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:

Step-by-step explanation:
Isolate the variable by doing operations to both sides of the equation.
The work is shown below:

Answer:
diabetes
Step-by-step explanation:
Notice that the given info consists of boundary lines for an area in the xy plane. We are interested ONLY in values of x and y that are 0 or greater (positive). Graph 5x + 3y

37 and 3x + 5y less than or equal to 35.
Find the points of intersection of all four straight lines (including the x- and y-axes). There will be 4 such points (incl. the origin).
Next, evaluate the objective function 2x + 14y at each of these 4 points. Which of the four results is the largest? the smallest? Label them as 'maximum' and 'minimum.'
Questions? Just comment on this discussion.
5022 ÷ 18 = 279
your tell you mom to collect 279