Answer:
2x - y - 3z = 0
Step-by-step explanation:
Since the set
{i, j} = {(1,0), (0,1)}
is a base in 
and F is linear, then
<em>{F(1,0), F(0,1)} </em>
would be a base of the plane generated by F.
F(1,0) = a+b = (2i-2j+k)+(i+2j+k) = 3i+2k
F(0,1) = a+c = (2i-2j+k)+(2i+j+2k) = 4i-j+3k
Now, we just have to find the equation of the plane that contains the vectors 3i+2k and 4i-j+3k
We need a normal vector which is the cross product of 3i+2k and 4i-j+3k
(3i+2k)X(4i-j+3k) = 2i-j-3k
The equation of the plane whose normal vector is 2i-j-3k and contains the point (3,0,2) (the end of the vector F(1,0)) is given by
2(x-3) -1(y-0) -3(z-2) = 0
or what is the same
2x - y - 3z = 0
Answer:
thank you for the points
Step-by-step explanation:
fr
Answer:
1st picture: (0,4)
The lines intersect at point (0,4).
2nd picture: Graph D
2x ≥ y - 1
2x - 5y ≤ 10
Set these inequalities up in standard form.
y ≤ 2x + 1
-5y ≤ 10 - 2x → y ≥ -2 + 2/5x → y ≥ 2/5x - 2
When you divide by a negative number, you switch the inequality sign.
Now you have:
y ≤ 2x + 1
y ≥ 2/5x - 2
Looking at the graphs, you first want to find the lines that intersect the y-axis at (0, 1) and (0, -2).
In this case, it is all of them.
Next, you would look at the shaded regions.
The first inequality says the values are less than or equal to. So you look for a shaded region below a line. The second inequality says the values are greater than or equal to. So you look for a shaded region above a line.
That would mean Graph B or D.
Then you look at the specific lines. You can see that the lower line is y ≥ 2/5x - 2. You need a shaded region above this line. You can see the above line is y ≤ 2x + 1. You need a shaded region below this line. That is Graph D.
Answer: ^^7 (76-87 x 834 =876*#+9647) - 974 (0 +142 = (8x 17/906385) )
Step-by-step explanation:
you are to nt9 smart to answer this question
T=90-Z
Because 9 solved correctly*10 points per question answered correctly=90 and each incorrect question is a point lost
