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Math Problems
Grade 6
Properties of multiplication
Name:
\newline
\qquad
\newline
Date:
\newline
5
/
151
5 / 151
5/151
\newline
B
=
T
B=T
B
=
T
\newline
Identify the Volume of Triangular Prism
\newline
A
=
1
2
B
⋅
h
A=\frac{1}{2} B \cdot h
A
=
2
1
B
⋅
h
\newline
Pac
\newline
1
1
1
What is the volume of the prism below?
\newline
Answer
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Find all the factor pairs for
75
75
75
and complete the equations.
\newline
75
=
_
_
_
×
75
75 = \_\_\_ \times 75
75
=
___
×
75
\newline
75
=
3
×
_
_
_
75 = 3 \times \_\_\_
75
=
3
×
___
\newline
75
=
5
×
_
_
_
75 = 5 \times \_\_\_
75
=
5
×
___
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78
78
78
、鸡兔同䇝,共
100
100
100
个头,
320
320
320
只脚,那么,鸡有()只,兔有()只。
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03
03
03
、同学们进行广搯操比赛, 全班正好排成相等的
6
6
6
行。小红排在第二行, 从头数, 她站在第
5
5
5
个位置,从后数她站在第
3
3
3
个位置,这个班共有()人
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03
03
03
、同学们进行广播操比赛, 全班正好排成相等的
6
6
6
行。小红排在第二行, 从头数, 她站在第
5
5
5
个位置, 从后数她站在第
3
3
3
个位置, 这个班共有()人
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Marie wants to buy a new car. She saves
25
%
25\%
25%
of her wage and is left with
$
600
\$600
$600
spending money. What is Marie's wage?
\newline
(A)
$
450
\$450
$450
\newline
(B)
$
625
\$625
$625
\newline
(C)
$
750
\$750
$750
\newline
(D)
$
800
\$800
$800
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Direct Messages - SEQTA Le
\newline
Assessment Master - Exam Wi
\newline
assess.scsa.wa.edu.au/engine/index.php/lms/index
\newline
School Curriculum
\newline
and Standards
\newline
Authority
\newline
Numeracy
\newline
31751624
31751624
31751624
\newline
()
0
:
17
:
14
0: 17: 14
0
:
17
:
14
/
0
0
0
:
50
50
50
:
00
00
00
\newline
Marie wants to buy a new car.
\newline
She saves
25
%
25 \%
25%
of her wage and is left with
$
600
\$ 600
$600
spending money. What is Marie's wage?
\newline
42
42
42
of
45
45
45
\newline
Question
37
37
37
\newline
Question
38
38
38
\newline
Question
39
39
39
\newline
$
450
\$ 450
$450
\newline
$
625
\$ 625
$625
\newline
$
750
\$ 750
$750
\newline
$
800
\$ 800
$800
\newline
Content copyright WA School Curriculum and Standards Authority
2024
2024
2024
\newline
a
a
a
\newline
Assessment
\newline
Copyright
⊕
\oplus
⊕
SoNET Systems Pty Ltd.
2024
2024
2024
\newline
(.)
\newline
Master
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已知正方形
A
B
C
D
ABCD
A
BC
D
的边长为
2
2
2
,点
Q
Q
Q
为边
B
C
BC
BC
的中点,点
P
P
P
在正方形边
A
D
AD
A
D
外部,且满足
∠
A
P
D
=
13
5
∘
\angle APD=135^\circ
∠
A
P
D
=
13
5
∘
,则
P
Q
PQ
PQ
的最大值
Get tutor help
已知正方形
A
B
C
D
ABCD
A
BC
D
的边长,点
Q
Q
Q
为边
B
C
BC
BC
的中点,点
P
P
P
在正方形外部,且满足
∠
A
P
D
=
13
5
∘
\angle APD=135^\circ
∠
A
P
D
=
13
5
∘
,则
P
Q
PQ
PQ
的最大值
Get tutor help
1
+
5
−
68
+
66666
−
55
=
9
1+ 5 - 68 + 66666 - 55 = 9
1
+
5
−
68
+
66666
−
55
=
9
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32
×
10
=
320
32 \times 10=320
32
×
10
=
320
(元)
\newline
答: 一张桌子
320
320
320
元, 一把椅子
32
32
32
元。
\newline
2
2
2
、
3
3
3
箱苹果重
45
45
45
千克。一箱梨比一箱苹果多
5
5
5
千克,
3
3
3
箱梨重多少千克?
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45
−
3
20
+
4
5
\sqrt{45} -3 \sqrt{20} +4\sqrt{5}
45
−
3
20
+
4
5
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41
=
3
+
x
41=3+x
41
=
3
+
x
\newline
1
1
1
□
\square
□
2
2
2
□
\square
□
\newline
□
\square
□
\newline
□
\square
□
\newline
42
42
42
\newline
71
71
71
\newline
44
44
44
\newline
38
38
38
Get tutor help
307
−
189
307-189
307
−
189
Get tutor help
Use the circle below for questions
7
−
10
7-10
7
−
10
.
\newline
7
7
7
. Find
N
K
N K
N
K
\newline
8
8
8
. Find
M
K
M K
M
K
to the nearest degree.
\newline
9
9
9
. Find
J
K
J K
J
K
\newline
10
10
10
. Find
J
P
K
J P K
J
P
K
\newline
1
1
1
. Solve for
x
x
x
.
\newline
62
−
5
x
+
133
=
180
62-5 x+133=180
62
−
5
x
+
133
=
180
\newline
19
19
19
)
5
5
5
−
5
-5
−
5
x
=
180
=180
=
180
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Pembahasan
\newline
4
4
4
\newline
5
5
5
\newline
6
6
6
\newline
7
7
7
\newline
8
8
8
\newline
9
9
9
\newline
10
10
10
\newline
11
11
11
\newline
TPS Pengetahuan Kuantitatif
\newline
Terdapat delapan buah kartu yang masing-masing diberi nomor
1
1
1
,
1
1
1
,
2
2
2
,
2
2
2
,
4
4
4
,
4
4
4
,
5
5
5
,
6
6
6
. Kemudian, kartu-kartu tersebut disusun sejajar.
\newline
Berdasarkan informasi tersebut, banyak pernyataan berikut yang bernilai benar adalah
\qquad
\newline
1
1
1
. Banyak keseluruhan susunan adalah
5
5
5
.
040
040
040
.
\newline
2
2
2
. Banyak keseluruhan susunan dengan syarat semua kartu-kartu bernomor genap bersebelahan adalah
180
180
180
.
\newline
3
3
3
. Banyak keseluruhan susunan dengan syarat semua kartu-kartu bernomor ganjil bersebelahan adalah
270
270
270
.
\newline
4
4
4
. Banyak keseluruhan susunan dengan syarat susunan kartu diawali dan diakhiri dengan kartu bernomor
1
1
1
adalah
90
90
90
.
\newline
A
0
0
0
\newline
B
1
1
1
\newline
C
2
2
2
\newline
D
3
3
3
\newline
ᄃ
4
4
4
Get tutor help
1
1
1
. triangle: base,
8
8
8
.
2
2
2
meters; height,
4
4
4
.
6
6
6
meters
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1
1
1
. Solve each inequality alg Please circle or box your
\newline
a)
5
−
t
>
3
5-t>3
5
−
t
>
3
\newline
b)
5
−
1073
5-1073
5
−
1073
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3
×
9
+
25
÷
5
×
98
−
24
−
321
÷
3
−
123
×
2
−
120
−
17
=
?
3\times9+25\div5\times98-24-321\div3 -123\times2-120-17=\,?
3
×
9
+
25
÷
5
×
98
−
24
−
321
÷
3
−
123
×
2
−
120
−
17
=
?
Get tutor help
1
1
1
crore =
\newline
\qquad
\newline
hundred
Get tutor help
23
23
23
. 已知
△
A
B
C
\triangle A B C
△
A
BC
中,
A
C
=
8
,
A
B
=
41
,
B
C
A C=8, A B=\sqrt{41}, B C
A
C
=
8
,
A
B
=
41
,
BC
边上的高
A
G
=
5
,
D
A G=5, D
A
G
=
5
,
D
为线段
A
C
A C
A
C
上的动点, 在
B
C
B C
BC
上截取
C
E
=
A
D
C E=A D
CE
=
A
D
, 连接
A
E
,
B
D
A E, B D
A
E
,
B
D
, 则
A
E
+
B
D
A E+B D
A
E
+
B
D
的最小值为
\qquad
.
\newline
22
22
22
题图
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y
=
−
3
(
x
−
2
)
2
+
12
y=-3(x-2)^{2}+12
y
=
−
3
(
x
−
2
)
2
+
12
\newline
Vertex:
□
\square
□
Get tutor help
1170.16
=
[
C
×
1
⋅
1
(
1
+
0.08
)
5
0.08
]
+
[
0.4
×
(
1000
+
2
C
)
×
1
(
1
+
0.08
)
5
]
+
[
0.6
×
(
C
×
1
−
1
(
1
+
0.12
)
15
0.12
+
1000
(
1
+
0.12
)
15
)
×
1
(
1
+
0.08
)
5
]
\begin{aligned} 1170.16= & {\left[C \times \frac{1 \cdot \frac{1}{(1+0.08)^{5}}}{0.08}\right]+\left[0.4 \times(1000+2 C) \times \frac{1}{(1+0.08)^{5}}\right]+} \\ & {\left[0.6 \times\left(C \times \frac{1-\frac{1}{(1+0.12)^{15}}}{0.12}+\frac{1000}{(1+0.12)^{15}}\right) \times \frac{1}{(1+0.08)^{5}}\right] }\end{aligned}
1170.16
=
[
C
×
0.08
1
⋅
(
1
+
0.08
)
5
1
]
+
[
0.4
×
(
1000
+
2
C
)
×
(
1
+
0.08
)
5
1
]
+
[
0.6
×
(
C
×
0.12
1
−
(
1
+
0.12
)
15
1
+
(
1
+
0.12
)
15
1000
)
×
(
1
+
0.08
)
5
1
]
Get tutor help
2
x
2
−
6
x
=
−
4
2 x^{2}-6 x=-4
2
x
2
−
6
x
=
−
4
\newline
Vertex:
\newline
)
\newline
\begin{tabular}{|c|c|}
\newline
\hline
x
x
x
&
y
y
y
\\
\newline
\hline & \\
\newline
\hline & \\
\newline
\hline & \\
\newline
\hline & \\
\newline
\hline & \\
\newline
\hline
\newline
\end{tabular}
\newline
Solution:
Get tutor help
1
1
1
.
39
+
281
=
\mathbf{3 9}+\mathbf{2 8 1}=
39
+
281
=
Textbox
2
2
2
∗
\mathrm{*}
∗
\newline
a)
319
319
319
\newline
b)
321
321
321
\newline
c)
320
320
320
\newline
d)
324
324
324
Get tutor help
Which property of multiplication is shown?
\newline
h
⋅
g
=
g
⋅
h
h \cdot g = g \cdot h
h
⋅
g
=
g
⋅
h
\newline
Choices:
\newline
(A)zero
\newline
(B)commutative
\newline
(C)distributive
\newline
(D)associative
Get tutor help
Which equation shows the associative property of multiplication?
\newline
Choices:
\newline
(A)
(
h
⋅
j
)
⋅
k
=
h
⋅
(
j
⋅
k
)
(h \cdot j) \cdot k = h \cdot (j \cdot k)
(
h
⋅
j
)
⋅
k
=
h
⋅
(
j
⋅
k
)
\newline
(B)
h
⋅
k
+
j
⋅
k
=
(
h
+
j
)
⋅
k
h \cdot k + j \cdot k = (h + j) \cdot k
h
⋅
k
+
j
⋅
k
=
(
h
+
j
)
⋅
k
\newline
(C)
h
=
h
⋅
1
h = h \cdot 1
h
=
h
⋅
1
\newline
(D)
0
⋅
h
=
0
0 \cdot h = 0
0
⋅
h
=
0
Get tutor help
Which property of multiplication is shown?
\newline
h
⋅
j
=
j
⋅
h
h \cdot j = j \cdot h
h
⋅
j
=
j
⋅
h
\newline
Choices:
\newline
(A)zero
\newline
(B)associative
\newline
(C)identity
\newline
(D)commutative
Get tutor help
Which equation shows the zero property of multiplication?
\newline
Choices:
\newline
(A)
t
=
1
⋅
t
t = 1 \cdot t
t
=
1
⋅
t
\newline
(B)
(
t
+
u
)
⋅
v
=
t
⋅
v
+
u
⋅
v
(t + u) \cdot v = t \cdot v + u \cdot v
(
t
+
u
)
⋅
v
=
t
⋅
v
+
u
⋅
v
\newline
(C)
0
=
0
⋅
t
0 = 0 \cdot t
0
=
0
⋅
t
\newline
(D)
v
+
w
=
t
⋅
u
v + w = t \cdot u
v
+
w
=
t
⋅
u
Get tutor help
Which equation shows the identity property of multiplication?
\newline
Choices:
\newline
(A)
w
⋅
y
=
u
⋅
v
w \cdot y = u \cdot v
w
⋅
y
=
u
⋅
v
\newline
(B)
0
⋅
u
=
0
0 \cdot u = 0
0
⋅
u
=
0
\newline
(C)
u
⋅
v
−
u
⋅
w
=
u
⋅
(
v
−
w
)
u \cdot v - u \cdot w = u \cdot (v - w)
u
⋅
v
−
u
⋅
w
=
u
⋅
(
v
−
w
)
\newline
(D)
u
=
1
⋅
u
u = 1 \cdot u
u
=
1
⋅
u
Get tutor help
Which property of multiplication is shown?
\newline
n
⋅
(
p
−
q
)
=
n
⋅
p
−
n
⋅
q
n \cdot (p - q) = n \cdot p - n \cdot q
n
⋅
(
p
−
q
)
=
n
⋅
p
−
n
⋅
q
\newline
Choices:
\newline
(A) identity
\newline
(B) zero
\newline
(C) commutative
\newline
(D) distributive
Get tutor help
Which property of multiplication is shown?
\newline
d
⋅
f
=
f
⋅
d
d \cdot f = f \cdot d
d
⋅
f
=
f
⋅
d
\newline
Choices:
\newline
(A) identity
\newline
(B) zero
\newline
(C) distributive
\newline
(D) commutative
Get tutor help
Which property of multiplication is shown?
\newline
d
⋅
f
=
f
⋅
d
d \cdot f = f \cdot d
d
⋅
f
=
f
⋅
d
\newline
Choices:
\newline
(A)commutative
\newline
(B)distributive
\newline
(C)associative
\newline
(D)identity
\newline
Get tutor help
Which equation shows the commutative property of multiplication?
\newline
Choices:
\newline
(A)
q
=
m
⋅
n
⋅
p
q = m \cdot n \cdot p
q
=
m
⋅
n
⋅
p
\newline
(B)
m
⋅
n
+
m
⋅
p
=
m
⋅
(
n
+
p
)
m \cdot n + m \cdot p = m \cdot (n + p)
m
⋅
n
+
m
⋅
p
=
m
⋅
(
n
+
p
)
\newline
(C)
m
⋅
n
=
n
⋅
m
m \cdot n = n \cdot m
m
⋅
n
=
n
⋅
m
\newline
(D)
0
⋅
m
=
0
0 \cdot m = 0
0
⋅
m
=
0
Get tutor help
Which property of multiplication is shown?
\newline
(
q
−
r
)
⋅
s
=
q
⋅
s
−
r
⋅
s
(q - r) \cdot s = q \cdot s - r \cdot s
(
q
−
r
)
⋅
s
=
q
⋅
s
−
r
⋅
s
\newline
Choices:
\newline
(A)zero
\newline
(B)identity
\newline
(C)commutative
\newline
(D)distributive
Get tutor help
Which equation shows the distributive property of multiplication?
\newline
Choices:
\newline
(A)
c
⋅
d
⋅
f
=
g
c \cdot d \cdot f = g
c
⋅
d
⋅
f
=
g
\newline
(B)
c
⋅
(
d
⋅
f
)
=
(
c
⋅
d
)
⋅
f
c \cdot (d \cdot f) = (c \cdot d) \cdot f
c
⋅
(
d
⋅
f
)
=
(
c
⋅
d
)
⋅
f
\newline
(C)
0
=
c
⋅
0
0 = c \cdot 0
0
=
c
⋅
0
\newline
(D)
c
⋅
d
−
c
⋅
f
=
c
⋅
(
d
−
f
)
c \cdot d - c \cdot f = c \cdot (d - f)
c
⋅
d
−
c
⋅
f
=
c
⋅
(
d
−
f
)
Get tutor help
Which property of multiplication is shown?
\newline
d
⋅
f
=
f
⋅
d
d \cdot f = f \cdot d
d
⋅
f
=
f
⋅
d
\newline
Choices:
\newline
(A) identity
\newline
(B) commutative
\newline
(C) zero
\newline
(D) distributive
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Which property of multiplication is shown?
\newline
q
⋅
s
+
r
⋅
s
=
(
q
+
r
)
⋅
s
q \cdot s + r \cdot s = (q + r) \cdot s
q
⋅
s
+
r
⋅
s
=
(
q
+
r
)
⋅
s
\newline
Choices:
\newline
(A)zero
\newline
(B)associative
\newline
(C)distributive
\newline
(D)commutative
\newline
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Which equation shows the distributive property of multiplication?
\newline
Choices:
\newline
(A)
b
⋅
c
=
c
⋅
b
b \cdot c = c \cdot b
b
⋅
c
=
c
⋅
b
\newline
(B)
(
b
+
c
)
⋅
d
=
b
⋅
d
+
c
⋅
d
(b + c) \cdot d = b \cdot d + c \cdot d
(
b
+
c
)
⋅
d
=
b
⋅
d
+
c
⋅
d
\newline
(C)
b
⋅
1
=
b
b \cdot 1 = b
b
⋅
1
=
b
\newline
(D)
b
⋅
0
=
0
b \cdot 0 = 0
b
⋅
0
=
0
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Which property of multiplication is shown?
\newline
q
⋅
p
=
p
⋅
q
q \cdot p = p \cdot q
q
⋅
p
=
p
⋅
q
\newline
Choices:
\newline
(A) identity
\newline
(B) commutative
\newline
(C) distributive
\newline
(D) zero
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Which equation shows the zero property of multiplication?
\newline
Choices:
\newline
(A)
a
⋅
b
=
c
a \cdot b = c
a
⋅
b
=
c
\newline
(B)
b
⋅
a
=
a
⋅
b
b \cdot a = a \cdot b
b
⋅
a
=
a
⋅
b
\newline
(C)
(
a
−
b
)
⋅
c
=
a
⋅
c
−
b
⋅
c
(a - b) \cdot c = a \cdot c - b \cdot c
(
a
−
b
)
⋅
c
=
a
⋅
c
−
b
⋅
c
\newline
(D)
a
⋅
0
=
0
a \cdot 0 = 0
a
⋅
0
=
0
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Which equation shows the commutative property of multiplication?
\newline
Choices:
\newline
(A)
k
⋅
1
=
k
k \cdot 1 = k
k
⋅
1
=
k
\newline
(B)
k
⋅
m
=
m
⋅
k
k \cdot m = m \cdot k
k
⋅
m
=
m
⋅
k
\newline
(C)
(
k
⋅
m
)
⋅
n
=
k
⋅
(
m
⋅
n
)
(k \cdot m) \cdot n = k \cdot (m \cdot n)
(
k
⋅
m
)
⋅
n
=
k
⋅
(
m
⋅
n
)
\newline
(D)
1
⋅
k
=
k
1 \cdot k = k
1
⋅
k
=
k
Get tutor help
Which equation shows the zero property of multiplication?
\newline
Choices:
\newline
(A)
j
⋅
(
k
+
m
)
=
j
⋅
k
+
j
⋅
m
j \cdot (k + m) = j \cdot k + j \cdot m
j
⋅
(
k
+
m
)
=
j
⋅
k
+
j
⋅
m
\newline
(B)
k
⋅
j
=
j
⋅
k
k \cdot j = j \cdot k
k
⋅
j
=
j
⋅
k
\newline
(C)
(
j
⋅
k
)
⋅
m
=
j
⋅
(
k
⋅
m
)
(j \cdot k) \cdot m = j \cdot (k \cdot m)
(
j
⋅
k
)
⋅
m
=
j
⋅
(
k
⋅
m
)
\newline
(D)
0
=
0
⋅
j
0 = 0 \cdot j
0
=
0
⋅
j
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Which equation shows the distributive property of multiplication?
\newline
Choices:
\newline
(A)
(
r
⋅
s
)
⋅
t
=
r
⋅
(
s
⋅
t
)
(r \cdot s) \cdot t = r \cdot (s \cdot t)
(
r
⋅
s
)
⋅
t
=
r
⋅
(
s
⋅
t
)
\newline
(B)
(
r
+
s
)
⋅
t
=
r
⋅
t
+
s
⋅
t
(r + s) \cdot t = r \cdot t + s \cdot t
(
r
+
s
)
⋅
t
=
r
⋅
t
+
s
⋅
t
\newline
(C)
r
⋅
s
=
s
⋅
r
r \cdot s = s \cdot r
r
⋅
s
=
s
⋅
r
\newline
(D)
0
=
r
⋅
0
0 = r \cdot 0
0
=
r
⋅
0
Get tutor help
Which property of multiplication is shown?
\newline
0
=
a
⋅
0
0 = a \cdot 0
0
=
a
⋅
0
\newline
Choices:
\newline
(A) identity
\newline
(B) associative
\newline
(C) commutative
\newline
(D) zero
Get tutor help
Which equation shows the zero property of multiplication?
\newline
Choices:
\newline
(A)
1
⋅
m
=
m
1 \cdot m = m
1
⋅
m
=
m
\newline
(B)
m
⋅
n
=
n
⋅
m
m \cdot n = n \cdot m
m
⋅
n
=
n
⋅
m
\newline
(C)
(
m
⋅
n
)
⋅
p
=
m
⋅
(
n
⋅
p
)
(m \cdot n) \cdot p = m \cdot (n \cdot p)
(
m
⋅
n
)
⋅
p
=
m
⋅
(
n
⋅
p
)
\newline
(D)
0
=
0
⋅
m
0 = 0 \cdot m
0
=
0
⋅
m
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Khan Academy
\newline
Get Al Tutoring
\newline
NEW
\newline
Donate
\newline
3
(
x
+
1
)
2
=
108
3(x+1)^{2}=108
3
(
x
+
1
)
2
=
108
\newline
Solution steps:
\newline
\begin{tabular}{|c|}
\newline
\hline Add
1
1
1
to both sides \\
\newline
\hline Divide both sides by
3
3
3
\\
\newline
\hline Multiply both sides by
3
3
3
\\
\newline
\hline Subtract
1
1
1
from both sides \\
\newline
\hline Square both sides \\
\newline
\hline Take the square root of both \\
\newline
sides \\
\newline
\hline
\newline
\end{tabular}
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Evaluate
\newline
78
+
5
+
9917
÷
100
78+5+9917\div100
78
+
5
+
9917
÷
100
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1
2
+
1
6
=
\frac{1}{2}+\frac{1}{6}=
2
1
+
6
1
=
Get tutor help
1
1
1
- Simplify anythi
\newline
2
2
2
- Simplify any Ex
\newline
3
3
3
- Multiplication
\newline
4
4
4
-Addition or Su
\newline
9
9
9
)
3
2
+
6
−
2
×
7
3^{2}+6-2 \times 7
3
2
+
6
−
2
×
7
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1
2
Next