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Math Problems
Algebra 2
Simplify variable expressions using properties
Simplify the expression:
\newline
(
−
4
+
j
)
+
5
=
(-4 + j) + 5 =
(
−
4
+
j
)
+
5
=
_____
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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
3
(
−
7
d
)
=
3(-7d) =
3
(
−
7
d
)
=
_____
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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
(
6
k
)
(
−
4
)
=
(6k)(-4) =
(
6
k
)
(
−
4
)
=
_____
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Simplify the expression:
\newline
−
7
(
−
4
r
)
=
-7(-4r) =
−
7
(
−
4
r
)
=
_____
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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
+
b
)
+
−
5
=
(-3 + b) + -5 =
(
−
3
+
b
)
+
−
5
=
_____
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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
(
2
+
c
)
+
−
2
=
(2 + c) + -2 =
(
2
+
c
)
+
−
2
=
_____
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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
−
sin
2
A
cos
2
A
=
1
−
tan
A
1
+
tan
A
\frac{1-\sin 2 A}{\cos 2 A}=\frac{1-\tan A}{1+\tan A}
c
o
s
2
A
1
−
s
i
n
2
A
=
1
+
t
a
n
A
1
−
t
a
n
A
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g
(
x
)
=
{
x
2
+
5
,
x
>
2
m
(
x
+
3
)
+
k
,
−
1
<
x
≤
2
2
x
3
+
x
+
7
,
x
≤
−
1
g(x)=\left\{\begin{array}{cl}x^{2}+5, & x>2 \\ m(x+3)+k, & -1<x \leq 2 \\ 2 x^{3}+x+7, & x \leq-1\end{array}\right.
g
(
x
)
=
⎩
⎨
⎧
x
2
+
5
,
m
(
x
+
3
)
+
k
,
2
x
3
+
x
+
7
,
x
>
2
−
1
<
x
≤
2
x
≤
−
1
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S
=
{
−
3
4
,
144
,
17.85
,
18
3
}
S=\left\{\frac{-3}{4}, \sqrt{144}, 17.85, \frac{18}{3}\right\}
S
=
{
4
−
3
,
144
,
17.85
,
3
18
}
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Let
g
(
x
)
=
1
8
x
3
+
1
2
x
−
1
4
g(x)=\frac{1}{8} x^{3}+\frac{1}{2} x-\frac{1}{4}
g
(
x
)
=
8
1
x
3
+
2
1
x
−
4
1
and let
h
h
h
be the inverse function of
g
g
g
. Notice that
g
(
2
)
=
7
4
g(2)=\frac{7}{4}
g
(
2
)
=
4
7
.
\newline
h
′
(
7
4
)
=
h^{\prime}\left(\frac{7}{4}\right)=
h
′
(
4
7
)
=
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If
x
2
+
x
y
−
3
y
=
3
x^{2}+x y-3 y=3
x
2
+
x
y
−
3
y
=
3
, then at the point
(
2
,
1
)
,
d
y
d
x
=
(2,1), \frac{d y}{d x}=
(
2
,
1
)
,
d
x
d
y
=
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(
2
sin
y
+
1
)
d
y
d
x
=
4
and
y
(
0
)
=
π
/
2.
\begin{array}{l} (2 \sin y+1) \frac{d y}{d x}=4 \text { and } \\ y(0)=\pi / 2 . \end{array}
(
2
sin
y
+
1
)
d
x
d
y
=
4
and
y
(
0
)
=
π
/2.
\newline
What is
x
x
x
when
y
=
π
y=\pi
y
=
π
?
\newline
x
=
x=
x
=
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d
y
d
x
=
3
y
\frac{d y}{d x}=3 y
d
x
d
y
=
3
y
and
y
(
0
)
=
3
y(0)=3
y
(
0
)
=
3
.
\newline
y
(
ln
2
)
=
y(\ln 2)=
y
(
ln
2
)
=
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d
y
d
x
=
2
y
2
\frac{d y}{d x}=2 y^{2}
d
x
d
y
=
2
y
2
and
y
(
1
)
=
−
1
y(1)=-1
y
(
1
)
=
−
1
.
\newline
y
(
3
)
=
y(3)=
y
(
3
)
=
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Let
y
=
cos
(
x
)
x
3
y=\cos (x) x^{3}
y
=
cos
(
x
)
x
3
.
\newline
d
y
d
x
=
\frac{d y}{d x}=
d
x
d
y
=
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d
d
x
(
x
3
sin
(
x
)
)
=
\frac{d}{d x}\left(x^{3} \sin (x)\right)=
d
x
d
(
x
3
sin
(
x
)
)
=
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y
=
3
x
[
1
,
6
]
y=\frac{3}{x} \quad[1,6]
y
=
x
3
[
1
,
6
]
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x
=
3
−
2
(
19
29
)
=
x=3-2\left(\frac{19}{29}\right)=
x
=
3
−
2
(
29
19
)
=
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4)
2
27
+
2
5
+
3
12
39
33
6
3
+
2
5
+
9
2
\text { 4) } \begin{array}{l} 2 \sqrt{27}+2 \sqrt{5}+3 \sqrt{12} \\ 39 \\ 33 \\ 6 \sqrt{3}+2 \sqrt{5}+9 \sqrt{2} \end{array}
4)
2
27
+
2
5
+
3
12
39
33
6
3
+
2
5
+
9
2
\newline
6
6
6
)
(
4
3
)
(
5
6
)
(4 \sqrt{3})(5 \sqrt{6})
(
4
3
)
(
5
6
)
\newline
20
18
20 \sqrt{18}
20
18
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−
2
x
+
y
=
4
y
=
−
3
x
+
9
\begin{array}{r}-2 x+y=4 \\ y=-3 x+9\end{array}
−
2
x
+
y
=
4
y
=
−
3
x
+
9
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Khan Academy
\newline
Get
\newline
Unit te
\newline
Google Cl
\newline
−
6
7
×
(
−
3
8
)
-\frac{6}{7} \times\left(-\frac{3}{8}\right)
−
7
6
×
(
−
8
3
)
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Example
2
2
2
: Fill in the blanks with suitable numbers to make the statements correct.
\newline
(i)
7
×
(
−
2
+
3
)
=
7
×
(
−
2
)
+
7
×
3
7 \times(-2+3)=7 \times(-2)+7 \times 3
7
×
(
−
2
+
3
)
=
7
×
(
−
2
)
+
7
×
3
\newline
(ii)
3
×
(
−
4
+
2
)
=
3
×
2
+
3
×
2
3 \times(-4+2)=3 \times 2+3 \times 2
3
×
(
−
4
+
2
)
=
3
×
2
+
3
×
2
\newline
Solutions:
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Example
1
1
1
: Simplify the following:
\newline
(i)
(
−
15
)
×
8
+
(
−
15
)
×
(
−
6
)
(-15) \times 8+(-15) \times(-6)
(
−
15
)
×
8
+
(
−
15
)
×
(
−
6
)
\newline
(ii)
20
×
(
−
13
)
−
(
−
10
)
×
(
−
13
)
20 \times(-13)-(-10) \times(-13)
20
×
(
−
13
)
−
(
−
10
)
×
(
−
13
)
\newline
Solutions:
\newline
(i)
\newline
(
−
15
)
×
8
+
(
−
15
)
×
(
−
6
)
=
(
−
15
)
×
8
+
(
−
15
)
×
(
−
16
)
\begin{array}{l} (-15) \times 8+(-15) \times(-6) \\ =(-15) \times 8+(-15) \times(-16) \end{array}
(
−
15
)
×
8
+
(
−
15
)
×
(
−
6
)
=
(
−
15
)
×
8
+
(
−
15
)
×
(
−
16
)
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7
7
7
.
y
=
6
x
−
15
;
(
4
,
9
)
y=6 x-15 ;(4,9)
y
=
6
x
−
15
;
(
4
,
9
)
Get tutor help
Funktio
f
(
x
)
=
2
x
2
+
x
f(x)=2 x^{2}+x
f
(
x
)
=
2
x
2
+
x
derivoidaan. Silloin
f
′
(
3
)
=
1
)
\left.f^{\prime}(3)=1\right)
f
′
(
3
)
=
1
)
Valitse tästä
ı
^
\hat{\imath}
^
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3
3
3
.
300
÷
[
375
−
{
440
−
(
150
−
60
−
50
)
}
]
300 \div[375-\{440-(150-60-50)\}]
300
÷
[
375
−
{
440
−
(
150
−
60
−
50
)}]
Get tutor help
Expand and simplify
−
4
n
+
5
{
3
(
−
2
n
+
7
)
+
8
(
n
−
3
)
}
-4 n+5\{3(-2 n+7)+8(n-3)\}
−
4
n
+
5
{
3
(
−
2
n
+
7
)
+
8
(
n
−
3
)}
Get tutor help
Solve for
x
x
x
.
\newline
x
−
2
x
2
−
18
=
−
3
x-\sqrt{2 x^{2}-18}=-3
x
−
2
x
2
−
18
=
−
3
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(
x
)
(
x
4
3
)
=
(x)\left(\sqrt[3]{x^{4}}\right)=
(
x
)
(
3
x
4
)
=
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Convert
24
24
24
celsius to faranheit.
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Converting in decimals
\newline
5
7
=
\frac{5}{7}=
7
5
=
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cos
x
(
x
3
+
y
2
)
+
cos
(
2
x
(
x
−
1
)
2
+
y
2
)
=
a
2
−
2
a
+
3
,
cos
x
(
x
2
+
y
2
)
−
cos
(
2
π
(
x
−
1
)
2
+
y
2
)
=
2
(
a
−
1
)
:
\cos x(x^{3}+y^{2})+\cos \left(\frac{2x}{(x-1)^{2}+y^{2}}\right)=a^{2}-2a+3,\ \cos x(x^{2}+y^{2})-\cos \left(\frac{2\pi}{(x-1)^{2}+y^{2}}\right)=2(a-1):
cos
x
(
x
3
+
y
2
)
+
cos
(
(
x
−
1
)
2
+
y
2
2
x
)
=
a
2
−
2
a
+
3
,
cos
x
(
x
2
+
y
2
)
−
cos
(
(
x
−
1
)
2
+
y
2
2
π
)
=
2
(
a
−
1
)
:
Get tutor help
g)
25
a
5
a
b
=
\frac{25 a}{5 a b}=
5
ab
25
a
=
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5
5
5
. Which two expressions are equivalent?
\newline
A.
4
(
2
+
∣
x
)
4(2+\mid x)
4
(
2
+
∣
x
)
\newline
B.)
\newline
4
+
2
+
1
x
(
4
+
2
)
+
x
6
+
1
x
\begin{array}{c} 4+2+1 x \\ (4+2)+x \\ 6+1 x \end{array}
4
+
2
+
1
x
(
4
+
2
)
+
x
6
+
1
x
\newline
C.
4
⋅
x
+
2
4 \cdot x+2
4
⋅
x
+
2
\newline
D.
4
÷
(
2
−
x
)
4 \div(2-x)
4
÷
(
2
−
x
)
\newline
4
⋅
(
x
+
2
)
4 \cdot(x+2)
4
⋅
(
x
+
2
)
\newline
4
−
2
÷
x
4-2 \div x
4
−
2
÷
x
Get tutor help
4
4
4
. Simplify
2
x
−
{
[
(
2
x
−
6
y
−
3
x
)
−
(
5
y
−
2
x
)
−
9
y
]
−
3
x
}
2 x-\{[(2 x-6 y-3 x)-(5 y-2 x)-9 y]-3 x\}
2
x
−
{[(
2
x
−
6
y
−
3
x
)
−
(
5
y
−
2
x
)
−
9
y
]
−
3
x
}
Get tutor help
FIND
x
,
y
,
z
−
3
3
+
4
y
=
1
x
−
z
3
x
+
y
−
1
=
−
z
−
1
6
z
5
y
+
8
=
−
1
\begin{array}{l}\text { FIND } x, y, z \\ -\frac{3}{3+4 y}=\frac{1}{x-z} \\ \frac{3}{x+y-1}=-z^{-1} \\ \frac{6 z}{5 y+8}=-1\end{array}
FIND
x
,
y
,
z
−
3
+
4
y
3
=
x
−
z
1
x
+
y
−
1
3
=
−
z
−
1
5
y
+
8
6
z
=
−
1
Get tutor help
Дано:
\newline
1
1
1
)
u
(
x
0
)
=
4
i
u
′
(
x
0
)
=
−
3
u\left(x_{0}\right)=4 \mathrm{i} u^{\prime}\left(x_{0}\right)=-3
u
(
x
0
)
=
4
i
u
′
(
x
0
)
=
−
3
;
\newline
2
2
2
)
v
(
x
0
)
=
−
3
v\left(x_{0}\right)=-3
v
(
x
0
)
=
−
3
і
v
′
(
x
0
)
=
3
v^{\prime}\left(x_{0}\right)=3
v
′
(
x
0
)
=
3
;
\newline
3
3
3
)
f
(
x
)
=
u
(
x
)
v
(
x
)
f(x)=u(x) v(x)
f
(
x
)
=
u
(
x
)
v
(
x
)
\newline
Обчисли значення
f
′
(
x
0
)
f^{\prime}\left(x_{0}\right)
f
′
(
x
0
)
:
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Simplify the fraction completely. If the fraction does not simplify, submit the fraction in its current form.
\newline
24
x
2
30
\frac{24 x^{2}}{30}
30
24
x
2
\newline
Answer:
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Perform the operation and express your answer as a single fraction in simplest form.
\newline
4
+
1
3
x
4+\frac{1}{3 x}
4
+
3
x
1
\newline
Answer:
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Simplify the fraction completely. If the fraction does not simplify, submit the fraction in its current form.
\newline
6
x
4
7
x
3
\frac{6 x^{4}}{7 x^{3}}
7
x
3
6
x
4
\newline
Answer:
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Given the definitions of
f
(
x
)
f(x)
f
(
x
)
and
g
(
x
)
g(x)
g
(
x
)
below, find the value of
f
(
g
(
−
2
)
)
f(g(-2))
f
(
g
(
−
2
))
.
\newline
f
(
x
)
=
3
x
2
+
x
+
7
g
(
x
)
=
2
x
+
8
\begin{array}{l} f(x)=3 x^{2}+x+7 \\ g(x)=2 x+8 \end{array}
f
(
x
)
=
3
x
2
+
x
+
7
g
(
x
)
=
2
x
+
8
\newline
Answer:
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(a)
\newline
4
x
+
5
y
=
9
4
y
3
−
x
2
=
−
7
\begin{array}{l} 4 x+5 y=9 \\ \frac{4 y}{3}-\frac{x}{2}=-7 \end{array}
4
x
+
5
y
=
9
3
4
y
−
2
x
=
−
7
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3.9279
(
1
−
(
−
0.3086
)
)
2
8
−
1
8
−
2
=
3.9279 \sqrt{(1-(-0.3086))^{2} \frac{8-1}{8-2}}=
3.9279
(
1
−
(
−
0.3086
)
)
2
8
−
2
8
−
1
=
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