X = 4% * 350
x = 4/100 * 350
x = 4/10 * 35
x = 0.4 * 35
x = 14
14 is 4 percent of 350.
___________________
x = 4/100 * 350
x = 4/10 * 35
x = 0.4 * 35
x = 14
14 is 4 percent of 350.
___________________
PLZ HELP ME I HAVE 193 POINTS MORE IF YOU ANSWER CORRECTLY DON'T SCAM ME
1 answer:
Answer:
1. b.(03)
2. A.(6,12)
3. C.(10,-1)
4. (-9,2)
Please mark me BrainIIest
Step-by-step explanation:
1.I will solve your system by substitution.
−x+2y=6;6y=x+18
Step: Solve−x+2y=6for x:
−x+2y+−2y=6+−2y(Add -2y to both sides)
−x=−2y+6
Divide by -1
x=2y−6\
Well I did this problem differently I substitute
Step: Substitute2y−6for x in6y=x+18:
6y=x+18
6y=2y−6+18
6y=2y+12
6y+−2y=2y+12+−2y
4y=12
y=3
Step: Substitute3 for y in x=2y−6:
x=2y−6
x=(2)(3)−6
x=0
Your final answer will be x=0 and y=3
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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