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Math Problems
Grade 6
Multiply using the distributive property
g
(
r
)
=
−
5
r
+
13
g
(
3
)
=
□
\begin{array}{l}g(r)=-5 r+13 \\ g(3)=\square\end{array}
g
(
r
)
=
−
5
r
+
13
g
(
3
)
=
□
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k
(
t
)
=
13
t
−
2
k
(
3
)
=
\begin{array}{l}k(t)=13 t-2 \\ k(3)=\end{array}
k
(
t
)
=
13
t
−
2
k
(
3
)
=
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g
(
r
)
=
−
1
−
7
r
g
(
6
)
=
□
\begin{array}{l}g(r)=-1-7 r \\ g(6)=\square\end{array}
g
(
r
)
=
−
1
−
7
r
g
(
6
)
=
□
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h
(
x
)
=
17
+
x
6
h
(
−
18
)
=
\begin{array}{l}h(x)=17+\frac{x}{6} \\ h(-18)=\end{array}
h
(
x
)
=
17
+
6
x
h
(
−
18
)
=
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Evaluate.
\newline
12
3
+
1
=
\frac{12}{3+1}=
3
+
1
12
=
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k
(
t
)
=
10
t
−
19
k
(
−
7
)
=
\begin{array}{l}k(t)=10 t-19 \\ k(-7)=\end{array}
k
(
t
)
=
10
t
−
19
k
(
−
7
)
=
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g
(
r
)
=
−
1
−
7
r
g
(
6
)
=
□
\begin{array}{l}g(r)=-1-7 r \\ g(6)=\square\end{array}
g
(
r
)
=
−
1
−
7
r
g
(
6
)
=
□
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h
(
t
)
=
−
20
+
11
t
h
(
11
)
=
□
\begin{array}{l}h(t)=-20+11 t \\ h(11)=\square\end{array}
h
(
t
)
=
−
20
+
11
t
h
(
11
)
=
□
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What is the midline equation of
\newline
y
=
−
4
sin
(
2
x
−
7
)
+
3
?
y
=
□
\begin{array}{l} y=-4 \sin (2 x-7)+3 ? \\ y=\square \end{array}
y
=
−
4
sin
(
2
x
−
7
)
+
3
?
y
=
□
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What is the midline equation of
\newline
y
=
8
cos
(
5
π
x
+
3
π
2
)
−
9
?
y
=
□
\begin{array}{l} y=8 \cos \left(5 \pi x+\frac{3 \pi}{2}\right)-9 ? \\ y=\square \end{array}
y
=
8
cos
(
5
π
x
+
2
3
π
)
−
9
?
y
=
□
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What is the midline equation of
\newline
y
=
−
5
cos
(
2
π
x
+
1
)
−
10
?
y
=
□
\begin{array}{l} y=-5 \cos (2 \pi x+1)-10 ? \\ y=\square \end{array}
y
=
−
5
cos
(
2
π
x
+
1
)
−
10
?
y
=
□
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What is the midline equation of
\newline
y
=
sin
(
3
x
+
5
)
?
y
=
□
\begin{array}{l} y=\sin (3 x+5) ? \\ y=\square \end{array}
y
=
sin
(
3
x
+
5
)?
y
=
□
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What is the midline equation of the function
\newline
g
(
x
)
=
2
cos
(
7
x
+
5
)
+
1
?
y
=
□
\begin{array}{l} g(x)=2 \cos (7 x+5)+1 ? \\ y=\square \end{array}
g
(
x
)
=
2
cos
(
7
x
+
5
)
+
1
?
y
=
□
Get tutor help
What is the midline equation of
\newline
y
=
cos
(
2
π
3
x
)
+
2
?
y
=
□
\begin{array}{l} y=\cos \left(\frac{2 \pi}{3} x\right)+2 ? \\ y=\square \end{array}
y
=
cos
(
3
2
π
x
)
+
2
?
y
=
□
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log
16
1
2
=
\log _{16} \frac{1}{2}=
lo
g
16
2
1
=
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z
=
4.1
i
+
85
Re
(
z
)
=
Im
(
z
)
=
\begin{array}{l}z=4.1 i+85 \\ \operatorname{Re}(z)= \\ \operatorname{Im}(z)=\end{array}
z
=
4.1
i
+
85
Re
(
z
)
=
Im
(
z
)
=
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z
=
6
−
3
i
Re
(
z
)
=
Im
(
z
)
=
\begin{array}{l}z=6-3 i \\ \operatorname{Re}(z)= \\ \operatorname{Im}(z)=\end{array}
z
=
6
−
3
i
Re
(
z
)
=
Im
(
z
)
=
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z
=
14
i
+
12.1
Re
(
z
)
=
Im
(
z
)
=
\begin{array}{l}z=14 i+12.1 \\ \operatorname{Re}(z)= \\ \operatorname{Im}(z)=\end{array}
z
=
14
i
+
12.1
Re
(
z
)
=
Im
(
z
)
=
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z
=
6.2
+
37
i
Re
(
z
)
=
Im
(
z
)
=
\begin{array}{l}z=6.2+37 i \\ \operatorname{Re}(z)= \\ \operatorname{Im}(z)=\end{array}
z
=
6.2
+
37
i
Re
(
z
)
=
Im
(
z
)
=
Get tutor help
z
=
4
−
2
i
Re
(
z
)
=
Im
(
z
)
=
\begin{array}{l}z=4-2 i \\ \operatorname{Re}(z)= \\ \operatorname{Im}(z)=\end{array}
z
=
4
−
2
i
Re
(
z
)
=
Im
(
z
)
=
Get tutor help
h
(
x
)
=
x
−
11
h
(
□
)
=
−
5
\begin{array}{l}h(x)=x-11 \\ h(\square)=-5\end{array}
h
(
x
)
=
x
−
11
h
(
□
)
=
−
5
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h
(
t
)
=
4
t
+
20
h
(
□
)
=
4
\begin{array}{l}h(t)=4 t+20 \\ h(\square)=4\end{array}
h
(
t
)
=
4
t
+
20
h
(
□
)
=
4
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f
(
t
)
=
−
2
t
+
5
f
(
□
)
=
13
\begin{array}{l}f(t)=-2 t+5 \\ f(\square)=13\end{array}
f
(
t
)
=
−
2
t
+
5
f
(
□
)
=
13
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g
(
x
)
=
8
x
+
2
g
(
□
)
=
−
62
\begin{array}{l}g(x)=8 x+2 \\ g(\square)=-62\end{array}
g
(
x
)
=
8
x
+
2
g
(
□
)
=
−
62
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Solve the equation.
\newline
5
13
=
t
−
6
13
t
=
□
\begin{array}{l} \frac{5}{13}=t-\frac{6}{13} \\ t=\square \end{array}
13
5
=
t
−
13
6
t
=
□
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Solve the equation.
\newline
11
15
=
w
−
8
15
w
=
□
\begin{array}{l} \frac{11}{15}=w-\frac{8}{15} \\ w=\square \end{array}
15
11
=
w
−
15
8
w
=
□
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Distribute to create an equivalent expression with the fewest symbols possible.
\newline
3
(
7
x
+
1
)
=
3(7 x+1)=
3
(
7
x
+
1
)
=
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