Answer:
1. Slope =0.667/2.000=0.333
2.c-intercept = 5/1=5.00000
3.p-intercept=5/-3= -1.66667
I think that's the element on the second row and the first column which is 15.
Solve each of the equations independently, then determine if the are continuous or discontinuous.
15≥-3x [start here]
-5≤x [divide both sides by (-3). *Dividing by a negative number means the direction of the sign changes!]
x≥-5 [just turned around for analysis]
Next equation:
2/3x≥-2 [start here]
x≥-2(3/2) [multiply both sides of the equation by the reciprocal, 3/2)
x≥-3
So, (according to the first equation) all values of x must be greater than, or equal to -5.
(According to the second equation) all values of x must be greater than, or equal to -3.
So, when graphed on a number line, both equations graph in the same direction, so they are continuous.
The cross product of the normal vectors of two planes result in a vector parallel to the line of intersection of the two planes.
Corresponding normal vectors of the planes are
<5,-1,-6> and <1,1,1>
We calculate the cross product as a determinant of (i,j,k) and the normal products
i j k
5 -1 -6
1 1 1
=(-1*1-(-6)*1)i -(5*1-(-6)1)j+(5*1-(-1*1))k
=5i-11j+6k
=<5,-11,6>
Check orthogonality with normal vectors using scalar products
(should equal zero if orthogonal)
<5,-11,6>.<5,-1,-6>=25+11-36=0
<5,-11,6>.<1,1,1>=5-11+6=0
Therefore <5,-11,6> is a vector parallel to the line of intersection of the two given planes.
