We have a three unknown, 4 equation homogeneous system. These always have at least (0,0,0) as a solution. Let's write the equations, one column at a time.
1a + 0b + 0c = 0
-1a + 1b +0c = 0
0a - 1b + c = 0
0a + 0b + -1 c = 0
We could do row reduction but these are easy enough not to bother.
Equation 1 says
a = 0
Equation 4 says
c = 0
Substituting in the two remaining,
-1(0) + 1b + 0c = 0
b = 0
0(0) - 1b + 0 = 0
b = 0
The only 3-tuple satisfying the vector equation is (a,b,c)=(0,0,0)
Answer:
less than
higher than
Step-by-step explanation:
I got it correct on edg
Answer:
O lies on the perpendicular bisector of AB. ... Then, O' will lie on the perpendicular bisector PQ and RS. We know that, two lines cannot intersect at more than one point, So O' must coincide with O. Hence, there is one and only one circle passing through three non collinear points
Step-by-step explanation:
Answer:
60
Step-by-step explanation:
you should divide 152.4 by 2.54
the result will be 60
The equation for slope is:
(y2 - y1) / (x2 - x1)
-2 -(-7) = 5
-9 -(-1) = -8
Slope of the line:
5/-8 or -5/8