Find an equation of the plane that passes through the points p, q, and r. p(7, 2, 1), q(6, 3, 0), r(0, 0, 0)
Alona [7]
Answer:
x - 2y - 3z = 0
Step-by-step explanation:
The cross product of vectors rp and rq will give a vector that is normal to the plane:
... rp × rq = (-3, 6, 9)
Dividing this by -3 (to reduce it and make the x-coefficient positive) gives a normal vector to the plane of (1, -2, -3). Usint point r as a point on the plane, we find the constant in the formula to be zero. Hence, your equation can be written ...
... x -2y -3z = 0
Answer:
Correct!
Step-by-step explanation:
First, I'll try to solve it myself.
2(2a-1)+3a=7a-2
7a-2
4a-2=7a-2
His is very correct!
Answer: Yes
Step-by-step explanation: A triangle can be formed because A+B>C
A and B are 2.9 and 4.8
Since those numbers have the least value
I added both and found that A and B are equal to 7.7
7.7>7.4
These sides could form a triangle
Answer:
The x intercept is (-5,0)
The y intercept is (0,-3)
Step-by-step explanation:
To find the x intercept we set y=0 and solve for x
3x+5y = -15
3x+0 = -15
3x = -15
Divide by 3
3x/3 = -15/3
x = -5
The x intercept is (-5,0)
To find the x intercept we set x=0 and solve for y
3x+5y = -15
0+5y = -15
5y = -15
Divide by 5
5y/5 = -15/5
y = -3
The y intercept is (0,-3)
