Answer:
Note that orthogonal to the plane means perpendicular to the plane.
Step-by-step explanation:
-1x+3y-3z=1 can also be written as -1x+3y-3z=0
The direction vector of the plane -1x+3y-3z-1=0 is (-1,3,-3).
Let us find a point on this line for which the vector from this point to (0,0,5) is perpendicular to the given line. The point is x-0,y-0 and z-0 respectively
Therefore, the vector equation is given as:
-1(x-0) + 3(y-0) + -3(z-5) = 0
-x + 3y + (-3z+15) = 0
-x + 3y -3z + 15 = 0
Multiply through by - to get a positive x coordinate to give
x - 3y + 3z - 15 = 0
Here is your Answer
Step-by-step explanation:
Ravi bought rice = 50kg
Rate it =40kg
Sold it at the rate = = 44kg
<h3>Answer: -34</h3><h2 /><h2 /><h2>Mark me as brainlist and give me 20 thx pls</h2>
A differential equation is one that contains the derivative of a function. For example f(x) + 3 = 4f'(x) - 2. Usually you will be given one of the functions and initial conditions if you have to solve.
We want to know the time, <em>t</em>, it takes the ball to reach a height (<em>y</em>) of 0.

We can factor out the GCF first. The largest number that will divide evenly into 16 and 24 is 8. Also, both terms have a <em>t</em>, so we can factor that out as well:

(-16/8 = -2 and 24/8 = 3)
Using the zero product property, we know that either 8t=0 or -2t+3=0. Solving the first equation, we would divide both sides by 8:
8t/8=0/8
t=0
This is at 0 seconds, before the ball is in the air at all.
Solving the second equation, we start by subtracting 3 from both sides:
-2t+3-3=0-3
-2t=-3
Now we divide both sides by -2
-2t/-2=-3/-2
t=1.5
After 1.5 seconds, the ball will hit the ground again.
* J,K,&L are coplanar
* K is the midpoint of JL
* J,K,&L are collinear
* Jk=KL