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Let’s check out your problem:
Determine the number of different values that
∠
x
\angle x
∠
x
could be based on the given information.
\newline
sin
x
=
0.4048
\sin x=0.4048
sin
x
=
0.4048
\newline
3
3
3
\newline
0
0
0
\newline
1
1
1
\newline
2
2
2
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Home
Math Problems
Algebra 2
Multiply radical expressions
Full solution
Q.
Determine the number of different values that
∠
x
\angle x
∠
x
could be based on the given information.
\newline
sin
x
=
0.4048
\sin x=0.4048
sin
x
=
0.4048
\newline
3
3
3
\newline
0
0
0
\newline
1
1
1
\newline
2
2
2
Identify sine value:
First, we know that
sin
x
=
0.4048
\sin x = 0.4048
sin
x
=
0.4048
. The sine function is positive in the first and second quadrants.
Calculate reference angle:
To find the reference angle, we use the inverse sine function on a calculator:
x
=
sin
−
1
(
0.4048
)
x = \sin^{-1}(0.4048)
x
=
sin
−
1
(
0.4048
)
.
Determine angle in first quadrant:
Calculating the reference angle gives us
x
≈
23.9
x \approx 23.9
x
≈
23.9
degrees. This is the angle in the first quadrant.
Find angle in second quadrant:
The second possible angle is in the second quadrant, which is
180
−
23.9
180 - 23.9
180
−
23.9
degrees
=
156.1
= 156.1
=
156.1
degrees.
Final possible values:
So, there are two possible values for
∠
x
\angle x
∠
x
:
23.9
23.9
23.9
degrees and
156.1
156.1
156.1
degrees.
More problems from Multiply radical expressions
Question
Order the expressions from least to greatest.
\newline
1
1
1
8
2
−
60
2
2
×
5
11^{1} \quad8^{2}-60 \quad 2^{2} \times 5
1
1
1
8
2
−
60
2
2
×
5
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Question
Order the expressions from least to greatest.
\newline
2
3
−
2
1
2
1
+
3
1
3
2
2^{3}-2^{1} \quad 2^{1}+3^{1} \quad 3^{2}
2
3
−
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1
2
1
+
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3
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Question
Let
y
=
e
x
y=\sqrt{e^{x}}
y
=
e
x
.
\newline
Find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
.
\newline
Choose
1
1
1
answer:
\newline
(A)
e
x
2
\frac{\sqrt{e^{x}}}{2}
2
e
x
\newline
(B)
e
x
\sqrt{e^{x}}
e
x
\newline
(C)
x
e
x
−
1
x \sqrt{e^{x-1}}
x
e
x
−
1
\newline
(D)
1
2
e
x
\frac{1}{2 \sqrt{e^{x}}}
2
e
x
1
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Posted 3 months ago
Question
What is the area of the region between the graphs of
f
(
x
)
=
−
x
2
+
2
x
+
12
f(x)=-x^{2}+2 x+12
f
(
x
)
=
−
x
2
+
2
x
+
12
and
g
(
x
)
=
x
2
−
12
g(x)=x^{2}-12
g
(
x
)
=
x
2
−
12
from
x
=
−
3
x=-3
x
=
−
3
to
x
=
4
x=4
x
=
4
?
\newline
Choose
1
1
1
answer:
\newline
(A)
7
7
7
\newline
(B)
343
3
\frac{343}{3}
3
343
\newline
(C)
83
3
\frac{83}{3}
3
83
\newline
(D)
52
3
13
−
1
−
32
3
\frac{52}{3} \sqrt{13}-1-32 \sqrt{3}
3
52
13
−
1
−
32
3
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Posted 3 months ago
Question
What is the average value of
1
x
\frac{1}{x}
x
1
on the interval
4
≤
x
≤
8
4 \leq x \leq 8
4
≤
x
≤
8
?
\newline
Choose
1
1
1
answer:
\newline
(A)
3
16
\frac{3}{16}
16
3
\newline
(B)
1
16
\frac{1}{16}
16
1
\newline
(C)
ln
(
32
)
4
\frac{\ln (32)}{4}
4
l
n
(
32
)
\newline
(D)
ln
(
2
)
4
\frac{\ln (2)}{4}
4
l
n
(
2
)
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Posted 3 months ago
Question
Part
4
4
4
of
6
6
6
\newline
K
K
K
\newline
For the data given below, answer parts (a) through (f).
\newline
\begin{array}{cc}
\newline
x & y (\newline\)
−
19
-19
−
19
&
102
102
102
(\newline\)
−
16
-16
−
16
&
122
122
122
(\newline\)
−
14
-14
−
14
&
120
120
120
(\newline\)
−
13
-13
−
13
&
132
132
132
(\newline\)
−
9
-9
−
9
&
142
142
142
(\newline\)\end{array}
\newline
(a) Draw a scatter plot.
\newline
Choose the correct graph below.
\newline
A.
\newline
B.
\newline
C.
\newline
D.
\newline
(b) Find the equation of the line containing the first and the last data points.
\newline
y
=
4
x
+
178
y=4x+178
y
=
4
x
+
178
\newline
(Type an equation. Type your answer in slope-intercept form. Use integers or fractions for any numbers in the equation.)
\newline
(c) Graph the line found in part (b) on the scatter plot.
\newline
Choose the correct graph below.
\newline
A.
\newline
B
\newline
C.
\newline
(d) Use a graphing utility to find the line of best fit.
\newline
y
=
□
x
+
□
y=\square x+\square
y
=
□
x
+
□
\newline
(Type integers or decimals rounded to four decimal places as needed)
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Posted 2 months ago
Question
noot: Additional Practice (LMS graded)
\newline
of
2
2
2
\newline
e surface area of a paper cup is defined by the function
\newline
s
(
h
)
=
6
π
16
+
h
2
s(h)=6\pi\sqrt{16+h^{2}}
s
(
h
)
=
6
π
16
+
h
2
, where
\newline
h
h
h
is the height of the cup. What is the domain and range of
\newline
s
(
h
)
s(h)
s
(
h
)
?
\newline
Describe the domain. Select the correct choice below and, if necessary, fill in the answer box within your choice.
\newline
A. The domain is
\newline
{
h
∣
h
>
−
}
\{h\mid h>-\}
{
h
∣
h
>
−
}
.
\newline
(Type an inequality or a compound inequality. Simplify your answer. Type an exact answer, using
\newline
π
\pi
π
as needed.)
\newline
B. The domain is all real numbers.
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Posted 2 months ago
Question
Determine the coefficient and degree of the monomial.
\newline
−
7
x
2
y
5
-7x^{2}y^{5}
−
7
x
2
y
5
\newline
The coefficient of the monomial is
\newline
□
\square
□
.
\newline
The degree of the monomial is
\newline
□
\square
□
.
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Posted 2 months ago
Question
a
2
a
−
5
a
−
3
a
0
⋅
(
a
2
a
3
)
−
4
=
a
x
\dfrac{a^2a^{-5}}{a^{-3}a^0}\cdot\left(\dfrac{a^2}{a^3}\right)^{-4}=a^x
a
−
3
a
0
a
2
a
−
5
⋅
(
a
3
a
2
)
−
4
=
a
x
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Posted 2 months ago
Question
x
3
+
4
x
5
=
1
3
\frac{x}{3} + \frac{4x}{5} = \frac{1}{3}
3
x
+
5
4
x
=
3
1
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Posted 2 months ago