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Math Problems
Algebra 2
Simplify rational expressions
Which of the following is a solution to the equation
(
x
−
2
)
2
=
25
?
(x-2)^{2}=25 ?
(
x
−
2
)
2
=
25
?
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Divide and simplify.
\newline
3
n
9
+
6
n
5
−
15
n
2
−
3
n
4
\frac{3 n^{9}+6 n^{5}-15 n^{2}}{-3 n^{4}}
−
3
n
4
3
n
9
+
6
n
5
−
15
n
2
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Choase the coroct ansurer
\newline
−
84
,
−
52
,
−
12
is
a)
18
b)
−
18
c)
\begin{array}{lll} -84,-52,-12 & \text { is } \\ \text { a) } \begin{array}{lll} 18 & \text { b) }-18 & \text { c) } \end{array} \end{array}
−
84
,
−
52
,
−
12
a)
18
b)
−
18
c)
is
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3
3
3
. The greatest common factor of
200
×
3
200 \times 3
200
×
3
and
2
×
300
2 \times 300
2
×
300
is
\newline
A)
2
×
3
2 \times 3
2
×
3
\newline
B)
20
×
30
20 \times 30
20
×
30
\newline
C)
23
×
100
23 \times 100
23
×
100
\newline
D)
200
×
300
200 \times 300
200
×
300
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K
M
h
undefined
KM^{\overrightarrow{h}}
K
M
h
and
N
P
h
undefined
NP^{\overrightarrow{h}}
N
P
h
are parallel lines. Which angles are adjacent angles?
\newline
/
M
L
O
/_{MLO}
/
M
L
O
and
/
M
L
J
/_{MLJ}
/
M
L
J
\newline
/
K
L
O
/_{KLO}
/
K
L
O
and
/
N
O
L
/_{NOL}
/
NO
L
\newline
/
K
U
/_{KU}
/
K
U
and
/
P
O
L
/_{POL}
/
PO
L
\newline
/
N
O
L
/_{NOL}
/
NO
L
and
/
M
L
O
/_{MLO}
/
M
L
O
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K
M
undefined
\overleftrightarrow{K M}
K
M
and
N
P
undefined
\overleftrightarrow{N P}
NP
are parallel lines.
\newline
Which angles are adjacent angles?
\newline
∠
M
L
O
\angle M L O
∠
M
L
O
and
∠
M
L
\angle M L
∠
M
L
\newline
∠
K
L
O
\angle K L O
∠
K
L
O
and
∠
N
O
L
\angle N O L
∠
NO
L
\newline
∠
K
U
\angle K U
∠
K
U
and
∠
P
O
L
\angle P O L
∠
PO
L
\newline
∠
N
O
L
\angle N O L
∠
NO
L
and
∠
M
L
O
\angle M L O
∠
M
L
O
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Figure I and Figure II are similar figures.
\newline
Figure I
\newline
Figure II
\newline
Which proportion must be true?
\newline
R
S
A
B
=
T
U
C
D
T
U
A
B
=
U
V
A
F
\frac{R S}{A B}=\frac{T U}{C D} \quad \frac{T U}{A B}=\frac{U V}{A F}
A
B
RS
=
C
D
T
U
A
B
T
U
=
A
F
U
V
\newline
Option
1
1
1
\newline
Option
2
2
2
\newline
S
T
E
F
=
W
R
C
D
W
V
A
B
=
S
T
E
F
\frac{S T}{E F}=\frac{W R}{C D} \quad \frac{W V}{A B}=\frac{S T}{E F}
EF
ST
=
C
D
W
R
A
B
WV
=
EF
ST
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For \#
5
5
5
\& \#
6
6
6
, graph the linear equation. Then des
\newline
5
5
5
.)
y
=
3.5
y=3.5
y
=
3.5
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Step
2
2
2
: If the rational number is not in its simplest form, first find Then divide both the numerator and denominator by their HCF to expres Example
1
1
1
: Express (i)
2
−
7
\frac{2}{-7}
−
7
2
\newline
(ii)
4
10
\frac{4}{10}
10
4
\newline
(iii)
2
3
\frac{2}{3}
3
2
and
\newline
(iv)
17
49
\frac{17}{49}
49
17
in their standard Solution: (i)
2
−
7
\frac{2}{-7}
−
7
2
is not in the standard form as its denominator is negat
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b) Find the general solution of
cos
θ
+
Cos
3
θ
+
cos
5
θ
=
0
\cos \theta+\operatorname{Cos} 3 \theta+\cos 5 \theta=0
cos
θ
+
Cos
3
θ
+
cos
5
θ
=
0
\newline
OR
\newline
Prove that:
tan
−
1
x
−
tan
−
1
y
=
tan
−
1
x
−
y
1
+
x
y
\tan ^{-1} x-\tan ^{-1} y=\tan ^{-1} \frac{x-y}{1+x y}
tan
−
1
x
−
tan
−
1
y
=
tan
−
1
1
+
x
y
x
−
y
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(a)
\newline
(
1
1
1
)
\newline
(c) (
0
0
0
) :
\newline
(D)
\newline
(a)
\newline
(
0
0
0
)
\newline
⋮
\vdots
⋮
\newline
1
1
1
\newline
→
I
\rightarrow I
→
I
\newline
Nilai dari
cos
7
5
∘
−
sin
16
\cos 75^{\circ}-\sin 16
cos
7
5
∘
−
sin
16
aq
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Simplify the rational expression.
\newline
2
q
2
−
q
r
−
3
r
2
2
q
2
−
9
q
r
+
9
r
2
\frac{2q^{2}-qr-3r^{2}}{2q^{2}-9qr+9r^{2}}
2
q
2
−
9
q
r
+
9
r
2
2
q
2
−
q
r
−
3
r
2
\newline
(
q
+
r
)
/
(
k
+
3
r
)
(q+r)/(k+3r)
(
q
+
r
)
/
(
k
+
3
r
)
\newline
Cannot be simplified
\newline
(
a
+
r
)
(
q
−
3
r
)
(a+r)(q-3r)
(
a
+
r
)
(
q
−
3
r
)
\newline
(
q
+
r
)
(
q
+
3
r
)
(q+r)(q+3r)
(
q
+
r
)
(
q
+
3
r
)
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d)
lim
x
→
0
e
x
−
e
−
x
−
2
x
x
−
sin
x
\lim _{x \rightarrow 0} \frac{e^{x}-e^{-x}-2 x}{x-\sin x}
lim
x
→
0
x
−
s
i
n
x
e
x
−
e
−
x
−
2
x
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Find the measures of
∠
E
A
F
,
∠
D
A
E
\angle E A F, \angle D A E
∠
E
A
F
,
∠
D
A
E
, and
∠
C
A
D
\angle C A D
∠
C
A
D
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Date
\newline
15
/
4
/
2024
15 / 4 / 2024
15/4/2024
\newline
Q
1
1
1
) Simplify by rationalising the denominator:
\newline
(a)
5
+
6
5
−
6
\frac{5+\sqrt{6}}{5-\sqrt{6}}
5
−
6
5
+
6
\newline
(b)
7
+
3
5
7
−
3
5
\frac{7+3 \sqrt{5}}{7-3 \sqrt{5}}
7
−
3
5
7
+
3
5
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(c) Use your graph to solve these simultaneous equations.
\newline
y
=
x
+
2
and
y
=
−
3
x
−
2
y=x+2 \text { and } y=-3 x-2
y
=
x
+
2
and
y
=
−
3
x
−
2
\newline
x
=
y
=
\begin{array}{l} x= \\ y= \end{array}
x
=
y
=
\newline
\qquad
\newline
1112102
/
M
/
1
/
10
1112102 / \mathrm{M} / 1 / 10
1112102/
M
/1/10
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8
c
20
32
c
4
\frac{\sqrt{8 c^{20}}}{\sqrt{32 c^{4}}}
32
c
4
8
c
20
\newline
Which of the following is equivalent to the given expression for all
c
≠
0
c \neq 0
c
=
0
?
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Which theorem or postulate will prove
△
Z
W
R
≅
△
Y
X
R
\triangle Z W R \cong \triangle Y X R
△
Z
W
R
≅
△
Y
XR
\newline
Given:
△
Z
W
X
≅
△
Y
X
W
,
Z
W
‾
∥
Y
X
‾
\triangle Z W X \cong \triangle Y X W, \overline{Z W} \| \overline{Y X}
△
Z
W
X
≅
△
Y
X
W
,
Z
W
∥
Y
X
\newline
Prove:
△
Z
W
R
≅
△
Y
X
R
\triangle Z W R \cong \triangle Y X R
△
Z
W
R
≅
△
Y
XR
\newline
SSS
\newline
ASA
\newline
AAS
\newline
SAS
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If
3
x
−
5
y
7
x
−
4
y
=
3
4
\frac{3 x-5 y}{7 x-4 y}=\frac{3}{4}
7
x
−
4
y
3
x
−
5
y
=
4
3
, find the value of
x
y
\frac{x}{y}
y
x
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Urutan pecahan
3
1
5
;
342
%
;
25
8
;
3
,
28
3 \frac{1}{5} ; 342 \% ; \frac{25}{8} ; 3,28
3
5
1
;
342%
;
8
25
;
3
,
28
mulai dari yang terbesar adalah....
\newline
A.
25
8
;
3
1
5
;
3
,
28
;
342
%
\frac{25}{8} ; 3 \frac{1}{5} ; 3,28 ; 342 \%
8
25
;
3
5
1
;
3
,
28
;
342%
\newline
B.
3
1
5
;
25
8
;
3
,
28
;
342
%
3 \frac{1}{5} ; \frac{25}{8} ; 3,28 ; 342 \%
3
5
1
;
8
25
;
3
,
28
;
342%
\newline
C.
342
%
;
3
,
28
;
3
1
5
;
25
8
342 \% ; 3,28 ; 3 \frac{1}{5} ; \frac{25}{8}
342%
;
3
,
28
;
3
5
1
;
8
25
\newline
D.
342
%
;
3
,
28
;
25
8
;
3
1
5
342 \% ; 3,28 ; \frac{25}{8} ; 3 \frac{1}{5}
342%
;
3
,
28
;
8
25
;
3
5
1
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she can solve the following
\newline
Id the system of equations bi
\newline
{
f
(
x
)
=
3
x
f
(
x
)
=
2
x
−
3
\left\{\begin{array}{l} f(x)=\frac{3}{x} \\ f(x)=\sqrt{2 x-3} \end{array}\right.
{
f
(
x
)
=
x
3
f
(
x
)
=
2
x
−
3
\newline
it
$
10
\$ 10
$10
to the store to buy sna
Get tutor help
Rewrite as equivalent rational expressions with denominator
(
3
x
+
3
)
(
x
−
6
)
(
x
−
9
)
(3 x+3)(x-6)(x-9)
(
3
x
+
3
)
(
x
−
6
)
(
x
−
9
)
.
\newline
2
3
x
2
−
15
x
−
18
,
8
x
3
x
2
−
24
x
−
27
\frac{2}{3 x^{2}-15 x-18}, \frac{8 x}{3 x^{2}-24 x-27}
3
x
2
−
15
x
−
18
2
,
3
x
2
−
24
x
−
27
8
x
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x
sin
37
=
b
sin
7.5
\frac{x}{\sin 37}=\frac{b}{\sin 7.5}
s
i
n
37
x
=
s
i
n
7.5
b
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b. If
d
y
d
x
+
2
y
tan
x
=
Sin
x
\frac{d y}{d x}+2 y \tan x=\operatorname{Sin} x
d
x
d
y
+
2
y
tan
x
=
Sin
x
, and
y
=
0
y=0
y
=
0
for
x
=
π
/
3
x=\pi / 3
x
=
π
/3
, show that the maximum value of
y
y
y
is
1
/
8
1 / 8
1/8
Get tutor help
Q
1
1
1
.) Identify their terms and their factors.
\newline
b)
4
a
2
b
2
−
4
a
2
b
2
c
2
+
c
2
4
a
2
b
2
−
4
a
2
b
2
c
2
+
c
2
4
a
2
b
2
\text { b) } \begin{array}{l} 4 a^{2} b^{2}-4 a^{2} b^{2} c^{2}+c^{2} \\ 4 a^{2} b^{2}-4 a^{2} b^{2} c^{2}+c^{2} \\ 4 a^{2} b^{2} \end{array}
b)
4
a
2
b
2
−
4
a
2
b
2
c
2
+
c
2
4
a
2
b
2
−
4
a
2
b
2
c
2
+
c
2
4
a
2
b
2
Get tutor help
1
1
1
)
csc
x
−
1
csc
x
=
1
−
sin
x
\frac{\csc x-1}{\csc x}=1-\sin x
c
s
c
x
c
s
c
x
−
1
=
1
−
sin
x
Get tutor help
Which value of
v
v
v
makes
12
=
12
+
v
−
8
2
12=12+\frac{v-8}{2}
12
=
12
+
2
v
−
8
a true statement?
\newline
Choose
1
1
1
answer:
\newline
(A)
v
=
8
v=8
v
=
8
\newline
(B)
v
=
10
v=10
v
=
10
\newline
(C)
v
=
12
v=12
v
=
12
\newline
(D)
v
=
14
v=14
v
=
14
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Factorise
48
a
4
−
243
b
4
48a^4-243b^4
48
a
4
−
243
b
4
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Factorise
x
4
−
y
4
x^{4}-y^{4}
x
4
−
y
4
.
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x
3
+
9
x
2
x
3
\frac{x^3+9x^2}{x^3}
x
3
x
3
+
9
x
2
which expression is equivalent to the given expression for all
x
>
1
x>1
x
>
1
?
\newline
(A)
9
x
2
9x^2
9
x
2
\newline
(B)
1
+
9
x
2
1+9x^2
1
+
9
x
2
\newline
(C)
x
+
9
x
x+\frac{9}{x}
x
+
x
9
\newline
(D)
1
+
9
x
x
1+\frac{9x}{x}
1
+
x
9
x
Get tutor help
Simplify the expression below and write your final expression.
\newline
6
x
2
−
384
2
x
2
−
36
x
+
160
⋅
x
−
10
x
2
+
12
x
+
32
\frac{6 x^{2}-384}{2 x^{2}-36 x+160} \cdot \frac{x-10}{x^{2}+12 x+32}
2
x
2
−
36
x
+
160
6
x
2
−
384
⋅
x
2
+
12
x
+
32
x
−
10
Get tutor help
simplify the following
(
p
+
q
)
2
−
36
q
2
2
p
2
−
50
q
2
(p+q)^2 - \frac{36q^2}{2p^2} - 50q^2
(
p
+
q
)
2
−
2
p
2
36
q
2
−
50
q
2
Get tutor help
Factorise:
\newline
(
3
x
−
4
y
)
4
−
x
4
(3x - 4y)^4 - x^4
(
3
x
−
4
y
)
4
−
x
4
Get tutor help
Simplify
\newline
2
3
+
2
−
3
2^{3}+2^{-3}
2
3
+
2
−
3
Get tutor help
Simplify
6
2
3
+
6
−
4
3
6
+
2
+
2
6
2
+
3
\frac{6 \sqrt{2}}{\sqrt{3}+\sqrt{6}}-\frac{4 \sqrt{3}}{\sqrt{6}+\sqrt{2}}+\frac{2 \sqrt{6}}{\sqrt{2}+\sqrt{3}}
3
+
6
6
2
−
6
+
2
4
3
+
2
+
3
2
6
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Select all the expressions that are equivalent to
(
1
1
3
)
0
(11^3)^0
(
1
1
3
)
0
.
\newline
Multi-select Choices:
\newline
(A)
1
1
1
\newline
(B)
1
1
0
11^0
1
1
0
\newline
(C)
11
11
11
\newline
(D)
1
1
1
0
\frac{1}{11^0}
1
1
0
1
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Select all the expressions that are equivalent to
8
5
4
5
\frac{8^5}{4^5}
4
5
8
5
.
\newline
Multi-select Choices:
\newline
(A)
1
2
5
\frac{1}{2^5}
2
5
1
\newline
(B)
1
2
−
5
\frac{1}{2^{-5}}
2
−
5
1
\newline
(C)
2
2
2
\newline
(D)
2
5
2^5
2
5
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Select all the expressions that are equivalent to
8
7
×
8
6
8^7 \times 8^6
8
7
×
8
6
.
\newline
Multi-select Choices:
\newline
(A)
1
8
13
\frac{1}{8^{13}}
8
13
1
\newline
(B)
1
8
42
\frac{1}{8^{42}}
8
42
1
\newline
(C)
8
42
8^{42}
8
42
\newline
(D)
1
8
−
13
\frac{1}{8^{-13}}
8
−
13
1
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Select all the expressions that are equivalent to
1
2
2
/
2
2
12^2/2^2
1
2
2
/
2
2
.
\newline
Multi-select Choices:
\newline
(A)
1
/
6
−
2
1/6^{-2}
1/
6
−
2
\newline
(B)
6
2
6^2
6
2
\newline
(C)
6
6
6
\newline
(D)
6
0
6^0
6
0
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Select all the expressions that are equivalent to
(
5
4
)
2
(5^4)^2
(
5
4
)
2
.
\newline
Multi-select Choices:
\newline
(A)
1
5
8
\frac{1}{5^8}
5
8
1
\newline
(B)
5
6
5^6
5
6
\newline
(C)
5
8
5^8
5
8
\newline
(D)
1
5
6
\frac{1}{5^6}
5
6
1
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Select all the expressions that are equivalent to
6
5
3
5
\frac{6^5}{3^5}
3
5
6
5
.
\newline
Multi-select Choices:
\newline
(A)
1
2
5
\frac{1}{2^5}
2
5
1
\newline
(B)
1
2
−
5
\frac{1}{2^{-5}}
2
−
5
1
\newline
(C)
2
2
2
\newline
(D)
2
0
2^0
2
0
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Select all the expressions that are equivalent to
1
2
5
3
5
\frac{12^5}{3^5}
3
5
1
2
5
.
\newline
Multi-select Choices:
\newline
(A)
1
4
5
\frac{1}{4^5}
4
5
1
\newline
(B)
4
5
4^5
4
5
\newline
(C)
4
4
4
\newline
(D)
4
0
4^0
4
0
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Select all the expressions that are equivalent to
5
5
×
5
9
5^5 \times 5^9
5
5
×
5
9
.
\newline
Multi-select Choices:
\newline
(A)
1
5
14
\frac{1}{5^{14}}
5
14
1
\newline
(B)
5
14
5^{14}
5
14
\newline
(C)
5
45
5^{45}
5
45
\newline
(D)
1
5
−
14
\frac{1}{5^{-14}}
5
−
14
1
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Select all the expressions that are equivalent to
6
4
×
6
8
6^4 \times 6^8
6
4
×
6
8
.
\newline
Multi-select Choices:
\newline
(A)
1
6
32
\frac{1}{6^{32}}
6
32
1
\newline
(B)
6
12
6^{12}
6
12
\newline
(C)
6
32
6^{32}
6
32
\newline
(D)
1
6
−
12
\frac{1}{6^{-12}}
6
−
12
1
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Golden beikale
\newline
1
1
1
) Find both solutions to the equa
\newline
100
+
(
n
−
2
)
2
=
116
100+(n-2)^{2}=116
100
+
(
n
−
2
)
2
=
116
\newline
Explain or show your reaso
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x
−
2
(
x
+
2
)
(
x
+
1
)
\frac{x-2}{(x+2)(x+1)}
(
x
+
2
)
(
x
+
1
)
x
−
2
=
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Simplify using Factorization:
(
65
y
3
(
50
y
2
−
98
)
)
/
(
26
y
2
(
5
y
+
7
)
)
(65y^3(50y^2-98))/(26y^2(5y+7))
(
65
y
3
(
50
y
2
−
98
))
/
(
26
y
2
(
5
y
+
7
))
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Simplify using Factorization:
(
39
y
3
(
50
y
2
−
98
)
)
/
(
52
y
2
(
5
y
+
7
)
)
(39y^3(50y^2-98))/(52y^2(5y+7))
(
39
y
3
(
50
y
2
−
98
))
/
(
52
y
2
(
5
y
+
7
))
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If
j
k
=
4
5
\frac{j}{k}=\frac{4}{5}
k
j
=
5
4
, which of the following correctly expresses
k
k
k
in terms of
j
j
j
?
\newline
Choose
1
1
1
answer:
\newline
(A)
k
=
4
5
j
k=\frac{4}{5 j}
k
=
5
j
4
\newline
(B)
k
=
5
4
j
k=\frac{5}{4 j}
k
=
4
j
5
\newline
(C)
k
=
4
j
5
k=\frac{4 j}{5}
k
=
5
4
j
\newline
(D)
k
=
5
j
4
k=\frac{5 j}{4}
k
=
4
5
j
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Khan Academy
\newline
Get Al Tutoring
\newline
NEW
\newline
C
π
C^{\pi}
C
π
\newline
You might need: Calculator
\newline
Solve for
x
x
x
.
\newline
12
x
+
7
<
−
11
OR
5
x
−
8
>
40
12 x+7<-11 \text { OR } 5 x-8>40
12
x
+
7
<
−
11
OR
5
x
−
8
>
40
\newline
Choose
1
1
1
answer:
\newline
(A)
x
<
−
3
2
x<-\frac{3}{2}
x
<
−
2
3
or
x
>
48
5
x>\frac{48}{5}
x
>
5
48
\newline
B)
−
3
2
<
x
<
48
5
-\frac{3}{2}<x<\frac{48}{5}
−
2
3
<
x
<
5
48
\newline
(C)
x
>
3
2
x>\frac{3}{2}
x
>
2
3
or
x
<
48
5
x<\frac{48}{5}
x
<
5
48
\newline
D There are no solutions
\newline
(E) All values of
x
x
x
are solutions
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