Answer:
0
Step-by-step explanation:
thanks for the points :)
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.
Answer:
Range: {-4, 3, 5}
Step-by-step explanation:
The range of a function includes all the set of possible output, or y-values in a given data set of a function.
Thus, the range of the function, {(2, 3), (-3, 5), (6, -4)} includes 3, 5, and -4.
This can be written as:
Range: {-4, 3, 5}
Answer:
its the first graph
Step-by-step explanation:
just did it
Answer:
Use trig ratios to find unknown sides in right triangles.
Step-by-step explanation:
