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Math Problems
Precalculus
Find the roots of factored polynomials
Solve for
x
\mathrm{x}
x
in simplest form.
\newline
10
=
1
2
(
x
−
4
)
10=\frac{1}{2}(x-4)
10
=
2
1
(
x
−
4
)
\newline
Answer:
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Solve for
x
x
x
:
\newline
−
(
7
x
+
4.8
)
=
−
x
−
(
x
−
1.3
)
-(7 x+4.8)=-x-(x-1.3)
−
(
7
x
+
4.8
)
=
−
x
−
(
x
−
1.3
)
\newline
Answer:
x
=
x=
x
=
Get tutor help
∠
1
\angle 1
∠1
and
∠
2
\angle 2
∠2
are vertical angles. If
m
∠
1
=
(
8
x
−
10
)
∘
\mathrm{m} \angle 1=(8 x-10)^{\circ}
m
∠1
=
(
8
x
−
10
)
∘
and
m
∠
2
=
(
6
x
+
26
)
∘
\mathrm{m} \angle 2=(6 x+26)^{\circ}
m
∠2
=
(
6
x
+
26
)
∘
, then find the value of
x
x
x
.
\newline
Answer:
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∠
1
\angle 1
∠1
and
∠
2
\angle 2
∠2
are vertical angles. If
m
∠
1
=
(
2
x
+
30
)
∘
\mathrm{m} \angle 1=(2 x+30)^{\circ}
m
∠1
=
(
2
x
+
30
)
∘
and
m
∠
2
=
(
3
x
−
10
)
∘
\mathrm{m} \angle 2=(3 x-10)^{\circ}
m
∠2
=
(
3
x
−
10
)
∘
, then find the measure of
∠
1
\angle 1
∠1
.
\newline
Answer:
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∠
1
\angle 1
∠1
and
∠
2
\angle 2
∠2
are vertical angles. If
m
∠
1
=
(
7
x
−
14
)
∘
\mathrm{m} \angle 1=(7 x-14)^{\circ}
m
∠1
=
(
7
x
−
14
)
∘
and
m
∠
2
=
(
4
x
+
25
)
∘
\mathrm{m} \angle 2=(4 x+25)^{\circ}
m
∠2
=
(
4
x
+
25
)
∘
, then find the value of
x
x
x
.
\newline
Answer:
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Solve for
b
\mathrm{b}
b
.
\newline
b
9
=
−
10
\frac{b}{9}=-10
9
b
=
−
10
\newline
Answer:
b
=
b=
b
=
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Solve for
t
\mathrm{t}
t
.
\newline
t
5
=
7
\frac{t}{5}=7
5
t
=
7
\newline
Answer:
t
=
t=
t
=
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Solve for
w
w
w
and simplify your answer.
\newline
−
3
2
w
=
−
12
-\frac{3}{2} w=-12
−
2
3
w
=
−
12
\newline
Answer:
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Solve for
u
u
u
and simplify your answer.
\newline
2
3
u
=
−
14
\frac{2}{3} u=-14
3
2
u
=
−
14
\newline
Answer:
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Solve for
b
b
b
.
\newline
−
b
2.5
−
2.8
=
−
4.04
-\frac{b}{2.5}-2.8=-4.04
−
2.5
b
−
2.8
=
−
4.04
\newline
Answer:
b
=
b=
b
=
Get tutor help
Solve for
s
s
s
and simplify your answer.
\newline
−
2
3
s
=
4
-\frac{2}{3} s=4
−
3
2
s
=
4
\newline
Answer:
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Solve for
y
y
y
. Express your answer as a proper or improper fraction in simplest terms.
\newline
1
12
y
−
1
2
=
−
1
2
\frac{1}{12} y-\frac{1}{2}=-\frac{1}{2}
12
1
y
−
2
1
=
−
2
1
\newline
Answer:
y
=
y=
y
=
Get tutor help
Solve for
x
x
x
. Express your answer as a proper or improper fraction in simplest terms.
\newline
−
2
7
=
−
1
6
x
−
5
7
-\frac{2}{7}=-\frac{1}{6} x-\frac{5}{7}
−
7
2
=
−
6
1
x
−
7
5
\newline
Answer:
x
=
x=
x
=
Get tutor help
Solve for
b
b
b
. Express your answer as a proper or improper fraction in simplest terms.
\newline
2
3
=
−
1
10
b
+
2
3
\frac{2}{3}=-\frac{1}{10} b+\frac{2}{3}
3
2
=
−
10
1
b
+
3
2
\newline
Answer:
b
=
b=
b
=
Get tutor help
Solve for
b
b
b
. Express your answer as a proper or improper fraction in simplest terms.
\newline
−
1
2
b
+
5
6
=
−
1
8
-\frac{1}{2} b+\frac{5}{6}=-\frac{1}{8}
−
2
1
b
+
6
5
=
−
8
1
\newline
Answer:
b
=
b=
b
=
Get tutor help
Solve for
a
a
a
. Express your answer as a proper or improper fraction in simplest terms.
\newline
−
7
8
=
7
11
a
+
1
8
-\frac{7}{8}=\frac{7}{11} a+\frac{1}{8}
−
8
7
=
11
7
a
+
8
1
\newline
Answer:
a
=
a=
a
=
Get tutor help
Find an expression which represents the sum of
(
−
10
x
−
1
)
(-10 x-1)
(
−
10
x
−
1
)
and
(
−
8
x
+
5
)
(-8 x+5)
(
−
8
x
+
5
)
in simplest terms.
\newline
Answer:
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Find an expression which represents the difference when
(
−
10
x
+
y
)
(-10 x+y)
(
−
10
x
+
y
)
is subtracted from
(
7
x
−
7
y
)
(7 x-7 y)
(
7
x
−
7
y
)
in simplest terms.
\newline
Answer:
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Find an expression which represents the difference when
(
x
−
7
)
(x-7)
(
x
−
7
)
is subtracted from
(
x
+
8
)
(x+8)
(
x
+
8
)
in simplest terms.
\newline
Answer:
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For the function
f
(
x
)
=
2
x
2
+
2
x
+
3
f(x)=2 x^{2}+2 x+3
f
(
x
)
=
2
x
2
+
2
x
+
3
, find the slope of the secant line between
x
=
−
4
x=-4
x
=
−
4
and
x
=
1
x=1
x
=
1
.
\newline
Answer:
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For the function
f
(
x
)
=
x
2
+
1
f(x)=x^{2}+1
f
(
x
)
=
x
2
+
1
, find the slope of the secant line between
x
=
−
4
x=-4
x
=
−
4
and
x
=
−
2
x=-2
x
=
−
2
.
\newline
Answer:
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For the function
f
(
x
)
=
x
2
+
5
x
+
4
f(x)=x^{2}+5 x+4
f
(
x
)
=
x
2
+
5
x
+
4
, find the slope of the secant line between
x
=
−
8
x=-8
x
=
−
8
and
x
=
1
x=1
x
=
1
.
\newline
Answer:
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For the function
f
(
x
)
=
x
2
+
3
f(x)=x^{2}+3
f
(
x
)
=
x
2
+
3
, find the slope of the secant line between
x
=
−
5
x=-5
x
=
−
5
and
x
=
−
3
x=-3
x
=
−
3
.
\newline
Answer:
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For the function
f
(
x
)
=
−
x
2
−
5
x
+
9
f(x)=-x^{2}-5 x+9
f
(
x
)
=
−
x
2
−
5
x
+
9
, find the slope of the secant line between
x
=
−
2
x=-2
x
=
−
2
and
x
=
2
x=2
x
=
2
.
\newline
Answer:
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For the function
f
(
x
)
=
−
x
2
+
4
x
−
8
f(x)=-x^{2}+4 x-8
f
(
x
)
=
−
x
2
+
4
x
−
8
, find the slope of the secant line between
x
=
2
x=2
x
=
2
and
x
=
5
x=5
x
=
5
.
\newline
Answer:
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For the function
f
(
x
)
=
−
x
2
+
2
x
−
6
f(x)=-x^{2}+2 x-6
f
(
x
)
=
−
x
2
+
2
x
−
6
, find the slope of the secant line between
x
=
−
4
x=-4
x
=
−
4
and
x
=
3
x=3
x
=
3
.
\newline
Answer:
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For the function
f
(
x
)
=
x
2
+
4
x
+
4
f(x)=x^{2}+4 x+4
f
(
x
)
=
x
2
+
4
x
+
4
, find the slope of the secant line between
x
=
−
7
x=-7
x
=
−
7
and
x
=
−
5
x=-5
x
=
−
5
.
\newline
Answer:
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For the function
f
(
x
)
=
2
x
2
−
2
x
−
1
f(x)=2 x^{2}-2 x-1
f
(
x
)
=
2
x
2
−
2
x
−
1
, find the slope of the secant line between
x
=
−
2
x=-2
x
=
−
2
and
x
=
3
x=3
x
=
3
.
\newline
Answer:
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For the function
f
(
x
)
=
x
2
+
4
x
−
8
f(x)=x^{2}+4 x-8
f
(
x
)
=
x
2
+
4
x
−
8
, find the slope of the secant line between
x
=
−
6
x=-6
x
=
−
6
and
x
=
−
1
x=-1
x
=
−
1
.
\newline
Answer:
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For the function
f
(
x
)
=
−
x
2
−
2
x
+
5
f(x)=-x^{2}-2 x+5
f
(
x
)
=
−
x
2
−
2
x
+
5
, find the slope of the secant line between
x
=
−
4
x=-4
x
=
−
4
and
x
=
2
x=2
x
=
2
.
\newline
Answer:
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Given that
y
=
2
u
3
+
1
y=2 u^{3}+1
y
=
2
u
3
+
1
, find
d
d
u
(
4
u
5
−
5
sin
y
)
\frac{d}{d u}\left(4 u^{5}-5 \sin y\right)
d
u
d
(
4
u
5
−
5
sin
y
)
in terms of only
u
u
u
.
\newline
Answer:
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Given that
y
=
2
x
4
+
1
y=2 x^{4}+1
y
=
2
x
4
+
1
, find
d
d
x
(
y
3
+
4
cos
x
)
\frac{d}{d x}\left(y^{3}+4 \cos x\right)
d
x
d
(
y
3
+
4
cos
x
)
in terms of only
x
x
x
.
\newline
Answer:
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Given that
x
=
4
v
5
+
3
x=4 v^{5}+3
x
=
4
v
5
+
3
, find
d
d
v
(
3
v
4
−
5
sin
x
)
\frac{d}{d v}\left(3 v^{4}-5 \sin x\right)
d
v
d
(
3
v
4
−
5
sin
x
)
in terms of only
v
v
v
.
\newline
Answer:
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Given that
y
=
4
x
2
−
5
y=4 x^{2}-5
y
=
4
x
2
−
5
, find
d
d
x
(
3
x
5
−
3
cos
y
)
\frac{d}{d x}\left(3 x^{5}-3 \cos y\right)
d
x
d
(
3
x
5
−
3
cos
y
)
in terms of only
x
x
x
.
\newline
Answer:
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Given that
x
=
3
y
4
+
1
x=3 y^{4}+1
x
=
3
y
4
+
1
, find
d
d
y
(
5
y
2
−
4
cos
x
)
\frac{d}{d y}\left(5 y^{2}-4 \cos x\right)
d
y
d
(
5
y
2
−
4
cos
x
)
in terms of only
y
y
y
.
\newline
Answer:
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Given that
x
=
3
y
3
+
1
x=3 y^{3}+1
x
=
3
y
3
+
1
, find
d
d
y
(
2
x
2
−
2
cos
y
)
\frac{d}{d y}\left(2 x^{2}-2 \cos y\right)
d
y
d
(
2
x
2
−
2
cos
y
)
in terms of only
y
y
y
.
\newline
Answer:
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Given that
x
=
3
v
3
+
2
x=3 v^{3}+2
x
=
3
v
3
+
2
, find
d
d
v
(
4
v
2
+
5
cos
x
)
\frac{d}{d v}\left(4 v^{2}+5 \cos x\right)
d
v
d
(
4
v
2
+
5
cos
x
)
in terms of only
v
v
v
.
\newline
Answer:
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Given that
y
=
2
v
5
−
2
y=2 v^{5}-2
y
=
2
v
5
−
2
, find
d
d
v
(
v
4
+
4
sin
y
)
\frac{d}{d v}\left(v^{4}+4 \sin y\right)
d
v
d
(
v
4
+
4
sin
y
)
in terms of only
v
v
v
.
\newline
Answer:
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Given that
u
=
3
v
4
+
1
u=3 v^{4}+1
u
=
3
v
4
+
1
, find
d
d
v
(
u
2
−
cos
v
)
\frac{d}{d v}\left(u^{2}-\cos v\right)
d
v
d
(
u
2
−
cos
v
)
in terms of only
v
v
v
.
\newline
Answer:
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Given that
v
=
2
y
4
−
5
v=2 y^{4}-5
v
=
2
y
4
−
5
, find
d
d
y
(
4
v
2
−
cos
y
)
\frac{d}{d y}\left(4 v^{2}-\cos y\right)
d
y
d
(
4
v
2
−
cos
y
)
in terms of only
y
y
y
.
\newline
Answer:
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Find the average value of the function
f
(
x
)
=
12
9
−
x
f(x)=\frac{12}{9-x}
f
(
x
)
=
9
−
x
12
from
x
=
1
x=1
x
=
1
to
x
=
7
x=7
x
=
7
. Express your answer as a constant times
ln
2
\ln 2
ln
2
.
\newline
Answer:
□
ln
2
\square\ln 2
□
ln
2
Get tutor help
Find the average value of the function
f
(
x
)
=
1
x
−
1
f(x)=\frac{1}{x-1}
f
(
x
)
=
x
−
1
1
from
x
=
5
x=5
x
=
5
to
x
=
9
x=9
x
=
9
. Express your answer as a constant times
ln
2
\ln 2
ln
2
.
\newline
Answer:
□
ln
2
\square\ln 2
□
ln
2
Get tutor help
Find the average value of the function
f
(
x
)
=
8
x
−
5
f(x)=\frac{8}{x-5}
f
(
x
)
=
x
−
5
8
from
x
=
1
x=1
x
=
1
to
x
=
3
x=3
x
=
3
. Express your answer as a constant times
ln
2
\ln 2
ln
2
.
\newline
Answer:
□
ln
2
\square\ln 2
□
ln
2
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Find the zeros of the function. Enter the solutions from least to greatest.
\newline
f
(
x
)
=
(
x
−
5
)
(
5
x
+
2
)
f(x)=(x-5)(5x+2)
f
(
x
)
=
(
x
−
5
)
(
5
x
+
2
)
\newline
lesser
x
=
□
x=\square
x
=
□
\newline
greater
x
=
□
x=\square
x
=
□
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The derivative of the function
f
f
f
is defined by
f
′
(
x
)
=
(
x
2
+
1
)
cos
(
2
x
+
1
)
f^{\prime}(x)=\left(x^{2}+1\right) \cos (2 x+1)
f
′
(
x
)
=
(
x
2
+
1
)
cos
(
2
x
+
1
)
. If
f
(
1
)
=
1
f(1)=1
f
(
1
)
=
1
, then use a calculator to find the value of
f
(
6
)
f(6)
f
(
6
)
to the nearest thousandth.
\newline
Answer:
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The derivative of the function
f
f
f
is defined by
f
′
(
x
)
=
(
x
2
−
3
x
)
cos
(
2
x
)
f^{\prime}(x)=\left(x^{2}-3 x\right) \cos (2 x)
f
′
(
x
)
=
(
x
2
−
3
x
)
cos
(
2
x
)
. If
f
(
0
)
=
5
f(0)=5
f
(
0
)
=
5
, then use a calculator to find the value of
f
(
6
)
f(6)
f
(
6
)
to the nearest thousandth.
\newline
Answer:
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The derivative of the function
f
f
f
is defined by
f
′
(
x
)
=
(
x
2
+
4
)
cos
(
x
+
2
)
f^{\prime}(x)=\left(x^{2}+4\right) \cos (x+2)
f
′
(
x
)
=
(
x
2
+
4
)
cos
(
x
+
2
)
. If
f
(
4
)
=
8
f(4)=8
f
(
4
)
=
8
, then use a calculator to find the value of
f
(
−
2
)
f(-2)
f
(
−
2
)
to the nearest thousandth.
\newline
Answer:
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The derivative of the function
f
f
f
is defined by
f
′
(
x
)
=
(
x
3
+
3
x
)
cos
(
2
x
+
5
)
f^{\prime}(x)=\left(x^{3}+3 x\right) \cos (2 x+5)
f
′
(
x
)
=
(
x
3
+
3
x
)
cos
(
2
x
+
5
)
. If
f
(
−
3
)
=
−
5
f(-3)=-5
f
(
−
3
)
=
−
5
, then use a calculator to find the value of
f
(
4
)
f(4)
f
(
4
)
to the nearest thousandth.
\newline
Answer:
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The derivative of the function
f
f
f
is defined by
f
′
(
x
)
=
(
x
3
+
1
)
cos
(
3
x
)
f^{\prime}(x)=\left(x^{3}+1\right) \cos (3 x)
f
′
(
x
)
=
(
x
3
+
1
)
cos
(
3
x
)
. If
f
(
5
)
=
−
8
f(5)=-8
f
(
5
)
=
−
8
, then use a calculator to find the value of
f
(
0
)
f(0)
f
(
0
)
to the nearest thousandth.
\newline
Answer:
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The derivative of the function
f
f
f
is defined by
f
′
(
x
)
=
(
x
2
−
x
)
cos
(
x
)
f^{\prime}(x)=\left(x^{2}-x\right) \cos (x)
f
′
(
x
)
=
(
x
2
−
x
)
cos
(
x
)
. If
f
(
−
3
)
=
−
6
f(-3)=-6
f
(
−
3
)
=
−
6
, then use a calculator to find the value of
f
(
6
)
f(6)
f
(
6
)
to the nearest thousandth.
\newline
Answer:
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