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Math Problems
Grade 6
Multiply using the distributive property
Simplify the expression:
\newline
2
(
2
+
3
k
)
=
2(2 + 3k) =
2
(
2
+
3
k
)
=
_____
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Simplify the expression:
\newline
2
(
2
+
s
)
=
2(2 + s) =
2
(
2
+
s
)
=
_____
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Simplify the expression:
\newline
(
w
+
2
)
(
2
)
=
(w + 2)(2) =
(
w
+
2
)
(
2
)
=
_____
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Simplify the expression:
\newline
3
(
v
+
1
)
=
3(v + 1) =
3
(
v
+
1
)
=
_____
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Simplify the expression:
\newline
2
(
1
+
m
)
=
2(1 + m) =
2
(
1
+
m
)
=
_____
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Simplify the expression:
\newline
4
(
1
+
7
g
)
=
4(1 + 7g) =
4
(
1
+
7
g
)
=
_____
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Simplify the expression:
\newline
4
(
2
+
4
r
)
=
4(2 + 4r) =
4
(
2
+
4
r
)
=
_____
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Simplify the expression:
\newline
7
(
8
t
+
1
)
7(8t + 1)
7
(
8
t
+
1
)
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Simplify the expression:
\newline
2
(
2
y
+
1
)
=
2(2y + 1) =
2
(
2
y
+
1
)
=
_____
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Simplify the expression:
\newline
2
(
2
f
+
3
)
=
2(2f + 3) =
2
(
2
f
+
3
)
=
_____
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Simplify the expression:
\newline
2
(
2
w
+
2
)
=
2(2w + 2) =
2
(
2
w
+
2
)
=
_____
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Simplify the expression:
\newline
2
(
1
+
v
)
2(1 + v)
2
(
1
+
v
)
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Simplify the expression:
\newline
2
(
1
+
2
t
)
=
2(1 + 2t) =
2
(
1
+
2
t
)
=
_____
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Simplify the expression:
\newline
3
(
2
+
4
c
)
=
3(2 + 4c) =
3
(
2
+
4
c
)
=
_____
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Simplify the expression:
\newline
2
(
p
+
1
)
=
2(p + 1) =
2
(
p
+
1
)
=
_____
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Simplify the expression:
\newline
2
(
9
k
+
4
)
=
2(9k + 4) =
2
(
9
k
+
4
)
=
_____
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Simplify the expression:
\newline
5
(
1
+
h
)
5(1 + h)
5
(
1
+
h
)
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Simplify the expression:
\newline
8
(
1
+
10
q
)
=
8(1 + 10q) =
8
(
1
+
10
q
)
=
_____
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Simplify the expression:
\newline
2
(
4
+
7
n
)
=
2(4 + 7n) =
2
(
4
+
7
n
)
=
_____
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Simplify the expression:
\newline
5
(
1
+
7
z
)
=
5(1 + 7z) =
5
(
1
+
7
z
)
=
_____
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Simplify the expression:
\newline
2
(
3
s
+
2
)
=
2(3s + 2) =
2
(
3
s
+
2
)
=
_____
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Simplify the expression:
\newline
(
5
+
y
)
(
2
)
=
(5 + y)(2) =
(
5
+
y
)
(
2
)
=
_____
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Simplify the expression:
\newline
(
2
+
6
b
)
(
2
)
=
(2 + 6b)(2) =
(
2
+
6
b
)
(
2
)
=
_____
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Simplify the expression:
\newline
(
3
+
3
n
)
(
3
)
=
(3 + 3n)(3) =
(
3
+
3
n
)
(
3
)
=
_____
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Simplify the expression:
\newline
6
(
4
q
+
1
)
=
6(4q + 1) =
6
(
4
q
+
1
)
=
_
_
_
_
_
\_\_\_\_\_
_____
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Simplify the expression:
\newline
5
(
2
+
3
k
)
=
_
_
_
_
_
5(2 + 3k) = \,\_\_\_\_\_
5
(
2
+
3
k
)
=
_____
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Simplify the expression:
\newline
2
(
9
p
+
5
)
=
2(9p + 5) =
2
(
9
p
+
5
)
=
_____
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Simplify the expression:
\newline
(
1
+
2
b
)
(
2
)
=
(1 + 2b)(2) =
(
1
+
2
b
)
(
2
)
=
_____
Get tutor help
Simplify the expression:
\newline
(
1
+
7
j
)
(
8
)
=
(1 + 7j)(8) =
(
1
+
7
j
)
(
8
)
=
_____
Get tutor help
Simplify the expression:
\newline
(
2
+
r
)
(
2
)
=
(2 + r)(2) =
(
2
+
r
)
(
2
)
=
_____
Get tutor help
Simplify the expression:
\newline
9
(
1
+
3
p
)
=
9(1 + 3p) =
9
(
1
+
3
p
)
=
_____
Get tutor help
Simplify the expression:
\newline
3
(
1
+
x
)
=
3(1+x)=
3
(
1
+
x
)
=
Get tutor help
Use the distributive property to write an equivalent expression.
\newline
7
(
7
w
+
2
x
−
1
)
7(7 w+2 x-1)
7
(
7
w
+
2
x
−
1
)
\newline
Answer:
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Use the distributive property to write an equivalent expression.
\newline
7
(
7
r
−
5
s
+
8
)
7(7 r-5 s+8)
7
(
7
r
−
5
s
+
8
)
\newline
Answer:
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Solve for
s
s
s
:
\newline
−
7
=
s
+
(
−
13
)
s
=
□
\begin{array}{l} -7=s+(-13) \\ s=\square \end{array}
−
7
=
s
+
(
−
13
)
s
=
□
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Solve for
m
m
m
:
\newline
(
−
11
6
)
+
m
=
−
2
9
m
=
□
\begin{array}{l} \left(-\frac{11}{6}\right)+m=-\frac{2}{9} \\ m=\square \end{array}
(
−
6
11
)
+
m
=
−
9
2
m
=
□
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Solve for
k
k
k
:
\newline
k
−
(
−
5
6
)
=
−
1
6
k
=
□
\begin{array}{l} k-\left(-\frac{5}{6}\right)=-\frac{1}{6} \\ k=\square \end{array}
k
−
(
−
6
5
)
=
−
6
1
k
=
□
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Solve for
z
z
z
:
\newline
14
=
z
−
(
−
12
)
z
=
□
\begin{array}{l} 14=z-(-12) \\ z=\square \end{array}
14
=
z
−
(
−
12
)
z
=
□
Get tutor help
Solve for
g
g
g
:
\newline
3
16
=
(
−
5
4
)
+
g
g
=
□
\begin{array}{l} \frac{3}{16}=\left(-\frac{5}{4}\right)+g \\ g=\square \end{array}
16
3
=
(
−
4
5
)
+
g
g
=
□
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Solve for
k
k
k
:
\newline
17
=
(
−
14
)
+
k
k
=
□
\begin{array}{l} 17=(-14)+k \\ k=\square \end{array}
17
=
(
−
14
)
+
k
k
=
□
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Solve for
b
b
b
:
\newline
b
−
7
4
=
−
2
3
b
=
□
\begin{array}{l} b-\frac{7}{4}=-\frac{2}{3} \\ b=\square \end{array}
b
−
4
7
=
−
3
2
b
=
□
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Solve for
v
v
v
:
\newline
−
9
8
=
v
−
1
2
v
=
□
\begin{array}{l} -\frac{9}{8}=v-\frac{1}{2} \\ v=\square \end{array}
−
8
9
=
v
−
2
1
v
=
□
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Solve for
t
t
t
:
\newline
17
20
=
t
+
(
−
13
20
)
t
=
□
\begin{array}{l} \frac{17}{20}=t+\left(-\frac{13}{20}\right) \\ t=\square \end{array}
20
17
=
t
+
(
−
20
13
)
t
=
□
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Solve for
r
r
r
:
\newline
7
10
=
r
−
(
−
8
5
)
r
=
□
\begin{array}{l} \frac{7}{10}=r-\left(-\frac{8}{5}\right) \\ r=\square \end{array}
10
7
=
r
−
(
−
5
8
)
r
=
□
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lim
x
→
2
x
−
2
1
−
3
x
−
5
=
\lim _{x \rightarrow 2} \frac{x-2}{1-\sqrt{3 x-5}}=
lim
x
→
2
1
−
3
x
−
5
x
−
2
=
Get tutor help
lim
x
→
3
x
−
3
2
−
x
+
1
=
\lim _{x \rightarrow 3} \frac{x-3}{2-\sqrt{x+1}}=
lim
x
→
3
2
−
x
+
1
x
−
3
=
Get tutor help
Solve for
x
x
x
.
\newline
−
30
=
5
(
x
+
1
)
x
=
□
\begin{array}{l} -30=5(x+1) \\ x=\square \end{array}
−
30
=
5
(
x
+
1
)
x
=
□
Get tutor help
Solve for
g
g
g
.
\newline
3
=
g
−
4
−
5
g
=
□
\begin{array}{l} 3=\frac{g}{-4}-5 \\ g=\square \end{array}
3
=
−
4
g
−
5
g
=
□
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Solve for
r
r
r
.
\newline
−
13
=
r
9
+
8
r
=
□
\begin{array}{l} -13=\frac{r}{9}+8 \\ r=\square \end{array}
−
13
=
9
r
+
8
r
=
□
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Solve for
y
y
y
.
\newline
6
=
2
(
y
+
2
)
y
=
□
\begin{array}{l} 6=2(y+2) \\ y=\square \end{array}
6
=
2
(
y
+
2
)
y
=
□
Get tutor help
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