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Math Problems
Algebra 2
Simplify variable expressions using properties
n
+
7
=
6
×
9
n+7=6\times 9
n
+
7
=
6
×
9
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(
t
)
{
[
x
y
]
,
[
−
2
,
8
]
,
[
−
1
,
5
]
,
[
1
,
4
,
−
3
]
,
[
9
,
4
]
,
[
1
,
5
+
1
]
,
[
2
,
8
3
+
3
]
,
[
2
]
\begin{array}{c} (t) \ \left\{\begin{array}{cc} [xy], & [-2,8], [-1,5], [1,4,-3], [9,4], [1,5+1], [2,8^{3}+3], [2] \end{array}\right. \end{array}
(
t
)
{
[
x
y
]
,
[
−
2
,
8
]
,
[
−
1
,
5
]
,
[
1
,
4
,
−
3
]
,
[
9
,
4
]
,
[
1
,
5
+
1
]
,
[
2
,
8
3
+
3
]
,
[
2
]
Get tutor help
{
−
5
,
−
1.5
,
−
1
3
,
4
,
π
,
4.
3
‾
}
\left\{-\sqrt{5},-1.5,-\frac{1}{3}, \sqrt{4}, \pi, 4 . \overline{3}\right\}
{
−
5
,
−
1.5
,
−
3
1
,
4
,
π
,
4.
3
}
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Consider the following system of equations,
\newline
−
2
x
−
9
y
=
12
−
2
x
−
7
y
=
8
\begin{array}{l} -2 \mathrm{x}-9 \mathrm{y}=12 \\ -2 \mathrm{x}-7 \mathrm{y}=8 \end{array}
−
2
x
−
9
y
=
12
−
2
x
−
7
y
=
8
\newline
The value of
x
x
x
is
\newline
□
\square
□
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ID: Di sebuah toples terdapat
65
65
65
permen dengan rincian:
\newline
-
15
15
15
permen cokelat,
\newline
-
7
7
7
permen stroberi,
\newline
-
10
10
10
permen vanila,
\newline
-
8
8
8
permen jeruk,
\newline
-
10
10
10
permen kopi,
\newline
-
15
15
15
permen karamel.
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Which of the following expressions has a coefficient of
10
10
10
and a constant of
5
5
5
?
\newline
10
−
5
10
x
+
5
10
+
5
10
+
5
x
\begin{array}{l} 10-5 \\ 10 x+5 \\ 10+5 \\ 10+5 x \\ \end{array}
10
−
5
10
x
+
5
10
+
5
10
+
5
x
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lim
x
→
0
tan
−
1
(
9
x
)
−
9
x
cos
(
9
x
)
−
243
2
x
3
x
5
=
\lim _{x \rightarrow 0} \frac{\tan ^{-1}(9 x)-9 x \cos (9 x)-\frac{243}{2} x^{3}}{x^{5}}=
lim
x
→
0
x
5
t
a
n
−
1
(
9
x
)
−
9
x
c
o
s
(
9
x
)
−
2
243
x
3
=
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−
22
−
36
=
5
−
(
−
8
)
=
\begin{array}{l}-22-36= \\ 5-(-8)=\end{array}
−
22
−
36
=
5
−
(
−
8
)
=
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−
2
+
x
3
+
x
=
6
\frac{-2+x}{3+x}=6
3
+
x
−
2
+
x
=
6
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Simplify.
\newline
Rewrite the expression in the form
3
n
3^{n}
3
n
.
\newline
3
4
⋅
3
2
=
3^{4} \cdot 3^{2}=
3
4
⋅
3
2
=
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6
2
+
3
x
−
8
÷
2
x
x
=
4
\begin{array}{r}6^{2}+3 x-8 \div 2 x \\ x=4\end{array}
6
2
+
3
x
−
8
÷
2
x
x
=
4
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Question
4
4
4
*
\newline
1
1
1
point
\newline
If
A
=
(
2
1
−
4
1
)
A=\left(\begin{array}{cc}2 & 1 \\ -4 & 1\end{array}\right)
A
=
(
2
−
4
1
1
)
, the
A
−
1
=
A^{-1}=
A
−
1
=
\newline
1
6
(
1
−
1
−
4
2
)
−
1
2
(
1
−
1
4
2
)
\frac{1}{6}\left(\begin{array}{cc} 1 & -1 \\ -4 & 2 \end{array}\right) \quad-\frac{1}{2}\left(\begin{array}{cc} 1 & -1 \\ 4 & 2 \end{array}\right)
6
1
(
1
−
4
−
1
2
)
−
2
1
(
1
4
−
1
2
)
\newline
A
\newline
B
\newline
1
6
(
1
−
1
4
2
)
1
2
(
1
−
1
4
2
)
\frac{1}{6}\left(\begin{array}{cc} 1 & -1 \\ 4 & 2 \end{array}\right) \quad \frac{1}{2}\left(\begin{array}{cc} 1 & -1 \\ 4 & 2 \end{array}\right)
6
1
(
1
4
−
1
2
)
2
1
(
1
4
−
1
2
)
\newline
C
\newline
D
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55.
x
−
6
x
2
−
5
x
+
6
=
A
x
−
2
+
B
x
−
3
⇒
5
A
−
B
=
?
\begin{array}{l}\text { 55. } \frac{x-6}{x^{2}-5 x+6}=\frac{A}{x-2}+\frac{B}{x-3} \\ \Rightarrow 5 A-B=? \\\end{array}
55.
x
2
−
5
x
+
6
x
−
6
=
x
−
2
A
+
x
−
3
B
⇒
5
A
−
B
=
?
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Simplify the complex fraction.
\newline
7
x
+
2
y
1
x
y
=
\frac{\frac{7}{x}+\frac{2}{y}}{\frac{1}{x y}}=
x
y
1
x
7
+
y
2
=
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=
2
π
⋅
1
+
e
−
V
2
=
=2 \pi \sqrt{ } \cdot 1+\frac{e^{-V}}{2}=
=
2
π
⋅
1
+
2
e
−
V
=
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{:
(
7
−
9
)
+
5
∗
∗
(
3
−
7
)
)
+
8
∗
∗
(
5
−
3
∗
∗
(
1
−
10
)
−
6
)
(7-9)+5^{**}(3-7))+8^{**}(5-3^{**}(1-10)-6)
(
7
−
9
)
+
5
∗∗
(
3
−
7
))
+
8
∗∗
(
5
−
3
∗∗
(
1
−
10
)
−
6
)
, cor
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Simplufy:
\newline
16
x
−
8
x
4
=
2
\frac{16 x-8 x}{4}=2
4
16
x
−
8
x
=
2
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2
2
2
.
d
d
x
(
x
−
ln
x
x
2
+
1
)
∣
x
=
1
=
\left.\frac{d}{d x}\left(\frac{x-\ln x}{x^{2}+1}\right)\right|_{x=1}=
d
x
d
(
x
2
+
1
x
−
l
n
x
)
∣
∣
x
=
1
=
Get tutor help
y
=
8
x
+
16
y
=
5
x
+
7
\begin{array}{l}y=8 x+16 \\ y=5 x+7\end{array}
y
=
8
x
+
16
y
=
5
x
+
7
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Simplify the following:
\newline
2
x
+
2
−
4
2
x
+
1
\frac{2}{x+2}-\frac{4}{2x+1}
x
+
2
2
−
2
x
+
1
4
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Simplify the expression:
\newline
3
(
3
p
)
=
3(3p) =
3
(
3
p
)
=
_____
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Simplify the expression:
\newline
(
5
+
z
)
+
−
2
=
(5 + z) + -2 =
(
5
+
z
)
+
−
2
=
_____
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Simplify the expression:
\newline
−
2
(
−
2
d
)
=
-2(-2d) =
−
2
(
−
2
d
)
=
_____
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Simplify the expression:
\newline
3
+
(
q
+
−
8
)
=
3 + (q + -8) =
3
+
(
q
+
−
8
)
=
_____
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Simplify the expression:
\newline
−
1
+
(
2
+
w
)
=
-1 + (2 + w) =
−
1
+
(
2
+
w
)
=
_____
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Simplify the expression:
\newline
(
1
+
d
)
+
1
=
(1 + d) + 1 =
(
1
+
d
)
+
1
=
_____
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Simplify the expression:
\newline
(
−
2
r
)
(
−
2
)
=
(-2r)(-2) =
(
−
2
r
)
(
−
2
)
=
_____
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Simplify the expression:
\newline
3
+
(
–
3
+
p
)
=
3 + (\text{–}3 + p) =
3
+
(
–
3
+
p
)
=
_____
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Simplify the expression:
\newline
−
7
(
−
4
r
)
=
-7(-4r) =
−
7
(
−
4
r
)
=
_____
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Simplify the expression:
\newline
(
6
k
)
(
−
4
)
=
(6k)(-4) =
(
6
k
)
(
−
4
)
=
_____
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Simplify the expression:
\newline
−
5
+
(
b
+
2
)
=
-5 + (b + 2) =
−
5
+
(
b
+
2
)
=
_____
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Simplify the expression:
\newline
−
4
+
(
−
4
+
f
)
=
-4 + (-4 + f) =
−
4
+
(
−
4
+
f
)
=
_____
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Simplify the expression:
\newline
9
+
(
z
+
−
7
)
=
9 + (z + -7) =
9
+
(
z
+
−
7
)
=
_____
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Simplify the expression:
\newline
(
−
3
+
b
)
+
−
5
=
(-3 + b) + -5 =
(
−
3
+
b
)
+
−
5
=
_____
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Simplify the expression:
\newline
8
+
(
q
+
−
1
)
=
8 + (q + -1) =
8
+
(
q
+
−
1
)
=
_____
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Simplify the expression:
\newline
(
−
4
+
j
)
+
5
=
(-4 + j) + 5 =
(
−
4
+
j
)
+
5
=
_____
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Simplify the expression:
\newline
(
2
+
c
)
+
−
2
=
(2 + c) + -2 =
(
2
+
c
)
+
−
2
=
_____
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Simplify the expression:
\newline
(
s
+
3
)
+
(
−
6
)
=
_
_
_
_
_
(s + 3) + (-6) = \_\_\_\_\_
(
s
+
3
)
+
(
−
6
)
=
_____
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Simplify the expression:
\newline
−
2
+
(
2
+
m
)
=
-2 + (2 + m) =
−
2
+
(
2
+
m
)
=
_____
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Simplify the expression:
\newline
−
6
+
(
4
+
s
)
=
-6 + (4 + s) =
−
6
+
(
4
+
s
)
=
_____
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Simplify the expression:
\newline
−
6
+
(
f
+
1
)
=
-6 + (f + 1) =
−
6
+
(
f
+
1
)
=
_____
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Simplify the expression:
\newline
(
3
+
z
)
+
−
9
=
(3 + z) + -9 =
(
3
+
z
)
+
−
9
=
_____
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Simplify the expression:
\newline
(
q
+
8
)
+
−
4
=
(q + 8) + -4 =
(
q
+
8
)
+
−
4
=
_____
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Simplify the expression:
\newline
6
+
(
–
2
+
u
)
=
6 + (\text{–}2 + u) =
6
+
(
–
2
+
u
)
=
_____
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Simplify the expression:
\newline
3
(
−
7
d
)
=
3(-7d) =
3
(
−
7
d
)
=
_____
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Simplify the expression:
\newline
−
1
+
(
n
+
1
)
=
-1 + (n + 1) =
−
1
+
(
n
+
1
)
=
_____
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Simplify the expression:
\newline
(
7
p
)
(
−
3
)
=
(7p)(-3) =
(
7
p
)
(
−
3
)
=
_____
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Simplify the expression:
\newline
(
−
5
z
)
(
−
3
)
=
(-5z)(-3) =
(
−
5
z
)
(
−
3
)
=
_____
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sin
6
x
−
sin
2
x
cos
6
x
−
cos
2
x
=
−
cot
4
x
\frac{\sin 6 x-\sin 2 x}{\cos 6 x-\cos 2 x}=-\cot 4 x
c
o
s
6
x
−
c
o
s
2
x
s
i
n
6
x
−
s
i
n
2
x
=
−
cot
4
x
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−
c
(
11
c
+
4
)
-c(11 c+4)
−
c
(
11
c
+
4
)
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1
2
3
...
7
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