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Let’s check out your problem:
Does
(
0
,
6
)
(0, 6)
(
0
,
6
)
make the equation
y
=
−
2
x
2
+
−
2
x
−
−
6
y = -2x^2 + -2x - -6
y
=
−
2
x
2
+
−
2
x
−
−
6
true?
\newline
Choices:
\newline
(A)yes
\newline
(B)no
View step-by-step help
Home
Math Problems
Grade 8
Is (x, y) a solution to the nonlinear equation?
Full solution
Q.
Does
(
0
,
6
)
(0, 6)
(
0
,
6
)
make the equation
y
=
−
2
x
2
+
−
2
x
−
−
6
y = -2x^2 + -2x - -6
y
=
−
2
x
2
+
−
2
x
−
−
6
true?
\newline
Choices:
\newline
(A)yes
\newline
(B)no
Substitute Values:
Step
1
1
1
: Substitute the
x
x
x
and
y
y
y
values from the point
(
0
,
6
)
(0, 6)
(
0
,
6
)
into the equation.
\newline
y
=
–
2
x
2
+
–
2
x
−–
6
y = –2x^2 + –2x − –6
y
=
–2
x
2
+
–2
x
−–6
\newline
Substitute
x
=
0
x = 0
x
=
0
and
y
=
6
y = 6
y
=
6
:
\newline
6
=
–
2
(
0
)
2
+
–
2
(
0
)
−–
6
6 = –2(0)^2 + –2(0) − –6
6
=
–2
(
0
)
2
+
–2
(
0
)
−–6
Simplify Equation:
Step
2
2
2
: Simplify the equation to check if both sides are equal.
\newline
6
=
−
2
(
0
)
+
−
2
(
0
)
+
6
6 = -2(0) + -2(0) + 6
6
=
−
2
(
0
)
+
−
2
(
0
)
+
6
\newline
6
=
0
+
0
+
6
6 = 0 + 0 + 6
6
=
0
+
0
+
6
\newline
6
=
6
6 = 6
6
=
6
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\newline
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then find
d
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d
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\frac{d y}{d x}
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(
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1
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y
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Question
If
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5
−
3
y
2
=
0
then find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
at the point
(
4
,
5
)
(4,5)
(
4
,
5
)
.
\newline
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d
y
d
x
∣
(
4
,
5
)
=
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Question
If
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2
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d
x
d
y
at the point
(
2
,
1
)
(2,1)
(
2
,
1
)
.
\newline
Answer:
d
y
d
x
∣
(
2
,
1
)
=
\left.\frac{d y}{d x}\right|_{(2,1)}=
d
x
d
y
∣
∣
(
2
,
1
)
=
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Question
If
y
2
+
3
x
3
=
2
y
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y
2
+
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x
3
=
2
y
then find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
at the point
(
−
1
,
3
)
(-1,3)
(
−
1
,
3
)
.
\newline
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y
d
x
∣
(
−
1
,
3
)
=
\left.\frac{d y}{d x}\right|_{(-1,3)}=
d
x
d
y
∣
∣
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−
1
,
3
)
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Question
If
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4
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3
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4
x
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4
−
4
x
2
=
y
2
−
y
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
)
=
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Question
If
0
=
4
+
5
x
3
−
y
2
0=4+5 x^{3}-y^{2}
0
=
4
+
5
x
3
−
y
2
then find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
at the point
(
1
,
−
3
)
(1,-3)
(
1
,
−
3
)
.
\newline
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d
y
d
x
∣
(
1
,
−
3
)
=
\left.\frac{d y}{d x}\right|_{(1,-3)}=
d
x
d
y
∣
∣
(
1
,
−
3
)
=
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Question
If
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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
)
=
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Question
Given
y
=
−
2
sin
(
2
x
)
y=-2 \sin (2 x)
y
=
−
2
sin
(
2
x
)
, find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
.
\newline
Answer:
d
y
d
x
=
\frac{d y}{d x}=
d
x
d
y
=
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Posted 9 months ago
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