Take an arbitrary vector (<em>x</em>, <em>y</em>, <em>z</em>), which goes from the origin to some point (<em>x</em>, <em>y</em>, <em>z</em>) on the plane we want to find.
Subtract from this vector, the vector that points to 
, which is (-3, 3, 1). This translates the first vector so that it starts at the point and is directed at some point (<em>x</em>, <em>y</em>, <em>z</em>). We get a new translated vector, (<em>x</em> + 3, <em>y</em> - 3, <em>z</em> - 1), which lies in the plane.
The normal vector to the plane is orthogonal to every vector in the plane. So taking the dot product of any vector in the plane with the normal to the plane will always result in 0. We use this to find the plane's equation:
and so the answer is D.
Step-by-step explanation:
if the 2 matrices are inverse, then their product must be the identity matrix
1 0
0 1
so,
m×3 + 2×-7 = 1
7×3 + 3×-7 = 0
m×-2 + 2×m = 0
7×-2 + 3×m = 1
that means we have to solve only
3m - 14 = 1
3m = 15
m = 5
Answer:
-6 ≤ y ≤ 9
Step-by-step explanation:
41º
1) Since we want to find the reference angle for a -139º angle, we need to resort to the following formula
2) Note that a -139 angle is in quadrant III, since negative angles are clockwise oriented
3) Thus the Reference Angle is 41º
Answer:
P [ K > 3.95] = 0.5633
Step-by-step explanation:
The interpretation of the given question goes thus;
Suppose that K is a random variable
P[-3.95 ≤ K ≤ 3.95] = 0.725
where; P [ + 3.95 < K ] = P [K < - 3.95]
P[K< 3.95] - P [K > - 3.95] =0.725
P [K < 3.95] - [ 1- P[K < 3.95]] = 0.725
P[k < 3.95] - 1 + P [ K < 3.95] = 0.725
3.95 P [ K < 3.95] -1 = 0.725
3.95 P [ K < 3.95] = 1.725
P [ K < 3.95] = 1.725/3.95
P [ K < 3.95] = 0.4367
P [ K > 3.95] = 1 - P[K< 3.95]
P [ K > 3.95] = 1 - 0.4367
P [ K > 3.95] = 0.5633
