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Math Problems
Algebra 1
Transformations of quadratic functions
Factor completely:
\newline
(
4
x
+
5
)
(
2
x
+
7
)
−
(
3
x
−
2
)
(
2
x
+
7
)
2
(4 x+5)(2 x+7)-(3 x-2)(2 x+7)^{2}
(
4
x
+
5
)
(
2
x
+
7
)
−
(
3
x
−
2
)
(
2
x
+
7
)
2
\newline
Answer:
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Factor completely:
\newline
(
7
x
−
10
)
2
−
(
x
+
1
)
2
(7 x-10)^{2}-(x+1)^{2}
(
7
x
−
10
)
2
−
(
x
+
1
)
2
\newline
Answer:
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Rewrite the expression as a product of four linear factors:
\newline
(
x
2
+
x
)
2
−
18
(
x
2
+
x
)
+
72
\left(x^{2}+x\right)^{2}-18\left(x^{2}+x\right)+72
(
x
2
+
x
)
2
−
18
(
x
2
+
x
)
+
72
\newline
Answer:
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Rewrite the expression as a product of four linear factors:
\newline
(
4
x
2
+
15
x
)
2
+
20
(
4
x
2
+
15
x
)
+
99
\left(4 x^{2}+15 x\right)^{2}+20\left(4 x^{2}+15 x\right)+99
(
4
x
2
+
15
x
)
2
+
20
(
4
x
2
+
15
x
)
+
99
\newline
Answer:
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In
△
K
L
M
,
K
M
‾
\triangle \mathrm{KLM}, \overline{K M}
△
KLM
,
K
M
is extended through point
M
\mathrm{M}
M
to point
N
,
m
∠
M
K
L
=
(
3
x
+
1
)
∘
\mathrm{N}, \mathrm{m} \angle M K L=(3 x+1)^{\circ}
N
,
m
∠
M
K
L
=
(
3
x
+
1
)
∘
,
m
∠
L
M
N
=
(
7
x
−
8
)
∘
\mathrm{m} \angle L M N=(7 x-8)^{\circ}
m
∠
L
MN
=
(
7
x
−
8
)
∘
, and
m
∠
K
L
M
=
(
2
x
+
11
)
∘
\mathrm{m} \angle K L M=(2 x+11)^{\circ}
m
∠
K
L
M
=
(
2
x
+
11
)
∘
. What is the value of
x
x
x
?
\newline
Answer:
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In
△
L
M
N
,
L
N
‾
\triangle \mathrm{LMN}, \overline{L N}
△
LMN
,
L
N
is extended through point
N
\mathrm{N}
N
to point
O
,
m
∠
L
M
N
=
(
x
+
12
)
∘
\mathrm{O}, \mathrm{m} \angle L M N=(x+12)^{\circ}
O
,
m
∠
L
MN
=
(
x
+
12
)
∘
,
m
∠
M
N
O
=
(
5
x
+
10
)
∘
\mathrm{m} \angle M N O=(5 x+10)^{\circ}
m
∠
MNO
=
(
5
x
+
10
)
∘
, and
m
∠
N
L
M
=
(
2
x
+
12
)
∘
\mathrm{m} \angle N L M=(2 x+12)^{\circ}
m
∠
N
L
M
=
(
2
x
+
12
)
∘
. Find
m
∠
M
N
O
\mathrm{m} \angle M N O
m
∠
MNO
.
\newline
Answer:
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In
△
P
Q
R
,
P
R
‾
\triangle \mathrm{PQR}, \overline{P R}
△
PQR
,
PR
is extended through point
R
\mathrm{R}
R
to point
S
,
m
∠
P
Q
R
=
(
x
+
12
)
∘
\mathrm{S}, \mathrm{m} \angle P Q R=(x+12)^{\circ}
S
,
m
∠
PQR
=
(
x
+
12
)
∘
,
m
∠
R
P
Q
=
(
2
x
+
7
)
∘
\mathrm{m} \angle R P Q=(2 x+7)^{\circ}
m
∠
RPQ
=
(
2
x
+
7
)
∘
, and
m
∠
Q
R
S
=
(
7
x
−
17
)
∘
\mathrm{m} \angle Q R S=(7 x-17)^{\circ}
m
∠
QRS
=
(
7
x
−
17
)
∘
. Find
m
∠
P
Q
R
\mathrm{m} \angle P Q R
m
∠
PQR
.
\newline
Answer:
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In
△
P
Q
R
,
P
R
‾
\triangle \mathrm{PQR}, \overline{P R}
△
PQR
,
PR
is extended through point
R
\mathrm{R}
R
to point
S
,
m
∠
R
P
Q
=
(
2
x
+
19
)
∘
\mathrm{S}, \mathrm{m} \angle R P Q=(2 x+19)^{\circ}
S
,
m
∠
RPQ
=
(
2
x
+
19
)
∘
,
m
∠
Q
R
S
=
(
7
x
+
13
)
∘
\mathrm{m} \angle Q R S=(7 x+13)^{\circ}
m
∠
QRS
=
(
7
x
+
13
)
∘
, and
m
∠
P
Q
R
=
(
x
+
18
)
∘
\mathrm{m} \angle P Q R=(x+18)^{\circ}
m
∠
PQR
=
(
x
+
18
)
∘
. Find
m
∠
P
Q
R
\mathrm{m} \angle P Q R
m
∠
PQR
.
\newline
Answer:
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In
Δ
J
K
L
,
J
L
‾
\Delta \mathrm{JKL}, \overline{J L}
Δ
JKL
,
J
L
is extended through point
L
\mathrm{L}
L
to point
M
,
m
∠
J
K
L
=
(
3
x
+
3
)
∘
\mathrm{M}, \mathrm{m} \angle J K L=(3 x+3)^{\circ}
M
,
m
∠
J
K
L
=
(
3
x
+
3
)
∘
,
m
∠
L
J
K
=
(
3
x
+
17
)
∘
\mathrm{m} \angle L J K=(3 x+17)^{\circ}
m
∠
L
J
K
=
(
3
x
+
17
)
∘
, and
m
∠
K
L
M
=
(
8
x
−
16
)
∘
\mathrm{m} \angle K L M=(8 x-16)^{\circ}
m
∠
K
L
M
=
(
8
x
−
16
)
∘
. Find
m
∠
K
L
M
\mathrm{m} \angle K L M
m
∠
K
L
M
.
\newline
Answer:
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In
△
M
N
O
,
M
O
‾
\triangle \mathrm{MNO}, \overline{M O}
△
MNO
,
MO
is extended through point
O
\mathrm{O}
O
to point
P
,
m
∠
M
N
O
=
(
2
x
+
17
)
∘
\mathrm{P}, \mathrm{m} \angle M N O=(2 x+17)^{\circ}
P
,
m
∠
MNO
=
(
2
x
+
17
)
∘
,
m
∠
O
M
N
=
(
3
x
+
18
)
∘
\mathrm{m} \angle O M N=(3 x+18)^{\circ}
m
∠
OMN
=
(
3
x
+
18
)
∘
, and
m
∠
N
O
P
=
(
8
x
−
13
)
∘
\mathrm{m} \angle N O P=(8 x-13)^{\circ}
m
∠
NOP
=
(
8
x
−
13
)
∘
. Find
m
∠
M
N
O
\mathrm{m} \angle M N O
m
∠
MNO
.
\newline
Answer:
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In
△
P
Q
R
,
P
R
‾
\triangle \mathrm{PQR}, \overline{P R}
△
PQR
,
PR
is extended through point
R
\mathrm{R}
R
to point
S
,
m
∠
R
P
Q
=
(
3
x
+
15
)
∘
\mathrm{S}, \mathrm{m} \angle R P Q=(3 x+15)^{\circ}
S
,
m
∠
RPQ
=
(
3
x
+
15
)
∘
,
m
∠
P
Q
R
=
(
2
x
−
9
)
∘
\mathrm{m} \angle P Q R=(2 x-9)^{\circ}
m
∠
PQR
=
(
2
x
−
9
)
∘
, and
m
∠
Q
R
S
=
(
7
x
−
12
)
∘
\mathrm{m} \angle Q R S=(7 x-12)^{\circ}
m
∠
QRS
=
(
7
x
−
12
)
∘
. Find
m
∠
R
P
Q
\mathrm{m} \angle R P Q
m
∠
RPQ
.
\newline
Answer:
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In
△
C
D
E
,
C
E
‾
\triangle \mathrm{CDE}, \overline{C E}
△
CDE
,
CE
is extended through point
E
\mathrm{E}
E
to point
F
,
m
∠
C
D
E
=
(
2
x
+
12
)
∘
\mathrm{F}, \mathrm{m} \angle C D E=(2 x+12)^{\circ}
F
,
m
∠
C
D
E
=
(
2
x
+
12
)
∘
,
m
∠
D
E
F
=
(
7
x
−
6
)
∘
\mathrm{m} \angle D E F=(7 x-6)^{\circ}
m
∠
D
EF
=
(
7
x
−
6
)
∘
, and
m
∠
E
C
D
=
(
2
x
+
3
)
∘
\mathrm{m} \angle E C D=(2 x+3)^{\circ}
m
∠
EC
D
=
(
2
x
+
3
)
∘
. Find
m
∠
C
D
E
\mathrm{m} \angle C D E
m
∠
C
D
E
.
\newline
Answer:
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In
Δ
G
H
I
,
G
I
‾
\Delta \mathrm{GHI}, \overline{G I}
Δ
GHI
,
G
I
is extended through point I to point J,
m
∠
I
G
H
=
(
x
−
2
)
∘
\mathrm{m} \angle I G H=(x-2)^{\circ}
m
∠
I
G
H
=
(
x
−
2
)
∘
,
m
∠
H
I
J
=
(
4
x
−
18
)
∘
\mathrm{m} \angle H I J=(4 x-18)^{\circ}
m
∠
H
I
J
=
(
4
x
−
18
)
∘
, and
m
∠
G
H
I
=
(
x
+
14
)
∘
\mathrm{m} \angle G H I=(x+14)^{\circ}
m
∠
G
H
I
=
(
x
+
14
)
∘
. Find
m
∠
H
I
J
\mathrm{m} \angle H I J
m
∠
H
I
J
.
\newline
Answer:
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In
△
\triangle
△
HIJ,
m
∠
H
=
(
6
x
+
8
)
∘
,
m
∠
I
=
(
x
+
4
)
∘
\mathrm{m} \angle H=(6 x+8)^{\circ}, \mathrm{m} \angle I=(x+4)^{\circ}
m
∠
H
=
(
6
x
+
8
)
∘
,
m
∠
I
=
(
x
+
4
)
∘
, and
m
∠
J
=
(
5
x
+
12
)
∘
\mathrm{m} \angle J=(5 x+12)^{\circ}
m
∠
J
=
(
5
x
+
12
)
∘
. Find
m
∠
I
\mathrm{m} \angle I
m
∠
I
.
\newline
Answer:
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∠
1
\angle 1
∠1
and
∠
2
\angle 2
∠2
are complementary angles. If
m
∠
1
=
(
4
x
+
7
)
∘
\mathrm{m} \angle 1=(4 x+7)^{\circ}
m
∠1
=
(
4
x
+
7
)
∘
and
m
∠
2
=
(
x
+
28
)
∘
\mathrm{m} \angle 2=(x+28)^{\circ}
m
∠2
=
(
x
+
28
)
∘
, then find the measure of
∠
2
\angle 2
∠2
.
\newline
Answer:
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∠
1
\angle 1
∠1
and
∠
2
\angle 2
∠2
are vertical angles. If
m
∠
1
=
(
2
x
+
13
)
∘
\mathrm{m} \angle 1=(2 x+13)^{\circ}
m
∠1
=
(
2
x
+
13
)
∘
and
m
∠
2
=
(
7
x
+
28
)
∘
\mathrm{m} \angle 2=(7 x+28)^{\circ}
m
∠2
=
(
7
x
+
28
)
∘
, then find the measure of
∠
2
\angle 2
∠2
.
\newline
Answer:
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∠
1
\angle 1
∠1
and
∠
2
\angle 2
∠2
are vertical angles. If
m
∠
1
=
(
2
x
+
20
)
∘
\mathrm{m} \angle 1=(2 x+20)^{\circ}
m
∠1
=
(
2
x
+
20
)
∘
and
m
∠
2
=
(
6
x
+
4
)
∘
\mathrm{m} \angle 2=(6 x+4)^{\circ}
m
∠2
=
(
6
x
+
4
)
∘
, then find the measure of
∠
2
\angle 2
∠2
.
\newline
Answer:
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∠
1
\angle 1
∠1
and
∠
2
\angle 2
∠2
are vertical angles. If
m
∠
1
=
(
x
+
2
)
∘
\mathrm{m} \angle 1=(x+2)^{\circ}
m
∠1
=
(
x
+
2
)
∘
and
m
∠
2
=
(
3
x
−
8
)
∘
\mathrm{m} \angle 2=(3 x-8)^{\circ}
m
∠2
=
(
3
x
−
8
)
∘
, 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
=
(
x
+
18
)
∘
\mathrm{m} \angle 1=(x+18)^{\circ}
m
∠1
=
(
x
+
18
)
∘
and
m
∠
2
=
(
4
x
−
24
)
∘
\mathrm{m} \angle 2=(4 x-24)^{\circ}
m
∠2
=
(
4
x
−
24
)
∘
, then find the measure of
∠
1
\angle 1
∠1
.
\newline
Answer:
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Will factored
7
x
6
7x^{6}
7
x
6
as
\newline
(
3
x
2
)
(
4
x
4
)
(3x^{2})(4x^{4})
(
3
x
2
)
(
4
x
4
)
.
\newline
Olivia factored
7
x
6
7x^{6}
7
x
6
as
\newline
(
7
x
2
)
(
x
3
)
(7x^{2})(x^{3})
(
7
x
2
)
(
x
3
)
.
\newline
Which of them factored
7
x
6
7x^{6}
7
x
6
correctly?
\newline
Choose
1
1
1
answer:
\newline
(A) Only Will
\newline
(B) Only Olivia
\newline
(C) Both Will and Olivia
\newline
(D) Neither Will nor Olivia
Get tutor help
Given the function
f
(
x
)
=
−
3
(
4
x
2
+
6
x
)
6
f(x)=-3\left(4 x^{2}+6 x\right)^{6}
f
(
x
)
=
−
3
(
4
x
2
+
6
x
)
6
, find
f
′
(
x
)
f^{\prime}(x)
f
′
(
x
)
in any form.
\newline
Answer:
f
′
(
x
)
=
f^{\prime}(x)=
f
′
(
x
)
=
Get tutor help
If
−
5
+
x
2
−
y
2
+
y
=
0
-5+x^{2}-y^{2}+y=0
−
5
+
x
2
−
y
2
+
y
=
0
then find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
in terms of
x
x
x
and
y
y
y
.
\newline
Answer:
d
y
d
x
=
\frac{d y}{d x}=
d
x
d
y
=
Get tutor help
If
−
2
x
3
−
y
3
−
4
x
2
−
4
−
3
x
=
0
-2 x^{3}-y^{3}-4 x^{2}-4-3 x=0
−
2
x
3
−
y
3
−
4
x
2
−
4
−
3
x
=
0
then find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
in terms of
x
x
x
and
y
y
y
.
\newline
Answer:
d
y
d
x
=
\frac{d y}{d x}=
d
x
d
y
=
Get tutor help
If
−
5
y
3
−
x
y
−
x
=
−
2
-5 y^{3}-x y-x=-2
−
5
y
3
−
x
y
−
x
=
−
2
then find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
in terms of
x
x
x
and
y
y
y
.
\newline
Answer:
d
y
d
x
=
\frac{d y}{d x}=
d
x
d
y
=
Get tutor help
If
x
y
2
=
−
y
3
−
x
3
x y^{2}=-y^{3}-x^{3}
x
y
2
=
−
y
3
−
x
3
then find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
in terms of
x
x
x
and
y
y
y
.
\newline
Answer:
d
y
d
x
=
\frac{d y}{d x}=
d
x
d
y
=
Get tutor help
If
5
y
=
−
x
y
2
−
x
5 y=-x y^{2}-x
5
y
=
−
x
y
2
−
x
then find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
in terms of
x
x
x
and
y
y
y
.
\newline
Answer:
d
y
d
x
=
\frac{d y}{d x}=
d
x
d
y
=
Get tutor help
If
−
x
2
−
x
3
y
+
1
=
−
y
-x^{2}-x^{3} y+1=-y
−
x
2
−
x
3
y
+
1
=
−
y
then find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
in terms of
x
x
x
and
y
y
y
.
\newline
Answer:
d
y
d
x
=
\frac{d y}{d x}=
d
x
d
y
=
Get tutor help
If
−
3
x
y
2
+
x
3
=
−
y
3
-3 x y^{2}+x^{3}=-y^{3}
−
3
x
y
2
+
x
3
=
−
y
3
then find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
in terms of
x
x
x
and
y
y
y
.
\newline
Answer:
d
y
d
x
=
\frac{d y}{d x}=
d
x
d
y
=
Get tutor help
If
−
y
3
+
4
x
2
−
5
x
y
=
1
-y^{3}+4 x^{2}-5 x y=1
−
y
3
+
4
x
2
−
5
x
y
=
1
then find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
in terms of
x
x
x
and
y
y
y
.
\newline
Answer:
d
y
d
x
=
\frac{d y}{d x}=
d
x
d
y
=
Get tutor help
If
−
5
x
y
2
−
4
y
=
4
+
x
-5 x y^{2}-4 y=4+x
−
5
x
y
2
−
4
y
=
4
+
x
then find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
in terms of
x
x
x
and
y
y
y
.
\newline
Answer:
d
y
d
x
=
\frac{d y}{d x}=
d
x
d
y
=
Get tutor help
If
y
3
−
2
−
3
x
=
2
x
y
y^{3}-2-3 x=2 x y
y
3
−
2
−
3
x
=
2
x
y
then find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
in terms of
x
x
x
and
y
y
y
.
\newline
Answer:
d
y
d
x
=
\frac{d y}{d x}=
d
x
d
y
=
Get tutor help
If
−
5
x
3
−
x
y
+
3
−
2
y
=
0
-5 x^{3}-x y+3-2 y=0
−
5
x
3
−
x
y
+
3
−
2
y
=
0
then find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
in terms of
x
x
x
and
y
y
y
.
\newline
Answer:
d
y
d
x
=
\frac{d y}{d x}=
d
x
d
y
=
Get tutor help
If
−
4
+
5
x
3
=
y
3
+
4
x
2
y
-4+5 x^{3}=y^{3}+4 x^{2} y
−
4
+
5
x
3
=
y
3
+
4
x
2
y
then find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
in terms of
x
x
x
and
y
y
y
.
\newline
Answer:
d
y
d
x
=
\frac{d y}{d x}=
d
x
d
y
=
Get tutor help
If
3
x
y
−
1
=
x
3
−
2
y
3 x y-1=x^{3}-2 y
3
x
y
−
1
=
x
3
−
2
y
then find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
in terms of
x
x
x
and
y
y
y
.
\newline
Answer:
d
y
d
x
=
\frac{d y}{d x}=
d
x
d
y
=
Get tutor help
If
−
y
3
+
2
y
−
x
3
−
3
=
0
-y^{3}+2 y-x^{3}-3=0
−
y
3
+
2
y
−
x
3
−
3
=
0
then find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
at the point
(
1
,
−
2
)
(1,-2)
(
1
,
−
2
)
.
\newline
Answer:
d
y
d
x
∣
(
1
,
−
2
)
=
\left.\frac{d y}{d x}\right|_{(1,-2)}=
d
x
d
y
∣
∣
(
1
,
−
2
)
=
Get tutor help
If
y
3
−
x
2
=
−
1
y^{3}-x^{2}=-1
y
3
−
x
2
=
−
1
then find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
at the point
(
3
,
2
)
(3,2)
(
3
,
2
)
.
\newline
Answer:
d
y
d
x
∣
(
3
,
2
)
=
\left.\frac{d y}{d x}\right|_{(3,2)}=
d
x
d
y
∣
∣
(
3
,
2
)
=
Get tutor help
If
y
−
x
3
+
2
y
2
=
5
y-x^{3}+2 y^{2}=5
y
−
x
3
+
2
y
2
=
5
then find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
at the point
(
1
,
−
2
)
(1,-2)
(
1
,
−
2
)
.
\newline
Answer:
d
y
d
x
∣
(
1
,
−
2
)
=
\left.\frac{d y}{d x}\right|_{(1,-2)}=
d
x
d
y
∣
∣
(
1
,
−
2
)
=
Get tutor help
If
−
5
x
2
−
y
3
+
4
y
2
=
3
-5 x^{2}-y^{3}+4 y^{2}=3
−
5
x
2
−
y
3
+
4
y
2
=
3
then find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
at the point
(
1
,
2
)
(1,2)
(
1
,
2
)
.
\newline
Answer:
d
y
d
x
∣
(
1
,
2
)
=
\left.\frac{d y}{d x}\right|_{(1,2)}=
d
x
d
y
∣
∣
(
1
,
2
)
=
Get tutor help
If
−
4
=
x
2
+
y
3
−
3
x
-4=x^{2}+y^{3}-3 x
−
4
=
x
2
+
y
3
−
3
x
then find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
at the point
(
4
,
−
2
)
(4,-2)
(
4
,
−
2
)
.
\newline
Answer:
d
y
d
x
∣
(
4
,
−
2
)
=
\left.\frac{d y}{d x}\right|_{(4,-2)}=
d
x
d
y
∣
∣
(
4
,
−
2
)
=
Get tutor help
If
y
3
+
5
y
−
x
2
+
1
=
0
y^{3}+5 y-x^{2}+1=0
y
3
+
5
y
−
x
2
+
1
=
0
then find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
in terms of
x
x
x
and
y
y
y
.
\newline
Answer:
d
y
d
x
=
\frac{d y}{d x}=
d
x
d
y
=
Get tutor help
If
y
3
−
5
x
2
+
x
3
=
−
5
y
2
y^{3}-5 x^{2}+x^{3}=-5 y^{2}
y
3
−
5
x
2
+
x
3
=
−
5
y
2
then find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
in terms of
x
x
x
and
y
y
y
.
\newline
Answer:
d
y
d
x
=
\frac{d y}{d x}=
d
x
d
y
=
Get tutor help
If
3
y
−
y
2
−
x
3
=
−
y
3
3 y-y^{2}-x^{3}=-y^{3}
3
y
−
y
2
−
x
3
=
−
y
3
then find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
in terms of
x
x
x
and
y
y
y
.
\newline
Answer:
d
y
d
x
=
\frac{d y}{d x}=
d
x
d
y
=
Get tutor help
If
−
x
2
−
x
−
5
y
2
+
y
3
=
2
y
-x^{2}-x-5 y^{2}+y^{3}=2 y
−
x
2
−
x
−
5
y
2
+
y
3
=
2
y
then find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
in terms of
x
x
x
and
y
y
y
.
\newline
Answer:
d
y
d
x
=
\frac{d y}{d x}=
d
x
d
y
=
Get tutor help
A curve is defined by the parametric equations
x
(
t
)
=
−
10
sin
(
10
t
)
x(t)=-10 \sin (10 t)
x
(
t
)
=
−
10
sin
(
10
t
)
and
y
(
t
)
=
3
t
2
y(t)=3 t^{2}
y
(
t
)
=
3
t
2
. Find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
.
\newline
Answer:
Get tutor help
If
z
=
5
−
2
i
1
+
i
z=\frac{5-2i}{1+i}
z
=
1
+
i
5
−
2
i
, which of the following options is equivalent to
z
z
z
, where the imaginary number
i
i
i
is such that
i
2
=
−
1
i^2=-1
i
2
=
−
1
?
\newline
(A)
0
0
0
\newline
(B)
i
i
i
\newline
(C)
7
(
1
−
i
)
2
\frac{7(1-i)}{2}
2
7
(
1
−
i
)
\newline
(D)
3
−
7
i
2
\frac{3-7i}{2}
2
3
−
7
i
Get tutor help
If
x
=
3
−
−
1
x=3-\sqrt{-1}
x
=
3
−
−
1
, what is the value of
(
x
2
+
2
x
−
2
)
(x^2+2x-2)
(
x
2
+
2
x
−
2
)
?
\newline
(A)
−
3
(
1
/
2
)
-3^{(1/2)}
−
3
(
1/2
)
\newline
(B)
0
0
0
\newline
(C)
3
(
1
/
2
)
3^{(1/2)}
3
(
1/2
)
\newline
(D)
3
3
3
Get tutor help
A curve is defined by the parametric equations
x
(
t
)
=
−
2
sin
(
−
9
t
)
x(t)=-2 \sin (-9 t)
x
(
t
)
=
−
2
sin
(
−
9
t
)
and
y
(
t
)
=
−
7
t
3
+
5
t
2
+
10
t
y(t)=-7 t^{3}+5 t^{2}+10 t
y
(
t
)
=
−
7
t
3
+
5
t
2
+
10
t
. Find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
.
\newline
Answer:
Get tutor help
(
−
5
x
2
+
x
+
3
)
−
(
4
x
2
−
7
x
+
1
)
=
a
x
2
+
b
x
+
c
\left(-5 x^{2}+x+3\right)-\left(4 x^{2}-7 x+1\right)=a x^{2}+b x+c
(
−
5
x
2
+
x
+
3
)
−
(
4
x
2
−
7
x
+
1
)
=
a
x
2
+
b
x
+
c
\newline
where
a
,
b
a, b
a
,
b
, and
c
c
c
are constants. What is the value of
b
b
b
?
\newline
Choose
1
1
1
answer:
\newline
(A)
−
9
-9
−
9
\newline
(B)
−
6
-6
−
6
\newline
(C)
7
7
7
\newline
(D)
8
8
8
Get tutor help
(
−
9
x
2
+
3
x
+
9
)
−
(
−
2
x
2
+
3
x
−
2
)
=
a
x
2
+
b
x
+
c
\left(-9 x^{2}+3 x+9\right)-\left(-2 x^{2}+3 x-2\right)=a x^{2}+b x+c
(
−
9
x
2
+
3
x
+
9
)
−
(
−
2
x
2
+
3
x
−
2
)
=
a
x
2
+
b
x
+
c
\newline
where
a
,
b
a, b
a
,
b
, and
c
c
c
are constants. What is the value of
b
b
b
?
\newline
Choose
1
1
1
answer:
\newline
(A)
−
7
-7
−
7
\newline
(B)
6
6
6
\newline
(C)
0
0
0
\newline
(D)
11
11
11
Get tutor help
Given the function
f
(
x
)
=
3
−
x
2
1
+
5
x
2
f(x)=\frac{3-x^{2}}{1+5 x^{2}}
f
(
x
)
=
1
+
5
x
2
3
−
x
2
, find
f
′
(
x
)
f^{\prime}(x)
f
′
(
x
)
in simplified form.
\newline
Answer:
f
′
(
x
)
=
f^{\prime}(x)=
f
′
(
x
)
=
Get tutor help
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