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Let’s check out your problem:
Does the point
(
3
,
0
)
(3, 0)
(
3
,
0
)
satisfy the equation
y
=
x
2
+
4
x
y = x^2 + 4x
y
=
x
2
+
4
x
?
\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 the point
(
3
,
0
)
(3, 0)
(
3
,
0
)
satisfy the equation
y
=
x
2
+
4
x
y = x^2 + 4x
y
=
x
2
+
4
x
?
\newline
Choices:
\newline
(A) yes
\newline
(B) no
Substitute x-value:
Substitute the
x
x
x
-value from the point
(
3
,
0
)
(3, 0)
(
3
,
0
)
into the equation
y
=
x
2
+
4
x
y = x^2 + 4x
y
=
x
2
+
4
x
.
\newline
Calculation:
y
=
3
2
+
4
×
3
=
9
+
12
=
21
y = 3^2 + 4\times3 = 9 + 12 = 21
y
=
3
2
+
4
×
3
=
9
+
12
=
21
.
Compare y-values:
Compare the calculated
y
y
y
-value with the
y
y
y
-value from the point
(
3
,
0
)
(3, 0)
(
3
,
0
)
.
\newline
Calculation: Calculated
y
=
21
y = 21
y
=
21
, Point's
y
=
0
y = 0
y
=
0
.
More problems from Is (x, y) a solution to the nonlinear equation?
Question
If
−
x
2
=
−
5
y
2
+
y
3
-x^{2}=-5 y^{2}+y^{3}
−
x
2
=
−
5
y
2
+
y
3
then find
d
y
d
x
\frac{d y}{d x}
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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Posted 9 months ago
Question
If
y
2
=
2
x
2
+
y
y^{2}=2 x^{2}+y
y
2
=
2
x
2
+
y
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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Posted 9 months ago
Question
If
−
3
−
3
x
2
−
5
y
=
−
y
3
−
y
2
-3-3 x^{2}-5 y=-y^{3}-y^{2}
−
3
−
3
x
2
−
5
y
=
−
y
3
−
y
2
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
Posted 9 months ago
Question
If
5
x
2
−
5
−
3
y
2
=
0
5 x^{2}-5-3 y^{2}=0
5
x
2
−
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
Answer:
d
y
d
x
∣
(
4
,
5
)
=
\left.\frac{d y}{d x}\right|_{(4,5)}=
d
x
d
y
∣
∣
(
4
,
5
)
=
Get tutor help
Posted 9 months ago
Question
If
−
2
x
2
+
y
2
+
y
3
+
3
x
=
0
-2 x^{2}+y^{2}+y^{3}+3 x=0
−
2
x
2
+
y
2
+
y
3
+
3
x
=
0
then find
d
y
d
x
\frac{d y}{d x}
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
)
=
Get tutor help
Posted 9 months ago
Question
If
y
2
+
3
x
3
=
2
y
y^{2}+3 x^{3}=2 y
y
2
+
3
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
Answer:
d
y
d
x
∣
(
−
1
,
3
)
=
\left.\frac{d y}{d x}\right|_{(-1,3)}=
d
x
d
y
∣
∣
(
−
1
,
3
)
=
Get tutor help
Posted 8 months ago
Question
If
4
x
−
4
−
4
x
2
=
y
2
−
y
3
4 x-4-4 x^{2}=y^{2}-y^{3}
4
x
−
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
)
=
Get tutor help
Posted 8 months ago
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
Answer:
d
y
d
x
∣
(
1
,
−
3
)
=
\left.\frac{d y}{d x}\right|_{(1,-3)}=
d
x
d
y
∣
∣
(
1
,
−
3
)
=
Get tutor help
Posted 8 months ago
Question
If
−
y
3
−
y
+
2
+
y
2
=
x
2
-y^{3}-y+2+y^{2}=x^{2}
−
y
3
−
y
+
2
+
y
2
=
x
2
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
Posted 8 months ago
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
=
Get tutor help
Posted 9 months ago
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