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Math Problems
Algebra 1
Transformations of quadratic functions
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
Given the function
y
=
(
3
x
−
3
+
9
)
(
−
6
−
2
x
−
7
x
3
)
y=\left(3 x^{-3}+9\right)\left(-6-2 x-7 x^{3}\right)
y
=
(
3
x
−
3
+
9
)
(
−
6
−
2
x
−
7
x
3
)
, find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
in any form.
\newline
Answer:
d
y
d
x
=
\frac{d y}{d x}=
d
x
d
y
=
Get tutor help
Given the function
y
=
(
−
10
+
7
x
−
2
+
x
)
(
4
x
−
2
+
3
)
y=\left(-10+7 x^{-2}+x\right)\left(4 x^{-2}+3\right)
y
=
(
−
10
+
7
x
−
2
+
x
)
(
4
x
−
2
+
3
)
, find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
in any form.
\newline
Answer:
d
y
d
x
=
\frac{d y}{d x}=
d
x
d
y
=
Get tutor help
Given the function
f
(
x
)
=
(
−
5
x
3
−
8
x
2
−
1
)
(
9
x
2
+
1
)
f(x)=\left(-5 x^{3}-8 x^{2}-1\right)\left(9 x^{2}+1\right)
f
(
x
)
=
(
−
5
x
3
−
8
x
2
−
1
)
(
9
x
2
+
1
)
, find
f
′
(
x
)
f^{\prime}(x)
f
′
(
x
)
in any form.
\newline
Answer:
f
′
(
x
)
=
f^{\prime}(x)=
f
′
(
x
)
=
Get tutor help
d
y
d
x
=
−
3
y
\frac{d y}{d x}=-3 y
d
x
d
y
=
−
3
y
, and
y
=
2
y=2
y
=
2
when
x
=
0
x=0
x
=
0
.
\newline
Solve the equation.
\newline
Choose
1
1
1
answer:
\newline
(A)
y
=
2
+
e
−
3
x
y=2+e^{-3 x}
y
=
2
+
e
−
3
x
\newline
(B)
y
=
1
+
e
−
3
x
y=1+e^{-3 x}
y
=
1
+
e
−
3
x
\newline
(C)
y
=
2
e
−
3
x
y=2 e^{-3 x}
y
=
2
e
−
3
x
\newline
(D)
y
=
e
−
3
x
y=e^{-3 x}
y
=
e
−
3
x
Get tutor help
d
y
d
x
=
−
4
y
\frac{d y}{d x}=-4 y
d
x
d
y
=
−
4
y
, and
y
=
3
y=3
y
=
3
when
x
=
2
x=2
x
=
2
.
\newline
Solve the equation.
\newline
Choose
1
1
1
answer:
\newline
(A)
y
=
3
e
8
−
4
x
y=3 e^{8-4 x}
y
=
3
e
8
−
4
x
\newline
(B)
y
=
3
e
−
4
x
y=3 e^{-4 x}
y
=
3
e
−
4
x
\newline
(C)
y
=
3
e
4
−
4
x
y=3 e^{4-4 x}
y
=
3
e
4
−
4
x
\newline
(D)
y
=
6
e
−
4
x
y=6 e^{-4 x}
y
=
6
e
−
4
x
Get tutor help
d
y
d
x
=
−
4
y
\frac{d y}{d x}=-4 y
d
x
d
y
=
−
4
y
, and
y
=
3
y=3
y
=
3
when
x
=
2
x=2
x
=
2
.
\newline
Solve the equation.
\newline
Choose
1
1
1
answer:
\newline
(A)
y
=
3
e
−
4
x
y=3 e^{-4 x}
y
=
3
e
−
4
x
\newline
(B)
y
=
6
e
−
4
x
y=6 e^{-4 x}
y
=
6
e
−
4
x
\newline
(C)
y
=
3
e
8
−
4
x
y=3 e^{8-4 x}
y
=
3
e
8
−
4
x
\newline
(D)
y
=
3
e
4
−
4
x
y=3 e^{4-4 x}
y
=
3
e
4
−
4
x
Get tutor help
Consider the curve given by the equation
y
2
−
6
y
−
9
x
2
−
144
x
=
576
.
y^{2}-6 y-9 x^{2}-144 x=576 \text {. }
y
2
−
6
y
−
9
x
2
−
144
x
=
576
.
It can be shown that
d
y
d
x
=
9
(
x
+
8
)
y
−
3
.
\frac{d y}{d x}=\frac{9(x+8)}{y-3} \text {. }
d
x
d
y
=
y
−
3
9
(
x
+
8
)
.
\newline
Find the
x
x
x
-coordinate of the point where the line tangent to the curve is the
x
x
x
-axis.
\newline
x
=
x=
x
=
Get tutor help
Consider the curve given by the equation
3
x
2
+
6
x
−
y
2
+
8
y
=
16.
3 x^{2}+6 x-y^{2}+8 y=16.
3
x
2
+
6
x
−
y
2
+
8
y
=
16.
It can be shown that
d
y
d
x
=
3
(
x
+
1
)
y
−
4
\frac{d y}{d x}=\frac{3(x+1)}{y-4}
d
x
d
y
=
y
−
4
3
(
x
+
1
)
.
\newline
Find the
y
y
y
-coordinate of the point where the line tangent to the curve is the
y
y
y
-axis.
\newline
y
=
y=
y
=
Get tutor help
We are given that
\newline
d
y
d
x
=
1
−
y
2
.
\frac{d y}{d x}=\sqrt{1-y^{2}} \text {. }
d
x
d
y
=
1
−
y
2
.
\newline
Find an expression for
d
2
y
d
x
2
\frac{d^{2} y}{d x^{2}}
d
x
2
d
2
y
in terms of
x
x
x
and
y
y
y
.
\newline
d
2
y
d
x
2
=
\frac{d^{2} y}{d x^{2}}=
d
x
2
d
2
y
=
Get tutor help
Find
g
(
x
)
g(x)
g
(
x
)
, where
g
(
x
)
g(x)
g
(
x
)
is the translation
5
5
5
units up of
f
(
x
)
=
x
2
f(x)=x^2
f
(
x
)
=
x
2
.
\newline
Write your answer in the form
a
(
x
–
h
)
2
+
k
a(x–h)^2+k
a
(
x
–
h
)
2
+
k
, where
a
a
a
,
h
h
h
, and
k
k
k
are integers.
\newline
g
(
x
)
=
g(x)=
g
(
x
)
=
____
Get tutor help
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