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Math Problems
Algebra 2
Simplify expressions using trigonometric identities
\newline
What is the particular solution to the differential equation
d
y
d
x
=
1
+
y
x
\frac{d y}{d x}=\frac{1+y}{x}
d
x
d
y
=
x
1
+
y
with the initial condition
y
(
e
)
=
1
y(e)=1
y
(
e
)
=
1
?
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Below is the graph of a trigonometric function. It has a maximum point at
(
1
4
π
,
−
2.2
)
(\frac{1}{4} \pi, -2.2)
(
4
1
π
,
−
2.2
)
and a minimum point at
(
3
4
π
,
−
8.5
)
(\frac{3}{4} \pi, -8.5)
(
4
3
π
,
−
8.5
)
what is the midline equation of the function
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1
+
tan
θ
1
−
tan
θ
+
1
+
cot
θ
1
−
cot
θ
\frac{1+\tan \theta}{1-\tan \theta}+\frac{1+\cot \theta}{1-\cot \theta}
1
−
t
a
n
θ
1
+
t
a
n
θ
+
1
−
c
o
t
θ
1
+
c
o
t
θ
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The potential energy of a particle varies the distance
x
x
x
from a fixed origin as
U
=
A
x
X
2
+
B
U=\frac{A\sqrt{x}}{X^{2}+B}
U
=
X
2
+
B
A
x
, where
A
A
A
and
B
B
B
are dimensional constants, then find the dimensional formula for
A
B
2
AB^{2}
A
B
2
.
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Make a complete graph of the following function. A graphing utility is useful in locating intercepts, local extreme values, and inflection points.
\newline
f
(
x
)
=
2
x
−
5
x
2
−
1
f(x)=\frac{2 x-5}{x^{2}-1}
f
(
x
)
=
x
2
−
1
2
x
−
5
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2
2
2
. Verify the identity
\newline
tan
α
−
1
tan
α
+
1
=
1
−
cot
α
1
+
cot
α
\frac{\tan \alpha-1}{\tan \alpha+1}=\frac{1-\cot \alpha}{1+\cot \alpha}
tan
α
+
1
tan
α
−
1
=
1
+
cot
α
1
−
cot
α
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1
−
sin
2
θ
1
−
cos
θ
=
−
cos
θ
1-\frac{\sin ^{2} \theta}{1-\cos \theta}=-\cos \theta
1
−
1
−
c
o
s
θ
s
i
n
2
θ
=
−
cos
θ
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sin
θ
cot
2
θ
−
sin
θ
cos
2
θ
\frac{\sin \theta}{\cot ^{2} \theta}-\frac{\sin \theta}{\cos ^{2} \theta}
c
o
t
2
θ
s
i
n
θ
−
c
o
s
2
θ
s
i
n
θ
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Solve for the angle
θ
\theta
θ
.
\newline
(a)
sin
2
θ
=
3
4
\sin^{2}\theta=\frac{3}{4}
sin
2
θ
=
4
3
.
\newline
(b)
sin
2
θ
−
cos
θ
=
0
\sin 2\theta-\cos \theta=0
sin
2
θ
−
cos
θ
=
0
.
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3
3
3
.
3
−
3
cos
2
θ
sin
θ
\frac{3-3 \cos ^{2} \theta}{\sin \theta}
s
i
n
θ
3
−
3
c
o
s
2
θ
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s centered at
−
2
-2
−
2
for
f
(
x
)
=
2
6
−
x
f(x)=\frac{2}{6-x}
f
(
x
)
=
6
−
x
2
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(
10
10
10
) The figure shows
2
2
2
identical quarter circles in a rectangle. Find the perimeter of the shaded part. (Take
π
=
22
7
\pi=\frac{22}{7}
π
=
7
22
.)
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Using implicit differentiation, find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
.
\newline
−
5
x
2
y
4
−
4
x
2
y
3
=
3
−
2
x
-5 x^{2} y^{4}-4 x^{2} y^{3}=3-2 x
−
5
x
2
y
4
−
4
x
2
y
3
=
3
−
2
x
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Solve the equation for all real solutions in simplest form.
\newline
−
3
r
2
−
r
−
10
=
−
4
r
2
-3 r^{2}-r-10=-4 r^{2}
−
3
r
2
−
r
−
10
=
−
4
r
2
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1
−
cos
α
sin
α
=
2
csc
α
\frac{1-\cos \alpha}{\sin \alpha}=2 \csc \alpha
s
i
n
α
1
−
c
o
s
α
=
2
csc
α
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Using implicit differentiation, find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
.
\newline
−
7
cos
(
3
x
)
sin
(
3
y
)
=
4
x
−
1
-7 \cos (3 x) \sin (3 y)=4 x-1
−
7
cos
(
3
x
)
sin
(
3
y
)
=
4
x
−
1
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Using implicit differentiation, find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
.
\newline
−
y
−
7
y
3
−
4
x
4
+
6
x
−
2
x
y
=
1
-y-7 y^{3}-4 x^{4}+6 x-2 x y=1
−
y
−
7
y
3
−
4
x
4
+
6
x
−
2
x
y
=
1
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Using implicit differentiation, find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
.
\newline
−
3
x
3
y
4
−
6
x
2
y
4
=
x
+
4
-3 x^{3} y^{4}-6 x^{2} y^{4}=x+4
−
3
x
3
y
4
−
6
x
2
y
4
=
x
+
4
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Simplify to a single trig function with no denominator.
\newline
cot
θ
csc
θ
\frac{\cot \theta}{\csc \theta}
csc
θ
cot
θ
\newline
Answer:
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Simplify to a single trig function with no denominator.
\newline
csc
θ
sin
θ
\frac{\csc \theta}{\sin \theta}
sin
θ
csc
θ
\newline
Answer:
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Simplify to a single trig function with no denominator.
\newline
sin
2
θ
⋅
csc
θ
\sin ^{2} \theta \cdot \csc \theta
sin
2
θ
⋅
csc
θ
\newline
Answer:
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Simplify to a single trig function with no denominator.
\newline
cot
2
θ
csc
2
θ
\frac{\cot ^{2} \theta}{\csc ^{2} \theta}
csc
2
θ
cot
2
θ
\newline
Answer:
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Simplify to a single trig function with no denominator.
\newline
sin
θ
tan
θ
\frac{\sin \theta}{\tan \theta}
tan
θ
sin
θ
\newline
Answer:
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Simplify to a single trig function with no denominator.
\newline
csc
θ
sec
θ
\frac{\csc \theta}{\sec \theta}
sec
θ
csc
θ
\newline
Answer:
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Simplify to a single trig function with no denominator.
\newline
cot
θ
cos
θ
\frac{\cot \theta}{\cos \theta}
cos
θ
cot
θ
\newline
Answer:
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Simplify to a single trig function with no denominator.
\newline
cos
θ
⋅
sec
θ
\cos \theta \cdot \sec \theta
cos
θ
⋅
sec
θ
\newline
Answer:
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Simplify to a single trig function with no denominator.
\newline
sin
2
θ
⋅
sec
2
θ
\sin ^{2} \theta \cdot \sec ^{2} \theta
sin
2
θ
⋅
sec
2
θ
\newline
Answer:
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Simplify to a single trig function with no denominator.
\newline
cos
θ
sec
θ
\frac{\cos \theta}{\sec \theta}
sec
θ
cos
θ
\newline
Answer:
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Simplify to a single trig function with no denominator.
\newline
tan
2
θ
sec
2
θ
\frac{\tan ^{2} \theta}{\sec ^{2} \theta}
sec
2
θ
tan
2
θ
\newline
Answer:
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Simplify to a single trig function with no denominator.
\newline
tan
2
θ
⋅
csc
2
θ
\tan ^{2} \theta \cdot \csc ^{2} \theta
tan
2
θ
⋅
csc
2
θ
\newline
Answer:
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Simplify to a single trig function with no denominator.
\newline
cos
θ
cot
θ
\frac{\cos \theta}{\cot \theta}
cot
θ
cos
θ
\newline
Answer:
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Simplify to a single trig function with no denominator.
\newline
sin
2
θ
⋅
cot
2
θ
\sin ^{2} \theta \cdot \cot ^{2} \theta
sin
2
θ
⋅
cot
2
θ
\newline
Answer:
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Using implicit differentiation, find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
.
\newline
−
x
2
y
4
−
4
x
2
y
3
=
2
x
+
5
-x^{2} y^{4}-4 x^{2} y^{3}=2 x+5
−
x
2
y
4
−
4
x
2
y
3
=
2
x
+
5
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Solve for the exact value of
x
x
x
.
\newline
4
ln
(
6
x
−
4
)
+
7
=
15
4 \ln (6 x-4)+7=15
4
ln
(
6
x
−
4
)
+
7
=
15
\newline
Answer:
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Using implicit differentiation, find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
.
\newline
−
6
y
3
−
5
x
4
+
3
x
+
6
y
−
4
x
y
=
5
-6 y^{3}-5 x^{4}+3 x+6 y-4 x y=5
−
6
y
3
−
5
x
4
+
3
x
+
6
y
−
4
x
y
=
5
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Linear Equations
\newline
Solving a proportion of the form
x
/
a
=
b
/
c
x / a=b / c
x
/
a
=
b
/
c
\newline
1
/
5
1 / 5
1/5
\newline
Español
\newline
Solve the following proportion for
x
x
x
.
\newline
12
x
=
11
17
\frac{12}{x}=\frac{11}{17}
x
12
=
17
11
\newline
Round your answer to the nearest
\newline
x
=
x=
x
=
\newline
Explanation
\newline
Check
\newline
AA
\newline
www-awu.aleks.com
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Using implicit differentiation, find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
.
\newline
−
3
cos
(
3
x
)
sin
(
2
y
)
=
5
−
x
-3 \cos (3 x) \sin (2 y)=5-x
−
3
cos
(
3
x
)
sin
(
2
y
)
=
5
−
x
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Using implicit differentiation, find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
.
\newline
−
2
x
y
4
−
6
x
y
2
=
3
x
+
4
-2 x y^{4}-6 x y^{2}=3 x+4
−
2
x
y
4
−
6
x
y
2
=
3
x
+
4
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Using implicit differentiation, find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
.
\newline
−
5
y
3
−
2
x
4
+
3
x
+
3
y
+
x
y
=
2
-5 y^{3}-2 x^{4}+3 x+3 y+x y=2
−
5
y
3
−
2
x
4
+
3
x
+
3
y
+
x
y
=
2
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Using implicit differentiation, find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
.
\newline
−
5
x
3
y
4
+
x
y
4
=
3
x
+
5
-5 x^{3} y^{4}+x y^{4}=3 x+5
−
5
x
3
y
4
+
x
y
4
=
3
x
+
5
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Using implicit differentiation, find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
.
\newline
(
−
x
3
−
4
y
2
)
3
=
−
3
x
y
\left(-x^{3}-4 y^{2}\right)^{3}=-3 x y
(
−
x
3
−
4
y
2
)
3
=
−
3
x
y
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TICE es
\newline
4
4
4
\newline
If
c
=
b
−
3
c=b^{-3}
c
=
b
−
3
, what is
b
b
b
in terms of
c
c
c
?
\newline
A.
c
1
3
c^{\frac{1}{3}}
c
3
1
\newline
B.
c
3
c^{3}
c
3
\newline
C.
1
c
3
\frac{1}{c^{3}}
c
3
1
\newline
D.
1
c
3
\frac{1}{\sqrt[3]{c}}
3
c
1
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Simplify the expression
\newline
tan
(
π
2
−
θ
)
sin
θ
\tan \left(\frac{\pi}{2}-\theta\right) \sin \theta
tan
(
2
π
−
θ
)
sin
θ
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Find the exact value of each of the remaining trigonometric functions of
θ
\theta
θ
.
tan
θ
=
−
1
5
,
sec
θ
<
0
\tan \theta = -\frac{1}{5},\quad \sec \theta < 0
tan
θ
=
−
5
1
,
sec
θ
<
0
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Using implicit differentiation, find
d
y
d
x
\frac{dy}{dx}
d
x
d
y
.
\newline
−
7
x
2
y
4
−
5
x
y
2
=
4
−
4
x
-7x^{2}y^{4}-5xy^{2}=4-4x
−
7
x
2
y
4
−
5
x
y
2
=
4
−
4
x
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Suppose that the functions
f
f
f
and
g
g
g
are defined as follows.
\newline
f
(
x
)
=
9
x
g
(
x
)
=
4
x
+
3
f(x)=\frac{9}{x} \quad g(x)=\frac{4}{x+3}
f
(
x
)
=
x
9
g
(
x
)
=
x
+
3
4
\newline
Find
g
f
\frac{g}{f}
f
g
. Then, give its domain using an interval or union of intervals.
\newline
Simplify your answers.
\newline
(
g
f
)
(
x
)
=
4
x
9
(
x
+
3
)
\left(\frac{g}{f}\right)(x)=\frac{4x}{9(x+3)}
(
f
g
)
(
x
)
=
9
(
x
+
3
)
4
x
\newline
Domain of
g
f
\frac{g}{f}
f
g
:
(
−
∞
,
−
3
)
∪
(
−
3
,
∞
)
(-\infty,-3) \cup (-3,\infty)
(
−
∞
,
−
3
)
∪
(
−
3
,
∞
)
\newline
−
∞
-\infty
−
∞
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Select the answer which is equivalent to the given expression using your calculator.
\newline
sin
(
arcsin
18
424
)
\sin \left(\arcsin \frac{18}{\sqrt{424}}\right)
sin
(
arcsin
424
18
)
\newline
18
424
\frac{18}{\sqrt{424}}
424
18
\newline
10
18
\frac{10}{18}
18
10
\newline
424
18
\frac{\sqrt{424}}{18}
18
424
\newline
18
10
\frac{18}{10}
10
18
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Select the answer which is equivalent to the given expression using your calculator.
\newline
cot
(
arccos
315
18
)
\cot \left(\arccos \frac{\sqrt{315}}{18}\right)
cot
(
arccos
18
315
)
\newline
3
315
\frac{3}{\sqrt{315}}
315
3
\newline
315
18
\frac{\sqrt{315}}{18}
18
315
\newline
315
3
\frac{\sqrt{315}}{3}
3
315
\newline
18
3
\frac{18}{3}
3
18
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Select the answer which is equivalent to the given expression using your calculator.
\newline
csc
(
arccos
15
4
)
\csc \left(\arccos \frac{\sqrt{15}}{4}\right)
csc
(
arccos
4
15
)
\newline
4
15
\frac{4}{\sqrt{15}}
15
4
\newline
4
4
4
\newline
15
4
\frac{\sqrt{15}}{4}
4
15
\newline
15
\sqrt{15}
15
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tan
(
π
4
x
)
+
3
=
0
\tan \left(\frac{\pi}{4}x\right)+\sqrt{3}=0
tan
(
4
π
x
)
+
3
=
0
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2
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