Because 3 because an input cannot be in 2 different outputs at the same time
The intersection line of two planes is the cross product of the normal vectors of the two planes.
p1: z=4x-y-13 => 4x-y-z=13
p2: z=6x+5y-13 => 6x+5y-z=13
The corresponding normal vectors are:
n1=<4,-1,-1>
n2=<6,5,-1>
The direction vector of the intersection line is the cross product of the two normals,
vl=
i j k
4 -1 -1
6 5 -1
=<1+5, -6+4, 20+6>
=<6,-2,26>
We simplify the vector by reducing its length by half, i.e.
vl=<3,-1,13>
To find the equation of the line, we need to find a point on the intersection line.
Equate z: 4x-y-13=6x+5y-13 => 2x+6y=0 => x+3y=0.
If x=0, then y=0, z=-13 => line passes through (0,0,-13)
Proceed to find the equation of the line:
L: (0,0,-13)+t(3,-1,13)
Convert to symmetric form:

=>
Answer:
-3n⁹
Step-by-step explanation:
= ![\sqrt[3]{-27n^(27)}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B-27n%5E%2827%29%7D)
= ![\sqrt[3]{-27} * \sqrt[3]{n^{27}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B-27%7D%20%2A%20%5Csqrt%5B3%5D%7Bn%5E%7B27%7D%7D)
= (-3) × n⁹
= -3n⁹
It’s 4.8
hope you have a good day
Answer:

Step-by-step explanation:
Given line = 30x - 5y = 5 or y = 6x + 1
so, the slope of the given line is 6.
now, let the line which is perpendicular to the given line be y = mx + c
where,
m = slope of the line
c = constant
As we know, if two lines are perpendicular to each other, the value of product of there slopes are -1.
so, slope of given line × slope of perpendicular line = -1
⇒ 6(m) = -1
⇒ m = 
By substitutiong the value of m in the equation, we get;
⇒ 
For c,
as the point (0,-3) passes through the line, we get;
⇒ 
⇒ 
Hence,
The line which is perpendicular to the given line and passes through (0,-3) is
.