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The answer could be 2b+3/b-2=3b/+2
Answer:
C and D
Step-by-step explanation:
Equating the line A and the parabola, we get
-3x + 2 = x² - 3x + 4
0 = x² - 3x + 4 +3x - 2
0 = x² + 2
-2 = x²
which has no real solutions. Then, the line A and the parabola don't intersect each other.
Equating the line B and the parabola, we get
-3x + 3 = x² - 3x + 4
0 = x² - 3x + 4 + 3x - 3
0 = x² + 1
-1 = x²
which has no real solutions. Then, the line B and the parabola don't intersect each other.
Equating the line C and the parabola, we get
-3x + 5 = x² - 3x + 4
0 = x² - 3x + 4 + 3x - 5
0 = x² - 1
1 = x²
√1 = x
which has 2 solutions, x = 1 and x = -1. Then, the line C and the parabola intersect each other.
Equating the line D and the parabola, we get
-3x + 6 = x² - 3x + 4
0 = x² - 3x + 4 + 3x - 6
0 = x² - 2
2 = x²
√2 = x
which has 2 solutions, x ≈ 1.41 and x ≈ -1.41. Then, the line D and the parabola intersect each other.
For
the function is undefined.
So, the limit doesn't exist.
Answer:
Here's a quick sketch of how to calculate the distance from a point P=(x1,y1,z1)
P
=
(
x
1
,
y
1
,
z
1
)
to a plane determined by normal vector N=(A,B,C)
N
=
(
A
,
B
,
C
)
and point Q=(x0,y0,z0)
Q
=
(
x
0
,
y
0
,
z
0
)
. The equation for the plane determined by N
N
and Q
Q
is A(x−x0)+B(y−y0)+C(z−z0)=0
A
(
x
−
x
0
)
+
B
(
y
−
y
0
)
+
C
(
z
−
z
0
)
=
0
, which we could write as Ax+By+Cz+D=0
A
x
+
B
y
+
C
z
+
D
=
0
, where D=−Ax0−By0−Cz0
D
=
−
A
x
0
−
B
y
0
−
C
z
0
.
This applet demonstrates the setup of the problem and the method we will use to derive a formula for the distance from the plane to the point P
P
.
Step-by-step explanation:
