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Math Problems
Grade 6
Multiply using the distributive property
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
)
=
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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
)
=
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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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