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Math Problems
Algebra 2
Solve complex trigonomentric equations
x
2
+
12
x
+
32
=
(
x^{2}+12 x+32=(
x
2
+
12
x
+
32
=
(
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f
(
x
)
=
3
x
2
−
10
+
18
x
f(x)=3 x^{2}-10+18 x
f
(
x
)
=
3
x
2
−
10
+
18
x
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6
6
6
.
m
2
=
0
m^{2}=0
m
2
=
0
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5
5
5
)
x
4
−
13
x
2
+
40
x^{4}-13 x^{2}+40
x
4
−
13
x
2
+
40
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Solve by completing the square
\newline
5
x
2
+
3
x
−
4
=
0
5 x^{2}+3 x-4=0
5
x
2
+
3
x
−
4
=
0
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Solve by Finding the
\newline
9
9
9
)
x
2
=
49
x^{2}=49
x
2
=
49
\newline
x
=
7
x=7
x
=
7
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find the area of the figure bonded by
y
=
x
y=\sqrt{x}
y
=
x
,
y
=
x
−
2
y=x-2
y
=
x
−
2
,
y
=
0
y=0
y
=
0
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x
+
2
y
=
5
x + 2y = 5
x
+
2
y
=
5
3
x
−
y
=
1
3x - y = 1
3
x
−
y
=
1
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Prove that
cot
2
θ
+
tan
2
θ
=
2
sin
4
θ
\cot 2\theta + \tan 2\theta = \frac{2}{\sin 4\theta}
cot
2
θ
+
tan
2
θ
=
sin
4
θ
2
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2
x
2
−
11
x
−
6
=
0
2 x^{2}-11 x-6=0
2
x
2
−
11
x
−
6
=
0
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x
2
+
α
⋅
x
+
β
=
0
x^{2}+\alpha \cdot x+\beta=0
x
2
+
α
⋅
x
+
β
=
0
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x
2
−
9
x
+
10
=
32
x^{2}-9 x+10=32
x
2
−
9
x
+
10
=
32
\newline
D.
6
x
(
x
−
8
)
=
29
6 x(x-8)=29
6
x
(
x
−
8
)
=
29
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Se
x
=
10
x=10
x
=
10
, então
1
(
x
−
4
)
=
1(x-4)=
1
(
x
−
4
)
=
□
\square
□
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Resolva para
p
p
p
:
\newline
−
4
p
−
6
=
34
p
=
?
\begin{array}{l} -4 p-6=34 \\ p=? \end{array}
−
4
p
−
6
=
34
p
=
?
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Resolva para
c
c
c
:
\newline
3
c
+
8
=
32
c
=
?
\begin{array}{l} 3 c+8=32 \\ c=? \end{array}
3
c
+
8
=
32
c
=
?
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1
1
1
. 已知集合
A
=
{
x
∣
x
−
1
⩽
2
}
⋅
B
=
{
y
∣
−
y
2
+
3
y
>
0
}
A=\{x \mid \sqrt{x-1} \leqslant 2\} \cdot B=\left\{y \mid-y^{2}+3 y>0\right\}
A
=
{
x
∣
x
−
1
⩽
2
}
⋅
B
=
{
y
∣
−
y
2
+
3
y
>
0
}
, 则
A
∩
B
=
A \cap B=
A
∩
B
=
\newline
C
3
\mathrm{C}_{3}
C
3
\newline
A.
8
8
8
\newline
B.
(
0.3
)
(0.3)
(
0.3
)
\newline
C.
(
0
,
5
]
(0,5]
(
0
,
5
]
\newline
D.
[
1
,
3
)
[1,3)
[
1
,
3
)
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24
24
24
.
75
g
3
=
147
g
75 g^{3}=147 g
75
g
3
=
147
g
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Which values of
x
x
x
are solutions of the equation
5
x
=
x
+
13
6
\frac{5}{x}=\frac{x+13}{6}
x
5
=
6
x
+
13
? Select all that apply.
\newline
A.
□
\square
□
−
15
-15
−
15
\newline
B.
□
\square
□
15
15
15
\newline
C.
□
\square
□
2
2
2
\newline
D.
□
\square
□
−
3
-3
−
3
\newline
E.
□
\square
□
−
10
-10
−
10
\newline
F.
□
\square
□
5
x
=
x
+
13
6
\frac{5}{x}=\frac{x+13}{6}
x
5
=
6
x
+
13
1
1
1
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1
1
1
)
x
2
+
x
−
2
=
0
x^{2}+x-2=0
x
2
+
x
−
2
=
0
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x
2
+
16
x
+
5
=
0
x^{2}+16 x+5=0
x
2
+
16
x
+
5
=
0
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10
w
2
+
45
w
=
7
10 w^{2}+45 w=7
10
w
2
+
45
w
=
7
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Solve
9
2
x
−
1
=
27
9^{2 x-1}=27
9
2
x
−
1
=
27
for
x
x
x
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\#
5
5
5
Solve
−
2
x
2
+
12
x
−
16
=
0
-2 x^{2}+12 x-16=0
−
2
x
2
+
12
x
−
16
=
0
by graphing.
\newline
Solutions:
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Which graph can be used to solve the equation
2
x
−
1
=
x
+
2
2^{x}-1=x+2
2
x
−
1
=
x
+
2
?
\newline
My Progress
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Solve each equation by takin
\newline
5
5
5
)
m
2
−
8
=
88
m^{2}-8=88
m
2
−
8
=
88
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8
8
8
Solve the equation
5
2
x
−
6
(
5
x
)
+
5
=
0
5^{2 x}-6\left(5^{x}\right)+5=0
5
2
x
−
6
(
5
x
)
+
5
=
0
.
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4
e
−
2
x
=
17
4 e^{-2 x}=17
4
e
−
2
x
=
17
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5
−
x
+
3
=
5
x
−
3
=
y
5^{-x}+3=5^{x}-3=y
5
−
x
+
3
=
5
x
−
3
=
y
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y
′
+
e
x
+
y
=
0
y^{\prime}+e^{x+y}=0
y
′
+
e
x
+
y
=
0
,
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15
15
15
)
6
x
2
+
37
x
+
6
6 x^{2}+37 x+6
6
x
2
+
37
x
+
6
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You Can Solve It
\newline
3
2
x
+
1
+
4
(
3
x
)
−
15
=
0
3^{2 x+1}+4\left(3^{x}\right)-15=0
3
2
x
+
1
+
4
(
3
x
)
−
15
=
0
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the polynomi
\newline
2
x
2
+
9
x
−
5
=
0
2 x^{2}+9 x-5=0
2
x
2
+
9
x
−
5
=
0
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ring.
\newline
2.
5
x
2
+
5
x
−
30
=
0
\text { 2. } 5 x^{2}+5 x-30=0
2.
5
x
2
+
5
x
−
30
=
0
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6
6
6
)
7
x
2
+
10
x
=
1
7 x^{2}+10 x=1
7
x
2
+
10
x
=
1
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4
4
4
Find the value of
x
3
x^{3}
x
3
when
\newline
a
x
=
2
x=2
x
=
2
\newline
b
x
=
3
x=3
x
=
3
\newline
c
x
=
5
\quad x=5
x
=
5
\newline
x
=
10
x=10
x
=
10
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D^\(4 -
0
0
0
^
3
3
3
-
9
9
9
D^
2
2
2
-
11
11
11
D -
4
4
4
) * y =
0
0
0
\
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how to make
x
x
x
the subject in the equation:
y
=
4
sin
(
2
x
+
π
6
)
y=4\sin(2x+\frac{\pi}{6})
y
=
4
sin
(
2
x
+
6
π
)
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9
x
2
−
37
x
−
40
=
0
9 x^{2}-37 x-40=0
9
x
2
−
37
x
−
40
=
0
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Solve the following equations
\newline
(a)
5
+
5
+
5
+
5
+
5
=
5
y
\sqrt{5}+\sqrt{5}+\sqrt{5}+\sqrt{5}+\sqrt{5}=5^{y}
5
+
5
+
5
+
5
+
5
=
5
y
,
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x
2
+
16
=
80
x^{2}+16=80
x
2
+
16
=
80
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x
2
+
y
2
+
18
x
+
6
y
+
86
=
0
x^{2}+y^{2}+18 x+6 y+86=0
x
2
+
y
2
+
18
x
+
6
y
+
86
=
0
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\begin{tabular}{|l|c|c|c|c|c|}
\newline
\hline
x
x
x
&
−
9
-9
−
9
&
−
8
-8
−
8
&
−
7
-7
−
7
&
−
4
-4
−
4
&
−
3
-3
−
3
\\
\newline
\hline
f
(
x
)
f(x)
f
(
x
)
&
−
18
-18
−
18
&
0
0
0
&
12
12
12
&
12
12
12
&
0
0
0
\\
\newline
\hline
\newline
\end{tabular}
\newline
A.
f
(
x
)
=
3
x
2
+
25
x
−
36
f(x)=3 x^{2}+25 x-36
f
(
x
)
=
3
x
2
+
25
x
−
36
\newline
B.
f
(
x
)
=
−
3
x
2
−
33
x
−
72
f(x)=-3 x^{2}-33 x-72
f
(
x
)
=
−
3
x
2
−
33
x
−
72
\newline
C.
f
(
x
)
=
5
x
2
+
8
x
−
17
f(x)=5 x^{2}+8 x-17
f
(
x
)
=
5
x
2
+
8
x
−
17
\newline
D.
f
(
x
)
=
−
5
x
2
−
42
x
+
11
f(x)=-5 x^{2}-42 x+11
f
(
x
)
=
−
5
x
2
−
42
x
+
11
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Next Q.uestior
\newline
Question
8
8
8
\newline
Solve
12
=
s
4
12=\frac{s}{4}
12
=
4
s
\newline
s
=
s=
s
=
\newline
\qquad
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(
x
+
1
)
(
x
−
4
1
x
+
1
x
−
7
x
2
+
147
x
−
378
−
7
(
x
2
+
=
21
\begin{array}{l}\frac{(x+1)(x-4}{1 x+1 x} \\ -7 x^{2}+147 x-378 \\ -7\left(x^{2}+=21\right.\end{array}
1
x
+
1
x
(
x
+
1
)
(
x
−
4
−
7
x
2
+
147
x
−
378
−
7
(
x
2
+
=
21
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(d) Use your graph to find the solutions of
2
x
2
+
5
x
−
8
=
0
2 x^{2}+\frac{5}{x}-8=0
2
x
2
+
x
5
−
8
=
0
.
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Calculate
(
22
+
m
)
÷
4
(22 + m) \div 4
(
22
+
m
)
÷
4
for
m
=
56
m = 56
m
=
56
.
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Write the expression in simplest form.
\newline
20
x
2
−
30
x
20 x^{2}-30 x
20
x
2
−
30
x
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Solve for
x
x
x
.
\newline
4
x
2
+
52
=
−
8
x
4 x^{2}+52=-8 x
4
x
2
+
52
=
−
8
x
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Solve for
x
x
x
.
\newline
x
2
−
x
=
1
x^{2}-x=1
x
2
−
x
=
1
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If
4
x
2
−
12
x
=
40
4 x^{2}-12 x=40
4
x
2
−
12
x
=
40
, what are the possible values for
x
x
x
?
\newline
A)
−
2
-2
−
2
and
5
5
5
\newline
B)
−
5
-5
−
5
and
2
2
2
\newline
C)
−
5
-5
−
5
and
−
8
-8
−
8
\newline
D)
−
5
-5
−
5
and
−
16
-16
−
16
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