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Math Problems
Algebra 2
Find the vertex of the transformed function
The equation
(
x
3
+
y
)
⋅
x
5
=
x
20
(x^{3+y}) \cdot x^5 = x^{20}
(
x
3
+
y
)
⋅
x
5
=
x
20
is true for all values of
x
x
x
. What is the value of
y
y
y
?
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Solve for
a
a
a
:
1
1
a
=
7
11
11^a=\frac{7}{\sqrt{11}}
1
1
a
=
11
7
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A function
f
(
x
)
f(x)
f
(
x
)
is defined by
f
(
x
)
=
2
x
2
+
7
f(x)=2x^2+7
f
(
x
)
=
2
x
2
+
7
. What is the value of
2
f
(
x
)
−
3
2f(x)-3
2
f
(
x
)
−
3
?
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f
(
x
)
=
−
x
2
+
5
f(x)=-x^2+ 5
f
(
x
)
=
−
x
2
+
5
\newline
g
(
x
)
=
3
g(x)=3
g
(
x
)
=
3
\newline
The functions
f
f
f
and
g
g
g
are defined above. What is the value of
g
(
−
1
)
g(-1)
g
(
−
1
)
?
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The function
h
h
h
is defined by
h
(
x
)
=
k
x
2
−
7
h(x)=kx^{2}-7
h
(
x
)
=
k
x
2
−
7
. If
h
(
5
)
=
−
57
h(5)=-57
h
(
5
)
=
−
57
, what is the value of
k
k
k
?
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Which type of conic section is defined by the equation
4
x
2
−
16
y
2
+
16
x
−
160
y
−
448
=
0
4 x^{2}-16 y^{2}+16 x-160 y-448=0
4
x
2
−
16
y
2
+
16
x
−
160
y
−
448
=
0
?
\newline
This is an equation of
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The function
f
f
f
is defined as
f
(
x
)
=
x
2
−
1
f(x)=x^{2}-1
f
(
x
)
=
x
2
−
1
.
\newline
What is the
x
x
x
-coordinate of the point on the function's graph that is closest to the origin?
\newline
Choose all answers that apply:
\newline
A
−
3
3
-\frac{\sqrt{3}}{3}
−
3
3
\newline
B
−
2
2
-\frac{\sqrt{2}}{2}
−
2
2
\newline
c.
0
0
0
\newline
D
3
3
\frac{\sqrt{3}}{3}
3
3
\newline
E
2
2
\frac{\sqrt{2}}{2}
2
2
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What is the inverse of the function
g
(
x
)
=
x
3
8
+
16
g(x)=\frac{x^{3}}{8}+16
g
(
x
)
=
8
x
3
+
16
?
\newline
A)
g
−
1
(
x
)
=
x
−
16
3
2
g^{-1}(x)=\frac{\sqrt[3]{x-16}}{2}
g
−
1
(
x
)
=
2
3
x
−
16
\newline
B)
g
−
1
(
x
)
=
2
x
3
−
16
g^{-1}(x)=2\sqrt[3]{x}-16
g
−
1
(
x
)
=
2
3
x
−
16
\newline
C)
g
−
1
(
x
)
=
2
x
+
16
3
g^{-1}(x)=2\sqrt[3]{x+16}
g
−
1
(
x
)
=
2
3
x
+
16
]
\newline
D)
g
−
1
(
x
)
=
2
x
−
16
3
g^{-1}(x)=2\sqrt[3]{x-16}
g
−
1
(
x
)
=
2
3
x
−
16
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The polynomial function
f
f
f
is defined as
f
(
m
)
=
(
m
3
−
m
2
−
17
m
−
15
)
(
m
+
1
)
f(m)=\left(m^{3}-m^{2}-17 m-15\right)(m+1)
f
(
m
)
=
(
m
3
−
m
2
−
17
m
−
15
)
(
m
+
1
)
. When
f
(
m
)
f(m)
f
(
m
)
is divided by
(
m
+
1
)
(m+1)
(
m
+
1
)
, what is the remainder?
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Given the function
f
(
x
)
=
1
−
3
x
2
f(x)=1-3 x^{2}
f
(
x
)
=
1
−
3
x
2
, then what is
f
(
x
)
−
3
f(x)-3
f
(
x
)
−
3
as a simplified polynomial?
\newline
Answer:
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Given the equation
n
=
sin
(
θ
+
α
)
sin
θ
n=\frac{\sin (\theta+\alpha)}{\sin \theta}
n
=
s
i
n
θ
s
i
n
(
θ
+
α
)
, where
α
=
4
5
∘
\alpha=45^{\circ}
α
=
4
5
∘
, then an equivalent expression for
n
n
n
is
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The function
f
f
f
is defined by
f
(
x
)
=
a
x
2
+
b
x
+
c
f(x)=a x^{2}+b x+c
f
(
x
)
=
a
x
2
+
b
x
+
c
, where
a
,
b
a, b
a
,
b
, and
c
c
c
are constants and
1
<
a
<
4
1<a<4
1
<
a
<
4
. The graph of
y
=
f
(
x
)
y=f(x)
y
=
f
(
x
)
in the
x
y
x y
x
y
-plane passes through points.
(
11
,
0
)
(11,0)
(
11
,
0
)
and
(
−
2
,
0
)
(-2,0)
(
−
2
,
0
)
. If
a
a
a
is an integer, what could be the value of
a
+
b
a+b
a
+
b
?
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If
y
=
−
1
2
x
2
−
9
y=-\frac{1}{2}x^{2}-9
y
=
−
2
1
x
2
−
9
is graphed in the
x
y
xy
x
y
-plane, which of the following characteristics of the graph are displayed as a constant or coefficient in the equation?
\newline
I.
x
x
x
-intercept(s)
\newline
II.
y
y
y
-intercept
\newline
III.
y
y
y
-coordinate of the vertex
\newline
Choose
1
1
1
answer:
\newline
(A) II only
\newline
(B) III only
\newline
(C) I and II only
\newline
(D) II and III only
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The functions
s
(
x
)
s(x)
s
(
x
)
and
t
(
x
)
t(x)
t
(
x
)
are differentiable. The function
u
(
x
)
u(x)
u
(
x
)
is defined as:
u
(
x
)
=
s
(
x
)
t
(
x
)
u(x)= \frac{s(x)}{t(x)}
u
(
x
)
=
t
(
x
)
s
(
x
)
If
s
(
6
)
=
7
s(6)= 7
s
(
6
)
=
7
,
s
′
(
6
)
=
5
s'(6)= 5
s
′
(
6
)
=
5
,
t
(
6
)
=
9
t(6)= 9
t
(
6
)
=
9
, and
t
′
(
6
)
=
2
t'(6)= 2
t
′
(
6
)
=
2
, what is
u
′
(
6
)
u'(6)
u
′
(
6
)
? Simplify any fractions.
u
′
(
6
)
=
u'(6)=
u
′
(
6
)
=
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Find
d
2
t
d
s
2
\frac{d^2 t}{ds^2}
d
s
2
d
2
t
if
t
=
2
s
(
1
−
4
s
)
2
t = 2s(1 - 4s)^2
t
=
2
s
(
1
−
4
s
)
2
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Find the inverse of
\newline
f
(
x
)
=
x
2
−
5
,
x
≤
0
f(x)=x^{2}-5,\,x \leq 0
f
(
x
)
=
x
2
−
5
,
x
≤
0
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Given the function
f
(
x
)
=
−
x
2
−
1
4
f(x)=-x^{2}-\frac{1}{4}
f
(
x
)
=
−
x
2
−
4
1
, find the value of
f
(
1
)
f(1)
f
(
1
)
in simplest form.
\newline
Answer:
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Given the function
f
(
x
)
=
−
1
+
7
6
x
2
f(x)=-1+\frac{7}{6} x^{2}
f
(
x
)
=
−
1
+
6
7
x
2
, find the value of
f
(
1
7
)
f\left(\frac{1}{7}\right)
f
(
7
1
)
in simplest form.
\newline
Answer:
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Given the function
f
(
x
)
=
7
6
x
2
−
4
f(x)=\frac{7}{6} x^{2}-4
f
(
x
)
=
6
7
x
2
−
4
, find the value of
f
(
−
2
)
f(-2)
f
(
−
2
)
in simplest form.
\newline
Answer:
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Given the function
f
(
x
)
=
x
2
+
3
f(x)=x^{2}+3
f
(
x
)
=
x
2
+
3
, find the value of
f
(
−
2
3
)
f\left(-\frac{2}{3}\right)
f
(
−
3
2
)
in simplest form.
\newline
Answer:
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Given the function
f
(
x
)
=
−
7
x
2
−
1
f(x)=-7 x^{2}-1
f
(
x
)
=
−
7
x
2
−
1
, find the value of
f
(
3
7
)
f\left(\frac{3}{7}\right)
f
(
7
3
)
in simplest form.
\newline
Answer:
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Given the function
f
(
x
)
=
−
1
4
x
2
+
4
f(x)=-\frac{1}{4} x^{2}+4
f
(
x
)
=
−
4
1
x
2
+
4
, find the value of
f
(
−
1
)
f(-1)
f
(
−
1
)
in simplest form.
\newline
Answer:
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Given the function
f
(
x
)
=
−
1
4
x
2
−
4
f(x)=-\frac{1}{4} x^{2}-4
f
(
x
)
=
−
4
1
x
2
−
4
, find the value of
f
(
2
)
f(2)
f
(
2
)
in simplest form.
\newline
Answer:
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Find the inverse function in slope-intercept form
(
m
x
+
b
)
(\mathrm{mx}+\mathrm{b})
(
mx
+
b
)
:
\newline
f
(
x
)
=
x
−
16
f(x)=x-16
f
(
x
)
=
x
−
16
\newline
Answer:
f
−
1
(
x
)
=
f^{-1}(x)=
f
−
1
(
x
)
=
Get tutor help
Find the inverse function in slope-intercept form
(
m
x
+
b
)
(\mathrm{mx}+\mathrm{b})
(
mx
+
b
)
:
\newline
f
(
x
)
=
4
5
x
−
16
f(x)=\frac{4}{5} x-16
f
(
x
)
=
5
4
x
−
16
\newline
Answer:
f
−
1
(
x
)
=
f^{-1}(x)=
f
−
1
(
x
)
=
Get tutor help
Find the inverse function in slope-intercept form
(
m
x
+
b
)
(\mathrm{mx}+\mathrm{b})
(
mx
+
b
)
:
\newline
f
(
x
)
=
−
2
x
−
4
f(x)=-2 x-4
f
(
x
)
=
−
2
x
−
4
\newline
Answer:
f
−
1
(
x
)
=
f^{-1}(x)=
f
−
1
(
x
)
=
Get tutor help
Find the inverse function in slope-intercept form
(
m
x
+
b
)
(\mathrm{mx}+\mathrm{b})
(
mx
+
b
)
:
\newline
f
(
x
)
=
2
x
+
2
f(x)=2 x+2
f
(
x
)
=
2
x
+
2
\newline
Answer:
f
−
1
(
x
)
=
f^{-1}(x)=
f
−
1
(
x
)
=
Get tutor help
Find the inverse function in slope-intercept form
(
m
x
+
b
)
(\mathrm{mx}+\mathrm{b})
(
mx
+
b
)
:
\newline
f
(
x
)
=
−
1
4
x
+
15
f(x)=-\frac{1}{4} x+15
f
(
x
)
=
−
4
1
x
+
15
\newline
Answer:
f
−
1
(
x
)
=
f^{-1}(x)=
f
−
1
(
x
)
=
Get tutor help
Find the inverse function in slope-intercept form
(
m
x
+
b
)
(\mathrm{mx}+\mathrm{b})
(
mx
+
b
)
:
\newline
f
(
x
)
=
−
5
x
+
20
f(x)=-5 x+20
f
(
x
)
=
−
5
x
+
20
\newline
Answer:
f
−
1
(
x
)
=
f^{-1}(x)=
f
−
1
(
x
)
=
Get tutor help
Find the inverse function in slope-intercept form
(
m
x
+
b
)
(\mathrm{mx}+\mathrm{b})
(
mx
+
b
)
:
\newline
f
(
x
)
=
−
x
−
16
f(x)=-x-16
f
(
x
)
=
−
x
−
16
\newline
Answer:
f
−
1
(
x
)
=
f^{-1}(x)=
f
−
1
(
x
)
=
Get tutor help
Find the inverse function in slope-intercept form
(
m
x
+
b
)
(\mathrm{mx}+\mathrm{b})
(
mx
+
b
)
:
\newline
f
(
x
)
=
3
2
x
+
9
f(x)=\frac{3}{2} x+9
f
(
x
)
=
2
3
x
+
9
\newline
Answer:
f
−
1
(
x
)
=
f^{-1}(x)=
f
−
1
(
x
)
=
Get tutor help
Find the inverse function in slope-intercept form
(
m
x
+
b
)
(\mathrm{mx}+\mathrm{b})
(
mx
+
b
)
:
\newline
f
(
x
)
=
−
1
2
x
−
7
f(x)=-\frac{1}{2} x-7
f
(
x
)
=
−
2
1
x
−
7
\newline
Answer:
f
−
1
(
x
)
=
f^{-1}(x)=
f
−
1
(
x
)
=
Get tutor help
Find the inverse function in slope-intercept form
(
m
x
+
b
)
(\mathrm{mx}+\mathrm{b})
(
mx
+
b
)
:
\newline
f
(
x
)
=
5
x
−
15
f(x)=5 x-15
f
(
x
)
=
5
x
−
15
\newline
Answer:
f
−
1
(
x
)
=
f^{-1}(x)=
f
−
1
(
x
)
=
Get tutor help
Find the inverse function in slope-intercept form
(
m
x
+
b
)
(\mathrm{mx}+\mathrm{b})
(
mx
+
b
)
:
\newline
f
(
x
)
=
2
x
−
8
f(x)=2 x-8
f
(
x
)
=
2
x
−
8
\newline
Answer:
f
−
1
(
x
)
=
f^{-1}(x)=
f
−
1
(
x
)
=
Get tutor help
Find the inverse function in slope-intercept form
(
m
x
+
b
)
(\mathrm{mx}+\mathrm{b})
(
mx
+
b
)
:
\newline
f
(
x
)
=
−
2
x
+
6
f(x)=-2 x+6
f
(
x
)
=
−
2
x
+
6
\newline
Answer:
f
−
1
(
x
)
=
f^{-1}(x)=
f
−
1
(
x
)
=
Get tutor help
Find the inverse function in slope-intercept form
(
m
x
+
b
)
(\mathrm{mx}+\mathrm{b})
(
mx
+
b
)
:
\newline
f
(
x
)
=
−
2
5
x
−
16
f(x)=-\frac{2}{5} x-16
f
(
x
)
=
−
5
2
x
−
16
\newline
Answer:
f
−
1
(
x
)
=
f^{-1}(x)=
f
−
1
(
x
)
=
Get tutor help
Find the inverse function in slope-intercept form
(
m
x
+
b
)
(\mathrm{mx}+\mathrm{b})
(
mx
+
b
)
:
\newline
f
(
x
)
=
−
3
x
+
15
f(x)=-3 x+15
f
(
x
)
=
−
3
x
+
15
\newline
Answer:
f
−
1
(
x
)
=
f^{-1}(x)=
f
−
1
(
x
)
=
Get tutor help
Find the inverse function in slope-intercept form
(
m
x
+
b
)
(\mathrm{mx}+\mathrm{b})
(
mx
+
b
)
:
\newline
f
(
x
)
=
−
5
2
x
−
10
f(x)=-\frac{5}{2} x-10
f
(
x
)
=
−
2
5
x
−
10
\newline
Answer:
f
−
1
(
x
)
=
f^{-1}(x)=
f
−
1
(
x
)
=
Get tutor help
Find the inverse function in slope-intercept form
(
m
x
+
b
)
(\mathrm{mx}+\mathrm{b})
(
mx
+
b
)
:
\newline
f
(
x
)
=
−
x
−
8
f(x)=-x-8
f
(
x
)
=
−
x
−
8
\newline
Answer:
f
−
1
(
x
)
=
f^{-1}(x)=
f
−
1
(
x
)
=
Get tutor help
Find the inverse function in slope-intercept form
(
m
x
+
b
)
(\mathrm{mx}+\mathrm{b})
(
mx
+
b
)
:
\newline
f
(
x
)
=
−
2
5
x
+
8
f(x)=-\frac{2}{5} x+8
f
(
x
)
=
−
5
2
x
+
8
\newline
Answer:
f
−
1
(
x
)
=
f^{-1}(x)=
f
−
1
(
x
)
=
Get tutor help
Find the inverse function in slope-intercept form
(
m
x
+
b
)
(\mathrm{mx}+\mathrm{b})
(
mx
+
b
)
:
\newline
f
(
x
)
=
5
x
−
10
f(x)=5 x-10
f
(
x
)
=
5
x
−
10
\newline
Answer:
f
−
1
(
x
)
=
f^{-1}(x)=
f
−
1
(
x
)
=
Get tutor help
Find the inverse function in slope-intercept form
(
m
x
+
b
)
(\mathrm{mx}+\mathrm{b})
(
mx
+
b
)
:
\newline
f
(
x
)
=
2
3
x
−
12
f(x)=\frac{2}{3} x-12
f
(
x
)
=
3
2
x
−
12
\newline
Answer:
f
−
1
(
x
)
=
f^{-1}(x)=
f
−
1
(
x
)
=
Get tutor help
Find the inverse function in slope-intercept form
(
m
x
+
b
)
(\mathrm{mx}+\mathrm{b})
(
mx
+
b
)
:
\newline
f
(
x
)
=
5
2
x
−
15
f(x)=\frac{5}{2} x-15
f
(
x
)
=
2
5
x
−
15
\newline
Answer:
f
−
1
(
x
)
=
f^{-1}(x)=
f
−
1
(
x
)
=
Get tutor help
Find the inverse function in slope-intercept form
(
m
x
+
b
)
(\mathrm{mx}+\mathrm{b})
(
mx
+
b
)
:
\newline
f
(
x
)
=
3
2
x
+
6
f(x)=\frac{3}{2} x+6
f
(
x
)
=
2
3
x
+
6
\newline
Answer:
f
−
1
(
x
)
=
f^{-1}(x)=
f
−
1
(
x
)
=
Get tutor help
Find the inverse function in slope-intercept form
(
m
x
+
b
)
(\mathrm{mx}+\mathrm{b})
(
mx
+
b
)
:
\newline
f
(
x
)
=
−
2
x
+
4
f(x)=-2 x+4
f
(
x
)
=
−
2
x
+
4
\newline
Answer:
f
−
1
(
x
)
=
f^{-1}(x)=
f
−
1
(
x
)
=
Get tutor help
Find the inverse function in slope-intercept form
(
m
x
+
b
)
(\mathrm{mx}+\mathrm{b})
(
mx
+
b
)
:
\newline
f
(
x
)
=
−
x
−
14
f(x)=-x-14
f
(
x
)
=
−
x
−
14
\newline
Answer:
f
−
1
(
x
)
=
f^{-1}(x)=
f
−
1
(
x
)
=
Get tutor help
Find the inverse function in slope-intercept form
(
m
x
+
b
)
(\mathrm{mx}+\mathrm{b})
(
mx
+
b
)
:
\newline
f
(
x
)
=
−
2
3
x
+
12
f(x)=-\frac{2}{3} x+12
f
(
x
)
=
−
3
2
x
+
12
\newline
Answer:
f
−
1
(
x
)
=
f^{-1}(x)=
f
−
1
(
x
)
=
Get tutor help
Find the inverse function in slope-intercept form
(
m
x
+
b
)
(\mathrm{mx}+\mathrm{b})
(
mx
+
b
)
:
\newline
f
(
x
)
=
−
3
2
x
+
9
f(x)=-\frac{3}{2} x+9
f
(
x
)
=
−
2
3
x
+
9
\newline
Answer:
f
−
1
(
x
)
=
f^{-1}(x)=
f
−
1
(
x
)
=
Get tutor help
Find the inverse function of the function
f
(
x
)
=
1
5
x
+
9
f(x)=\frac{1}{5} x+9
f
(
x
)
=
5
1
x
+
9
.
\newline
f
−
1
(
x
)
=
5
x
−
9
f^{-1}(x)=5 x-9
f
−
1
(
x
)
=
5
x
−
9
\newline
f
−
1
(
x
)
=
5
x
−
45
f^{-1}(x)=5 x-45
f
−
1
(
x
)
=
5
x
−
45
\newline
f
−
1
(
x
)
=
1
5
x
−
9
f^{-1}(x)=\frac{1}{5} x-9
f
−
1
(
x
)
=
5
1
x
−
9
\newline
f
−
1
(
x
)
=
1
5
x
−
45
f^{-1}(x)=\frac{1}{5} x-45
f
−
1
(
x
)
=
5
1
x
−
45
Get tutor help
Given
f
(
x
)
=
−
x
2
−
9
x
f(x)=-x^{2}-9 x
f
(
x
)
=
−
x
2
−
9
x
, find
f
(
4
)
f(4)
f
(
4
)
\newline
Answer:
Get tutor help
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