOLUTION:
Since −10 • −10 • −10 = −1000, the cube root of
−1000 is −10.
ANSWER:
−10
11.
SOLUTION:
Since −7 • −7 • −7 = −343, the cube root of −343 is
−7.
ANSWER:
−7
Estimate each cube root to the nearest integer.
12.
SOLUTION:
The first perfect cube less than 74 is 64.
The first perfect cube greater than 74 is 125.
Approximate the placement of onanumber
line relative to 4 and 5. Since 74 is closer to 64 than
to 125, willbecloserto4.
.
ANSWER:
4
13.
SOLUTION:
The first perfect cube less than 39 is 27.
The first perfect cube greater than 39 is 64.
Approximate the placement of onanumber
line relative to 3 and 4. Since 39 is closer to 27 than
to 64, willbecloserto3.
ANSWER:
3
14.
SOLUTION:
The first perfect cube less than –636 is –729.
The first perfect cube greater than –636 is –512.
Approximate the placement of onanumber
line relative to –9 and –8. Since –636 is closer to –
729 than to –512, willbecloserto–9.
ANSWER:
−9
15.
SOLUTION:
The first perfect cube less than –879 is –1000.
The first perfect cube greater than –879 is –729.
Approximate the placement of onanumber
line relative to –10 and –9. Since –879 is closer to –
1000 than to –729, willbecloserto–10.
ANSWER:
−10
Find each square root.
16.
SOLUTION:
= 6
ANSWER:
6
17.
SOLUTION:
= 3
ANSWER:
3
18.
SOLUTION:
= −13
ANSWER:
−13
19.
SOLUTION:
= −12
ANSWER:
−12
20.
SOLUTION:
There is no real square root because no number
times itself is equal to −25.
ANSWER:
no real solution
21.
SOLUTION:
= ±1
ANSWER:
±1
Estimate each square root to the nearest
integer.
22.
SOLUTION:
The first perfect square less than 83 is 81. =9
The first perfect square greater than 83 is 100.
=10
The square root of 83 is between the integers 9 and
10. Since 83 is closer to 81 than to 100, iscloser
to 9 than to 10.
ANSWER:
9
23.
SOLUTION:
The first perfect square less than 34 is 25. =5
The first perfect square greater than 34 is 36. =
6
The square root of 34 is between the integers 5 and
6. Since 34 is closer to 36 than to 25, iscloser
to 6 than to 5.
ANSWER:
6
24.
SOLUTION:
The first perfect square less than 102 is 100. =
10
The first perfect square greater than 102 is 121.
=11
The negative square root of 102 is between the
integers –10 and –11. Since 102 is closer to 100 than
to 121, iscloserto–10 than to –11.
ANSWER:
−10
25.
SOLUTION:
The first perfect square less than 14 is 9. =3
The first perfect square greater than 14 is 16. =
4
The negative square root of 14 is between the
integers –3 and –4. Since 14 is closer to 16 than to 9,
iscloserto–4 than to –3.
ANSWER:
−4
26.
SOLUTION:
The first perfect square less than 78 is 64. =8
The first perfect square greater than 78 is 81. =
9
The positive and negative square roots of 78 are
between the integers ±8 and ±9. Since78 is closer to
81 than to 64, ± is closer to ±9 than to ±8.
ANSWER:
±9
27.
SOLUTION:
The first perfect square less than 146 is 144. =
12
The first perfect square greater than 146 is 169.
=13
The positive and negative square roots of 146 are
between the integers ±12 and ±13. Since 146 is
closer to 144 than to 169, ± is closer to ±12
than to ±13.
ANSWER:
±12
28.The table shows the heights of the tallest roller
coasters at Cedar Point. Use the formula from
Example 3 to determine how far a rider can see
from the highest point of each ride. Round to the
nearest tenth.
a.MillenniumForce
b.MeanStreak
c.HowmuchfarthercanariderseeontheTop
Thrill Dragster than on the Magnum XL-200?
SOLUTION:
a.
A rider can see about 21.5 miles from the highest
point on the Millennium Force roller coaster.
b.
A rider can see about 15.5 miles from the highest
point on the Mean Streak roller coaster.
c.
Top Thrill Dragster:
Magnum XL-200:
A rider can see 25.0 −17.5 or 7.5 miles farther on
the Top Thrill Dragster than on the Magnum XL-200.
ANSWER:
a.21.5mi
b.15.5mi
c.7.5mi
Find each cube root.
29.
SOLUTION:
Since −12 • −12 • −12 = −1728, the cube root of
−1728 is −12.
ANSWER:
−12
30.
SOLUTION:
Since −14 • −14 • −14 = −2744, the cube root of
−2744 is −14.
ANSWER:
−14
31.
SOLUTION:
Since 6 •6 •6 = 216, the cube root of 216 is 6.
ANSWER:
6
32.
SOLUTION:
Since 11 •11 •11 = 1331, the cube root of 1331 is
11.
ANSWER:
11
Estimate each cube root to the nearest integer.
Do not use a calculator.
33.
SOLUTION:
The first perfect cube less than 499 is 343.
The first perfect cube greater than 499 is 512.
Approximate the placement of onanumber
line relative to 7 and 8. Since 499 is closer to 512
than to 343, willbecloserto8.
ANSWER:
8
34.
SOLUTION:
The first perfect cube less than 576 is 512.
The first perfect cube greater than 576 is 729.
Approximate the placement of onanumber
line relative to 8 and 9. Since 576 is closer to 512
than to 729, willbecloserto8.
ANSWER:
8
35.
SOLUTION:
The first perfect cube less than –79 is –125.
The first perfect cube greater than –79 is –64.
Approximate the placement of onanumber
line relative to –5 and –4. Since –79 is closer to –64
than to –125, willbecloserto–4.
ANSWER:
–4
36.
SOLUTION:
The first perfect cube less than –1735 is –2197.
The first perfect cube greater than –1735 is –1728.
Approximate the placement of ona
number line relative to –13 and –12. Since –1735 is
closer to –1728 than to –2197, willbe
closer to –12.
ANSWER:
–12
37.The area of a square is 215 square centimeters. Find
the length of a side to the nearest tenth. Then find its
approximate perimeter.
SOLUTION:
The length of each side of the square is equal to the
square root of its area.
A side length of the square is about 14.7 centimeters.
Use the formula to find the perimeter of the
square.
The approximate perimeter of the square is 58.8
centimeters.
ANSWER:
14.7 cm; 58.8 cm
38.Order , −8, , 9, −10, , from
least to greatest.
SOLUTION:
Write the given integers as square roots, and then
order the numbers from least to greatest.
−8 =
9 =
−10 =
< , < < < < <
So, the correct order is −10, , , −8,
, , 9.
ANSWER:
−10, , , −8, , , 9
39.Reason Abstractly Write a number that completes
the analogy. x2 is to 121 as x3 is to ?.
SOLUTION:
121 is a perfect square of 11 since 11 •11 = 121.
1331 is a perfect cube of 11 since 11 •11 •11 =
1331.
So,x2isto121asx3 is to 1331.
ANSWER:
1331
40.Identify Structure Find a square root that lies
between 17 and 18.
SOLUTION:
and . Any number between 289
and 324 will have a square root between 17 and 18.
For example, .
ANSWER:
Sample answer:
41.PerseverewithProblemsUse inverse operations
to evaluate the following.
a.
b.
c.
SOLUTION:
a.
b.
c.
ANSWER:
a. 246
b.811
c.732
42.Building on the Essential Question Describe the
difference between an exact value and an
approximation when finding square roots of numbers
that are not perfect squares. Give an example of
each.
SOLUTION:
Sample answer: The exact value of a square root is
given using the square root symbol, such as . An
approximation is a decimal value, such as ≈6.
ANSWER:
Sample answer: The exact value of a square root is
given using the square root symbol, such as .
An approximation is a decimal value, such as
≈6.
43.Which point on the number line best represents
?
AA
BB
CC
DD
SOLUTION:
Use a calculator to estimate the square root of 210.
Point C best represents .
Choice C is the correct answer.
ANSWER:
C
44.Short Response Estimate the cube root of 65 to the
nearest integer.
SOLUTION:
The first perfect cube less than 65 is 64.
The first perfect cube greater than 65 is 125.
Approximate the placement of onanumber
line relative to 4 and 5. Since 65 is closer to 64 than
to 125, willbecloserto4.
ANSWER:
4
45.The new gymnasium at Oakdale Middle School has a
hardwood floor in the shape of a square. If the area
of the floor is 62,500 square feet, what is the length
of one side of the square floor?
F200ft
G225ft
H250ft
J275ft
SOLUTION:
The length of a side of the gymnasium floor is equal
to the square root of its area.
Each side has length 250 units.
Choice H is the correct answer.
ANSWER:
H
46.A surveyor determined the distance across a field
was feet.Whatistheapproximatedistance?
A25.6ft
B30.6ft
C39.6ft
D42.6ft
SOLUTION:
Use a calculator to estimate the square root of 1568
to the nearest tenth.
Choice C is the correct answer.
ANSWER:
C
Solve.
47.Joseph bought four books at a book sale. Each book
cost $4.50. He paid with $20.00. How much change
did he receive?
SOLUTION:
To find the cost of four books, multiply the number of
books by the cost for each book.
4×4.50=18.00
So, the four books cost $18.00.
To find the amount of change he received, subtract
the cost of the books from the amount that Joseph
paid.
20 – 18 = 2
So, Joseph received $2.00 in change.
ANSWER:
$2.00
48.Mrs. Tanner paid $18.00 for six boxes of pencils.
How much did each box of pencils cost?
SOLUTION:
To find the cost of each box of pencils, divide the
total cost by the number of boxes of pencils.
18÷6=3
So, each box of pencils cost $3.00.
ANSWER:
$3.00
49.Kathryn had $367.50 in her bank account. She wrote
a check for $25.00, and then withdrew $50.00 in
cash. She made a deposit of $100.00. How much
money is in her bank account now?
SOLUTION:
Kathryn has $392.50 in her bank account now.
Kathrynhad$367.50
in her bank account. $367.50
Shewroteacheckfor
$25.00. $367.50– $25.00 = $342.50
Shewithdrew$50.00
in cash. $342.50– $50.00 = $292.50
Shemadeadepositof
$100.00. $292.50+$100.00=$392.50
ANSWER:
$392.50
Name the property shown by each statement.
50.3 + 6 = 6 + 3
SOLUTION:
The order of the numbers changed. This is the
Commutative Property of Addition.
ANSWER:
Commutative Property (+)
51.13 + 0 = 13
SOLUTION:
When zero is added to any number, the sum is the
number. This is the Additive Identity Property.
ANSWER:
Additive Identity Property
52.(2 •5) + 6 = 6 + (2 •5)
SOLUTION:
The order of the numbers changed. This is the
Commutative Property of Addition.
ANSWER:
Commutative Property (+)
53.(x + 3) + 9 = x + (3 + 9)
SOLUTION:
The grouping of the numbers changed. This is the
Associative Property of Addition.
ANSWER:
Associative Property (+)
54.(y •2) •3 = y •(2 •3)
SOLUTION:
The grouping of the numbers changed. This is the
Associative Property of Multiplication.
ANSWER:
AssociativeProperty(×)
55.28 •1 = 28
SOLUTION:
When 1 is multiplied by any number, the product is
the number. This is the Multiplicative Identity
Property.
ANSWER:
Multiplicative Identity Property
56.n + t = t + n
SOLUTION:
The order of the variables has changed. This is the
Commutative Property of Addition.
ANSWER:
Commutative Property (+)
57.1652 •0 = 0
SOLUTION:
When zero is multiplied by any number, the product is
zero. This is the Zero Property of Multiplication.
ANSWER:
Zero Property of Multiplication
58.Suppose that four-tenths of the rectangle below is
shaded.
a. What fraction of the rectangle is not shaded?
b. What fraction of the rectangle would still need to
be shaded for half of the rectangle to be shaded?
c. If an additional of the original rectangle were
to be shaded, what fraction of the rectangle would be
shaded?
SOLUTION:
a. To find the fraction of the rectangle that is not
shaded, subtract the fraction that is shaded from one
whole.
of the rectangle is not shaded.
b. To find the fraction of the rectangle that would still
need to be shaded, subtract the fraction that is
already shaded from one-half.
of the rectangle would still need to be shaded for
half of the rectangle to be shaded.
c.
If an additional of the original rectangle were
shaded, of the rectangle would be shaded.
ANSWER:
a.
b.
c.
Find each product or quotient.
59.8 •(–8)
SOLUTION:
The product of two integers with different signs is
negative. So, 8 •(–8) = –64.
ANSWER:
–64
60.–4 •(–12)
SOLUTION:
The product of two integers with the same sign is
positive. So, –4 •(–12) = 48.
ANSWER:
48
61.40÷(–5)
SOLUTION:
The quotient of two integers with different signs is
negative.So,40÷(–5) = –8.
ANSWER:
–8
62.–150÷(–25)
SOLUTION:
The quotient of two integers with the same sign is
positive. So, –150÷(–25) = 6.
ANSWER:
6
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4-6 Square Roots and Cube Roots
