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Math Problems
Algebra 1
Solve advanced linear inequalities
Select the expressions that are equivalent to
2
(
m
+
7
)
2(m + 7)
2
(
m
+
7
)
.
\newline
Multi-select Choices:
\newline
(A)
9
m
9m
9
m
\newline
(B)
7
(
m
+
2
)
7(m + 2)
7
(
m
+
2
)
\newline
(C)
(
m
×
7
)
2
(m \times 7)2
(
m
×
7
)
2
\newline
(D)
2
m
+
14
2m + 14
2
m
+
14
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h
(
x
)
=
0.000371
(
x
2
−
1
,
280
x
)
+
152
h(x)=0.000371(x^2-1{,}280x)+152
h
(
x
)
=
0.000371
(
x
2
−
1
,
280
x
)
+
152
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x
2
−
36
>
0
x^2-36>0
x
2
−
36
>
0
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Solve for
x
x
x
.
\newline
3
x
+
10
=
25
3x+10 = 25
3
x
+
10
=
25
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Solve
2
+
log
3
(
7
x
)
=
4
2+\log _{3}(7 x)=4
2
+
lo
g
3
(
7
x
)
=
4
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Integrate
ln
[
(
x
)
(
1
−
x
)
]
\ln[(x)(1-x)]
ln
[(
x
)
(
1
−
x
)]
in the limits
0
0
0
to
1
1
1
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5
(
6
+
3
r
)
+
7
≥
127
5(6+3 r)+7 \geq 127
5
(
6
+
3
r
)
+
7
≥
127
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−
2
x
<
10
x
>
10
−
2
\begin{aligned} -2 x & <10 \\ x & >\frac{10}{-2} \end{aligned}
−
2
x
x
<
10
>
−
2
10
\newline
321
321
321
.
3
p
−
7
2
≤
0
\frac{3 p-7}{2} \leq 0
2
3
p
−
7
≤
0
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The inequality
6
−
2
3
6-\frac{2}{3}
6
−
3
2
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8
x
−
2
(
3
x
−
8
)
=
24
8x-2(3x-8)=24
8
x
−
2
(
3
x
−
8
)
=
24
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Solve for
a
a
a
.
\newline
a
−
1
5
>
1
a - \frac{1}{5} > 1
a
−
5
1
>
1
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Solve for
b
b
b
.
\newline
b
+
18
16
>
1
b + \frac{18}{16} > 1
b
+
16
18
>
1
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Solve for
r
r
r
.
3
(
r
+
7
)
+
7
<
1
3(r + 7) + 7 < 1
3
(
r
+
7
)
+
7
<
1
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Solve for
y
y
y
.
\newline
4
(
y
+
2
)
≥
12
4(y + 2) \geq 12
4
(
y
+
2
)
≥
12
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Solve for
d
d
d
.
\newline
2
(
d
−
16
)
−
6
≥
2
2(d - 16) - 6 \geq 2
2
(
d
−
16
)
−
6
≥
2
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Solve for
b
b
b
.
\newline
3
(
b
−
17
)
+
14
>
8
3(b - 17) + 14 > 8
3
(
b
−
17
)
+
14
>
8
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Solve for
s
s
s
.
\newline
7
(
s
−
19
)
+
19
≥
5
7(s - 19) + 19 \geq 5
7
(
s
−
19
)
+
19
≥
5
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−
2
(
3
n
+
1
)
−
3
n
>
25
-2(3 n+1)-3 n>25
−
2
(
3
n
+
1
)
−
3
n
>
25
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x
−
2
[
x
−
3
(
x
+
4
)
−
6
]
=
1
x-2[x-3(x+4)-6]=1
x
−
2
[
x
−
3
(
x
+
4
)
−
6
]
=
1
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9
x
7
−
x
<
2
\frac{9 x}{7}-x<2
7
9
x
−
x
<
2
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egin{equation}-\frac{
6
6
6
}{x} < x
−
1
-1
−
1
egin{equation}
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S. Solve
−
129
≤
3
(
−
6
v
+
5
)
\begin{array}{c}\text { S. Solve } \\ -129 \leq 3(-6 v+5)\end{array}
S. Solve
−
129
≤
3
(
−
6
v
+
5
)
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x
2
(
x
−
2
)
(
x
+
3
)
2
<
0
x^{2}(x-2)(x+3)^{2}<0
x
2
(
x
−
2
)
(
x
+
3
)
2
<
0
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3
3
3
.
1
+
2
(
x
−
1
)
>
−
11
1+2(x-1)>-11
1
+
2
(
x
−
1
)
>
−
11
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x
2
(
x
−
2
)
(
x
+
3
)
2
<
0
x^2(x-2)(x+3)^2<0
x
2
(
x
−
2
)
(
x
+
3
)
2
<
0
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Solve the inequality,
4
−
3
(
2
−
x
)
<
10
+
x
4-3(2-x)<10+x
4
−
3
(
2
−
x
)
<
10
+
x
.
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Solve for
z
\mathrm{z}
z
.
\newline
5
z
+
8
=
3
z
+
16
5z+8=3z+16
5
z
+
8
=
3
z
+
16
\newline
z
=
□
z = \square
z
=
□
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56
<
8
(
x
+
4
)
<
120
56<8(x+4)<120
56
<
8
(
x
+
4
)
<
120
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x
−
2
x
+
3
>
−
2
\frac{x-2}{x+3}>-2
x
+
3
x
−
2
>
−
2
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Prove that
1
2
−
1
+
2
3
+
1
⇒
2
+
3
\frac{1}{\sqrt{2}-1}+\frac{2}{\sqrt{3}+1} \Rightarrow \sqrt{2}+\sqrt{3}
2
−
1
1
+
3
+
1
2
⇒
2
+
3
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ities
x
2
−
5
≤
3
(
x
+
2
)
<
9
\frac{x}{2}-5 \leq 3(x+2)<9
2
x
−
5
≤
3
(
x
+
2
)
<
9
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2
p
+
5
>
2
(
p
−
3
)
2 p+5>2(p-3)
2
p
+
5
>
2
(
p
−
3
)
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1
<
e
x
−
1
ln
(
x
−
1
)
<
(
x
+
1
)
=
x
1<\frac{e^{x}-1}{\ln (x-1)}<(x+1)=x
1
<
l
n
(
x
−
1
)
e
x
−
1
<
(
x
+
1
)
=
x
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Exercise: G.M.V.T.
\newline
1
<
e
x
−
1
ln
(
x
+
1
)
<
(
x
+
1
)
e
x
1<\frac{e^{x}-1}{\ln (x+1)}<(x+1) e^{x}
1
<
ln
(
x
+
1
)
e
x
−
1
<
(
x
+
1
)
e
x
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Exercise: G.M.L.T.
\newline
1
<
e
x
−
1
ln
(
x
+
1
)
<
(
x
+
1
)
e
x
1<\frac{e^{x}-1}{\ln (x+1)}<(x+1) e^{x}
1
<
ln
(
x
+
1
)
e
x
−
1
<
(
x
+
1
)
e
x
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−
5
≤
3
n
−
2
4
≤
4
-5 \leq \frac{3 n-2}{4} \leq 4
−
5
≤
4
3
n
−
2
≤
4
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10
÷
43
10 \div 43
10
÷
43
\newline
<
<
<
Notes
\newline
15
15
15
−
6
-6
−
6
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1037
1037
1037
\newline
< Notes
\newline
15
−
6
15-6
15
−
6
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What is the sum of the solutions to the equation
(
t
+
3
)
(
t
−
357
)
=
0
(t+3)(t-357)=0
(
t
+
3
)
(
t
−
357
)
=
0
?
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17
17
17
)
n
+
5
−
16
=
−
1
\frac{n+5}{-16}=-1
−
16
n
+
5
=
−
1
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Solve for
q
q
q
.
\newline
1
<
q
−
15
−
18
1 < \frac{q - 15}{-18}
1
<
−
18
q
−
15
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solve for
x
x
x
if
log
10
(
3
x
+
1
)
+
log
10
(
2
x
−
5
)
=
0
\log_{10}(3x+1) + \log_{10}(2x-5)=0
lo
g
10
(
3
x
+
1
)
+
lo
g
10
(
2
x
−
5
)
=
0
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y
−
4
=
7
(
x
−
6
)
y-4=7(x-6)
y
−
4
=
7
(
x
−
6
)
ordered pairs
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Solve for
c
c
c
.
\newline
−
16
<
c
−
14
<
1
-16 < c - 14 < 1
−
16
<
c
−
14
<
1
\newline
Write your answer as a compound inequality with integers.
\newline
Choices:
\newline
(A)
c
<
−
2
c < -2
c
<
−
2
or
c
≥
15
c \geq 15
c
≥
15
\newline
(B)
−
2
<
c
<
15
-2 < c < 15
−
2
<
c
<
15
\newline
(C)
−
2
<
c
≤
15
-2 < c \leq 15
−
2
<
c
≤
15
\newline
(D)
c
<
−
2
c < -2
c
<
−
2
or
c
>
15
c > 15
c
>
15
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Solve for
p
p
p
.
\newline
0
<
p
+
16
<
13
0 < p + 16 < 13
0
<
p
+
16
<
13
\newline
Write your answer as a compound inequality with integers.
\newline
Choices:
\newline
(A)
p
<
−
16
p < -16
p
<
−
16
or
p
≥
−
3
p \geq -3
p
≥
−
3
\newline
(B)
−
16
<
p
≤
−
3
-16 < p \leq -3
−
16
<
p
≤
−
3
\newline
(C)
−
16
<
p
<
−
3
-16 < p < -3
−
16
<
p
<
−
3
\newline
(D)
p
<
−
16
p < -16
p
<
−
16
or
p
>
−
3
p > -3
p
>
−
3
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Solve for
k
k
k
.
4
<
k
−
9
−
3
4 < \frac{k - 9}{-3}
4
<
−
3
k
−
9
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Solve for
u
u
u
.
\newline
−
14
≤
2
(
u
+
5
)
−
10
-14 \leq 2(u + 5) - 10
−
14
≤
2
(
u
+
5
)
−
10
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Solve for
d
d
d
.
\newline
10
≤
5
(
d
−
17
)
10 \leq 5(d - 17)
10
≤
5
(
d
−
17
)
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Solve for
y
y
y
.
\newline
−
2
(
y
−
11
)
≤
2
-2(y - 11) \leq 2
−
2
(
y
−
11
)
≤
2
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Solve for
h
h
h
.
\newline
−
5
≥
5
(
h
−
10
)
-5 \geq 5(h - 10)
−
5
≥
5
(
h
−
10
)
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1
2
3
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