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:
4x - x - 10. 10 subtracted from the difference of quadruple a number and the same number means you are subtracting 10 from the difference of quadruple a number and the same number. This means you are finding the difference of quadruple a number and the number, and then subtracting 10 from that.
Answer: 7.5
Step-by-step explanation:
Answer:
chapter name and then I explained your questions
Answer:
Rock D.
Step-by-step explanation:
We can assume that the force that the catapult does is always the same.
So, here we need to remember Newton's second law:
F = m*a
force equals mass times acceleration.
Where acceleration is the rate of change of the velocity.
So, if we want the rock to hit closer to the catapult, the rock must be less accelerated than rock B.
So, we can rewrite:
a = F/m
So, as larger is the mass of the rock, smaller will be the acceleration of the rock after it leaves the catapult (because the mass is in the denominator). So if we want to have a smaller acceleration, we need to choose a rock with a larger mass than rock B.
Assuming that the mass depends on the size, the only one that has a mass larger than rock B is rock D.
So we can assume that rock D is the correct option.