The intersection point of the two lines is (2,1,0).
The respective direction vectors are
L1: <1,2,4>
L2: <4,1,15>
Since the normal vector of the required plane is perpendicular to both direction vectors, the normal vector of the plane is obtained by the cross product of L1 and L2.
i j k
1 2 4
4 1 15
=<30-4, 16-15, 1-8>
=<26, 1, -7>
We know that the plane must pass through (2,1,0), the equation of the plane is
26(x-2)+1(y-1)-7(z-0)=0
simplifying,
26x+y-7z=52+1+0=53
or
26x+y-7z=53
Check:
Put points on L1 in the plane
26(t+2)+(2t+1)-7(4t+0)=53 ok
For L2,
26(4t+2)+(t+1)-7(15t+0)=53 ok
Answer:
The end behavior of a function [f] describes the behavior of the graph of the function [f] at the ends of the x-axis.
Step-by-step explanation:
I've done this last year and this is what I can come up with
Use this website: https://www.angio.net/personal/climb/speed.html
The second equation is y=5x+2
Two angles are coterminal if they differentiate in a factor of k*(2pi). This is a multiple of 360 degrees.
In this case 21 [pi /10] = [pi / 10] + 20 [pi/10] = [pi/10] + 2pi, then 21pi/10 is coterminal with pi/10.
In the case of -pi/10 you cannot get to pi/10 by adding a multiple of 2pi , then they are not coternminals.
Conclusion: you would not be right.