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Math Problems
Precalculus
Solve trigonometric equations
The functions
y
=
3
(
x
+
2
)
2
−
4
y=3(x+2)^{2}-4
y
=
3
(
x
+
2
)
2
−
4
and
y
=
−
3
(
x
+
2
)
2
−
4
y=-3(x+2)^{2}-4
y
=
−
3
(
x
+
2
)
2
−
4
are graphed in the
x
y
x y
x
y
-plane. Which of the following must be true of the graphs of the vertexes and axes of symmetry of the two functions?
\newline
Choose
1
1
1
answer:
\newline
(A) The functions will have different vertexes.
\newline
(B) The functions will have different axes of symmetry.
\newline
(C) The function
y
=
3
(
x
+
2
)
2
−
4
y=3(x+2)^{2}-4
y
=
3
(
x
+
2
)
2
−
4
will have a minimum value, and the function
y
=
−
3
(
x
+
2
)
2
−
4
y=-3(x+2)^{2}-4
y
=
−
3
(
x
+
2
)
2
−
4
will have a maximum value.
\newline
(D) The function
y
=
3
(
x
+
2
)
2
−
4
y=3(x+2)^{2}-4
y
=
3
(
x
+
2
)
2
−
4
will have a maximum value, and the graph of
y
=
−
3
(
x
+
2
)
2
−
4
y=-3(x+2)^{2}-4
y
=
−
3
(
x
+
2
)
2
−
4
will have a minimum value.
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Let
x
1
x_1
x
1
and
x
2
x_2
x
2
be solutions to the equation shown, with
x
1
>
x
2
x_1 > x_2
x
1
>
x
2
. What is the value of
x
1
+
x
2
x_1+x_2
x
1
+
x
2
?
16
x
2
−
8
x
−
3
=
0
16x^2-8x-3=0
16
x
2
−
8
x
−
3
=
0
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Find all solutions with
−
π
2
≤
θ
≤
π
2
-\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}
−
2
π
≤
θ
≤
2
π
. Give the exact answer(s) in simplest form. If there are multiple answers, separate them with commas.
\newline
sin
(
θ
)
=
0
\sin(\theta)=0
sin
(
θ
)
=
0
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d
y
d
t
=
3
t
\frac{d y}{d t}=3 t
d
t
d
y
=
3
t
and
y
(
2
)
=
3
y(2)=3
y
(
2
)
=
3
.
\newline
What is
t
t
t
when
y
=
6
y=6
y
=
6
?
\newline
Choose all answers that apply:
\newline
(A)
t
=
−
2
t=-2
t
=
−
2
\newline
(B)
t
=
6
t=\sqrt{6}
t
=
6
\newline
(C)
t
=
6
t=6
t
=
6
\newline
(D)
t
=
−
6
t=-\sqrt{6}
t
=
−
6
\newline
(E)
t
=
2
t=2
t
=
2
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Alison tried to find all the points on the curve given by
x
2
+
4
y
2
=
7
+
3
x
y
x^{2}+4 y^{2}=7+3 x y
x
2
+
4
y
2
=
7
+
3
x
y
where the line tangent to the curve is horizontal. This is her solution:
\newline
Step
1
1
1
: Finding an expression for
d
y
d
x
\frac{d y}{d x}
d
x
d
y
.
\newline
d
y
d
x
=
3
y
−
2
x
8
y
−
3
x
\frac{d y}{d x}=\frac{3 y-2 x}{8 y-3 x}
d
x
d
y
=
8
y
−
3
x
3
y
−
2
x
\newline
Step
2
2
2
: Forming a system of equations.
\newline
{
x
2
+
4
y
2
=
7
+
3
x
y
3
y
−
2
x
=
0
8
y
−
3
x
≠
0
\left\{\begin{array}{l} x^{2}+4 y^{2}=7+3 x y \\ 3 y-2 x=0 \\ 8 y-3 x \neq 0 \end{array}\right.
⎩
⎨
⎧
x
2
+
4
y
2
=
7
+
3
x
y
3
y
−
2
x
=
0
8
y
−
3
x
=
0
\newline
Step
3
3
3
: Solving the system.
\newline
(
3
,
2
)
(3,2)
(
3
,
2
)
and
(
−
3
,
−
2
)
(-3,-2)
(
−
3
,
−
2
)
\newline
Is Alison's solution correct? If not, at which step did she make a mistake?
\newline
Choose
1
1
1
answer:
\newline
(A) The solution is correct.
\newline
(B) Step
1
1
1
is incorrect.
\newline
(C) Step
2
2
2
is incorrect.
\newline
(D) Step
3
3
3
is incorrect.
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Let
g
(
x
)
=
2
x
g(x)=2^{x}
g
(
x
)
=
2
x
.
\newline
Can we use the mean value theorem to say the equation
g
′
(
x
)
=
16
g^{\prime}(x)=16
g
′
(
x
)
=
16
has a solution where
3
<
x
<
5
3<x<5
3
<
x
<
5
?
\newline
Choose
1
1
1
answer:
\newline
(A) No, since the function is not differentiable on that interval.
\newline
(B) No, since the average rate of change of
g
g
g
over the interval
3
≤
x
≤
5
3 \leq x \leq 5
3
≤
x
≤
5
isn't equal to
1
6
‾
1 \overline{6}
1
6
.
\newline
(C) Yes, both conditions for using the mean value theorem have been met.
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Let
h
(
x
)
=
x
⋅
sin
(
x
)
h(x)=x \cdot \sin (x)
h
(
x
)
=
x
⋅
sin
(
x
)
.
\newline
Below is Eliza's attempt to write a formal justification for the fact that the equation
h
(
x
)
=
1
h(x)=1
h
(
x
)
=
1
has a solution where
0
≤
x
≤
2
0 \leq x \leq 2
0
≤
x
≤
2
.
\newline
Is Eliza's justification complete? If not, why?
\newline
Eliza's justification:
\newline
h
h
h
is defined for all real numbers, and polynomial and trigonometric functions are continuous at all points in their domains. Furthermore,
h
(
0
)
=
0
h(0)=0
h
(
0
)
=
0
and
h
(
2
)
≈
1.82
h(2) \approx 1.82
h
(
2
)
≈
1.82
, so
1
1
1
is between
h
(
0
)
h(0)
h
(
0
)
and
h
(
2
)
h(2)
h
(
2
)
.
\newline
So, according to the intermediate value theorem,
h
(
x
)
=
1
h(x)=1
h
(
x
)
=
1
must have a solution somewhere between
x
=
0
x=0
x
=
0
and
x
=
2
x=2
x
=
2
.
\newline
Choose
1
1
1
answer:
\newline
(A) Yes, Eliza's justification is complete.
\newline
(B) No, Eliza didn't establish that
1
1
1
is between
h
(
0
)
h(0)
h
(
0
)
and
h
(
2
)
h(2)
h
(
2
)
.
\newline
(C) No, Eliza didn't establish that
h
h
h
is continuous.
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The following are all angle measures (in degrees, rounded to the nearest tenth) whose cosine is
0
0
0
.
10
10
10
.
\newline
Which is the principal value of
arccos
(
0.10
)
\arccos (0.10)
arccos
(
0.10
)
?
\newline
Choose
1
1
1
answer:
\newline
(A)
−
635.
7
∘
-635.7^{\circ}
−
635.
7
∘
\newline
(B)
−
275.
7
∘
-275.7^{\circ}
−
275.
7
∘
\newline
(C)
84.
3
∘
84.3^{\circ}
84.
3
∘
\newline
(D)
444.
3
∘
444.3^{\circ}
444.
3
∘
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The following are all angle measures (in degrees, rounded to the nearest tenth) whose tangent is
−
44
-44
−
44
.
\newline
Which is the principal value of
tan
−
1
(
−
44
)
\tan ^{-1}(-44)
tan
−
1
(
−
44
)
?
\newline
Choose
1
1
1
answer:
\newline
(A)
−
448.
7
∘
-448.7^{\circ}
−
448.
7
∘
\newline
(B)
−
268.
7
∘
-268.7^{\circ}
−
268.
7
∘
\newline
(C)
−
88.
7
∘
-88.7^{\circ}
−
88.
7
∘
\newline
(D)
91.
3
∘
91.3^{\circ}
91.
3
∘
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The following are all angle measures (in degrees, rounded to the nearest tenth) whose cosine is
0
0
0
.
69
69
69
.
\newline
Which is the principal value of
cos
−
1
(
0.69
)
\cos ^{-1}(0.69)
cos
−
1
(
0.69
)
?
\newline
Choose
1
1
1
answer:
\newline
(A)
−
1033.
6
∘
-1033.6^{\circ}
−
1033.
6
∘
\newline
(B)
−
673.
6
∘
-673.6^{\circ}
−
673.
6
∘
\newline
(C)
−
313.
6
∘
-313.6^{\circ}
−
313.
6
∘
\newline
(D)
46.
4
∘
46.4^{\circ}
46.
4
∘
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