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Math Problems
Grade 6
Multiply two decimals: where does the decimal point go?
Solve the equation.
\newline
42
=
7
r
r
=
\begin{array}{l} 42=7 r \\ r= \end{array}
42
=
7
r
r
=
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Solve the equation.
\newline
10
=
p
10
p
=
□
\begin{array}{l} 10=\frac{p}{10} \\ p=\square \end{array}
10
=
10
p
p
=
□
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Solve the equation.
\newline
6
=
3
s
s
=
□
\begin{array}{l} 6=3 s \\ s=\square \end{array}
6
=
3
s
s
=
□
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Solve the equation.
\newline
9
x
=
108
x
=
□
\begin{array}{l} 9 x=108 \\ x=\square \end{array}
9
x
=
108
x
=
□
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Solve the equation.
\newline
9
=
h
9
h
=
□
\begin{array}{l} 9=\frac{h}{9} \\ h=\square \end{array}
9
=
9
h
h
=
□
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Solve the equation.
\newline
2
e
=
2
e
=
□
\begin{array}{l} 2 e=2 \\ e=\square \end{array}
2
e
=
2
e
=
□
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Solve the equation.
\newline
11
=
7
v
v
=
□
\begin{array}{l} 11=7 v \\ v=\square \end{array}
11
=
7
v
v
=
□
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Solve the equation.
\newline
6
=
x
5
x
=
□
\begin{array}{l} 6=\frac{x}{5} \\ x=\square \end{array}
6
=
5
x
x
=
□
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Solve the equation.
\newline
6
g
=
48
g
=
□
\begin{array}{l} 6 g=48 \\ g=\square \end{array}
6
g
=
48
g
=
□
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Solve for
k
k
k
.
\newline
5
k
−
2
3
k
=
7
k
=
\begin{array}{l} \frac{5 k-2}{3 k}=7 \\ k= \end{array}
3
k
5
k
−
2
=
7
k
=
Get tutor help
Solve for
k
k
k
.
\newline
4
k
−
6
3
k
−
9
=
1
3
k
=
\begin{array}{l} \frac{4 k-6}{3 k-9}=\frac{1}{3} \\ k= \end{array}
3
k
−
9
4
k
−
6
=
3
1
k
=
Get tutor help
Solve for
z
z
z
.
\newline
z
−
6
2
z
+
1
=
10
z
=
\begin{array}{l} \frac{z-6}{2 z+1}=10 \\ z= \end{array}
2
z
+
1
z
−
6
=
10
z
=
Get tutor help
Solve for
q
q
q
.
\newline
10
2
q
+
8
=
10
q
=
□
\begin{array}{l} \frac{10}{2 q+8}=10 \\ q=\square \end{array}
2
q
+
8
10
=
10
q
=
□
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Solve for
n
n
n
.
\newline
4
n
−
2
2
n
=
7
n
=
\begin{array}{l} \frac{4 n-2}{2 n}=7 \\ n= \end{array}
2
n
4
n
−
2
=
7
n
=
Get tutor help
Solve for
k
k
k
.
\newline
3
2
k
=
5
k
=
\begin{array}{l} \frac{3}{2 k}=5 \\ k= \end{array}
2
k
3
=
5
k
=
Get tutor help
Solve for
k
k
k
.
\newline
4
k
−
10
k
=
8
k
=
\begin{array}{l} \frac{4 k-10}{k}=8 \\ k= \end{array}
k
4
k
−
10
=
8
k
=
Get tutor help
Solve for
r
r
r
.
\newline
4
r
−
6
r
−
7
=
1
2
r
=
□
\begin{array}{l} \frac{4 r-6}{r-7}=\frac{1}{2} \\ r=\square \end{array}
r
−
7
4
r
−
6
=
2
1
r
=
□
Get tutor help
Evaluate.
\newline
1
3
4
⋅
48
4
=
\sqrt[4]{\frac{1}{3}} \cdot \sqrt[4]{48}=
4
3
1
⋅
4
48
=
Get tutor help
Evaluate.
\newline
32
4
1
4
⋅
1
4
4
=
324^{\frac{1}{4}} \cdot \sqrt[4]{\frac{1}{4}}=
32
4
4
1
⋅
4
4
1
=
Get tutor help
Evaluate.
\newline
6
4
7
10
6
4
1
5
=
\frac{64^{\frac{7}{10}}}{64^{\frac{1}{5}}}=
6
4
5
1
6
4
10
7
=
Get tutor help
Evaluate.
\newline
25
6
4
7
2
4
7
=
\frac{256^{\frac{4}{7}}}{2^{\frac{4}{7}}}=
2
7
4
25
6
7
4
=
Get tutor help
Evaluate.
\newline
(
1
81
)
1
2
⋅
(
1
81
)
−
3
4
=
\left(\frac{1}{81}\right)^{\frac{1}{2}} \cdot\left(\frac{1}{81}\right)^{-\frac{3}{4}}=
(
81
1
)
2
1
⋅
(
81
1
)
−
4
3
=
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lim
x
→
4
2
x
+
1
−
3
x
+
3
=
\lim _{x \rightarrow 4} \frac{2 x+1}{-3 x+3}=
lim
x
→
4
−
3
x
+
3
2
x
+
1
=
Get tutor help
Subtract.
\newline
The numerator should be expanded and simplified. The denominator should be either expanded or factored.
\newline
7
x
−
2
x
+
8
=
□
\frac{7}{x}-\frac{2}{x+8}=\square
x
7
−
x
+
8
2
=
□
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f
(
x
)
=
14
−
0.5
x
f
(
30
)
=
□
\begin{array}{l}f(x)=14-0.5 x \\ f(30)=\square\end{array}
f
(
x
)
=
14
−
0.5
x
f
(
30
)
=
□
Get tutor help
Evaluate.
\newline
14
−
4
2
=
\frac{14-4}{2}=
2
14
−
4
=
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g
(
r
)
=
25
−
3
r
g
(
4
)
=
\begin{array}{l}g(r)=25-3 r \\ g(4)=\end{array}
g
(
r
)
=
25
−
3
r
g
(
4
)
=
Get tutor help
8
⋅
5
−
t
9
=
346
8 \cdot 5^{\frac{-t}{9}}=346
8
⋅
5
9
−
t
=
346
\newline
Which of the following is the solution of the equation?
\newline
Choose
1
1
1
answer:
\newline
(A)
t
=
9
log
5
(
43.25
)
t=9 \log _{5}(43.25)
t
=
9
lo
g
5
(
43.25
)
\newline
(B)
t
=
−
9
log
40
(
346
)
t=-9 \log _{40}(346)
t
=
−
9
lo
g
40
(
346
)
\newline
(C)
t
=
−
9
log
5
(
43.25
)
t=-9 \log _{5}(43.25)
t
=
−
9
lo
g
5
(
43.25
)
\newline
(D)
t
=
9
log
346
(
40
)
t=9 \log _{346}(40)
t
=
9
lo
g
346
(
40
)
Get tutor help
Evaluate the following expression.
\newline
9
+
8
3
2
=
9+\frac{8^{3}}{2}=
9
+
2
8
3
=
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g
(
t
)
=
−
9
t
−
4
g
(
□
)
=
23
\begin{array}{l}g(t)=-9 t-4 \\ g(\square)=23\end{array}
g
(
t
)
=
−
9
t
−
4
g
(
□
)
=
23
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f
(
t
)
=
5
t
−
7
f
(
□
)
=
48
\begin{array}{l}f(t)=5 t-7 \\ f(\square)=48\end{array}
f
(
t
)
=
5
t
−
7
f
(
□
)
=
48
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f
(
t
)
=
−
3
t
+
4
f
(
□
)
=
19
\begin{array}{l}f(t)=-3 t+4 \\ f(\square)=19\end{array}
f
(
t
)
=
−
3
t
+
4
f
(
□
)
=
19
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y
=
1
5
x
−
9
5
y
2
+
15
=
7
x
\begin{aligned} y & =\frac{1}{5} x-9 \\ 5 y^{2}+15 & =7 x \end{aligned}
y
5
y
2
+
15
=
5
1
x
−
9
=
7
x
\newline
If
(
a
,
b
)
(a, b)
(
a
,
b
)
is a solution to the system of equations shown and
b
>
0
b>0
b
>
0
, what is the value of
b
b
b
?
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s
(
s
−
1
)
=
2
s(s-1)=2
s
(
s
−
1
)
=
2
\newline
What are the solutions to the given equation?
\newline
Choose
1
1
1
answer:
\newline
(A)
s
=
−
2
s=-2
s
=
−
2
and
s
=
1
s=1
s
=
1
\newline
(B)
s
=
0
s=0
s
=
0
and
s
=
−
1
s=-1
s
=
−
1
\newline
(C)
s
=
0
s=0
s
=
0
and
s
=
1
s=1
s
=
1
\newline
(D)
s
=
2
s=2
s
=
2
and
s
=
−
1
s=-1
s
=
−
1
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(
x
−
7
)
2
−
25
=
0
(x-7)^{2}-25=0
(
x
−
7
)
2
−
25
=
0
\newline
What are the solutions to the given equation?
\newline
Choose
1
1
1
answer:
\newline
(A)
x
=
12
,
x
=
2
x=12, x=2
x
=
12
,
x
=
2
\newline
(B)
x
=
12
,
x
=
−
2
x=12, x=-2
x
=
12
,
x
=
−
2
\newline
(C)
x
=
−
12
,
x
=
2
x=-12, x=2
x
=
−
12
,
x
=
2
\newline
(D)
x
=
−
12
,
x
=
−
2
x=-12, x=-2
x
=
−
12
,
x
=
−
2
Get tutor help
Solve the equation.
\newline
3
2
+
b
=
7
4
b
=
\begin{array}{l} \frac{3}{2}+b=\frac{7}{4} \\ b= \end{array}
2
3
+
b
=
4
7
b
=
Get tutor help
Solve the equation.
\newline
g
−
3
7
=
2
7
g
=
□
\begin{array}{l} g-\frac{3}{7}=\frac{2}{7} \\ g=\square \end{array}
g
−
7
3
=
7
2
g
=
□
Get tutor help
Solve the equation.
\newline
j
−
5
=
3.4
j
=
□
\begin{array}{l} j-5=3.4 \\ j=\square \end{array}
j
−
5
=
3.4
j
=
□
Get tutor help
Solve the equation.
\newline
h
−
3.5
=
9
h
=
\begin{array}{l} h-3.5=9 \\ h= \end{array}
h
−
3.5
=
9
h
=
Get tutor help
Solve the equation.
\newline
11
6
=
n
+
7
9
n
=
□
\begin{array}{l} \frac{11}{6}=n+\frac{7}{9} \\ n=\square \end{array}
6
11
=
n
+
9
7
n
=
□
Get tutor help
Solve the equation.
\newline
4.4
=
x
−
5.6
x
=
□
\begin{array}{l} 4.4=x-5.6 \\ x=\square \end{array}
4.4
=
x
−
5.6
x
=
□
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Solve the equation.
\newline
d
+
0.5
=
0.75
d
=
□
\begin{array}{l} d+0.5=0.75 \\ d=\square \end{array}
d
+
0.5
=
0.75
d
=
□
Get tutor help
Solve the equation.
\newline
17
=
r
+
10
r
=
□
\begin{array}{r} 17=r+10 \\ r=\square \end{array}
17
=
r
+
10
r
=
□
Get tutor help
Solve the equation.
\newline
27
=
15
+
n
n
=
□
\begin{array}{l} 27=15+n \\ n=\square \end{array}
27
=
15
+
n
n
=
□
Get tutor help
Solve the equation.
\newline
2
+
y
=
11
y
=
□
\begin{aligned} & 2+y=11 \\ y= & \square \end{aligned}
y
=
2
+
y
=
11
□
Get tutor help
Solve the equation.
\newline
n
=
+
n
=
30
n
=
□
\begin{array}{l} n=+n=30 \\ n=\square \end{array}
n
=
+
n
=
30
n
=
□
Get tutor help
Solve the equation.
\newline
k
+
10
=
27
k
=
□
\begin{array}{l} k+10=27 \\ k=\square \end{array}
k
+
10
=
27
k
=
□
Get tutor help
Solve the equation.
\newline
e
+
1.2
=
2
e
=
\begin{array}{l} e+1.2=2 \\ e= \end{array}
e
+
1.2
=
2
e
=
Get tutor help
Solve the equation.
\newline
5
7
=
p
+
4
7
p
=
□
\begin{array}{l} \frac{5}{7}=p+\frac{4}{7} \\ p=\square \end{array}
7
5
=
p
+
7
4
p
=
□
Get tutor help
Solve the equation.
\newline
24
=
y
+
19
y
=
□
\begin{array}{l} 24=y+19 \\ y=\square \end{array}
24
=
y
+
19
y
=
□
Get tutor help
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