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:
No, it does satisfy the Pythagorean Theorem
Step-by-step explanation:
In right triangle a² + b² = c²
Where a and b = legs and c = hypotenuse
The given measures of the legs are 3 and 7 and the given measure of the hypotenuse is √57
Which means that if this is a right triangle then 3² + 7² = √57²
3² = 9
7² = 49
√57² = 57
9 + 49 = 58
we're left with 57 ≠ 58 which is not true, meaning that , that is not a right triangle
Answer:
v = 48
Step-by-step explanation:
8 = v/6
Step 1: Multiply both sides by 6.
8 × 6 = v/6 × 6
v/6 cancels out and becomes v
48 = v
Step 2: Flip the equation.
v = 48
2(x-3)+9=3(x+1)+x
2x-6+9=3x+3+x
2x+3=4x+3
-2x -2x
3=2x+3
-3 -3
0=2x
x=0
Answer:
A
Step-by-step explanation:
A does. There is a gap at zero. A and D are your only real choices. Both show what happens at zero with minor variations and the devil is in the details. D shows that x is open ended when x = 0. That is not correct. The value should be x≥0 so y=2x is closed at x=0.
D is incorrect.
The answer is A.
