Answer:
-0.2084
Step-by-step explanation:
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:
Answer:
y=1/2x
Step-by-step explanation:
m=1/2
y=mx+b
-1=1/2*-2+b
-1=-1+b
b=0
y=1/2x
You multiply 65 to the x and negative 5
Answer:
the answer would be 16 then you square it and it 256
