The answer will be 37 yd. Hope it help!
Answer:
Ok, as i understand it:
for a point P = (x, y)
The values of x and y can be randomly chosen from the set {1, 2, ..., 10}
We want to find the probability that the point P lies on the second quadrant:
First, what type of points are located in the second quadrant?
We should have a value negative for x, and positive for y.
But in our set; {1, 2, ..., 10}, we have only positive values.
So x can not be negative, this means that the point can never be on the second quadrant.
So the probability is 0.
X= 6
Y= 0
Z= -2
CHECK
3x + y -2z =22
3(6) + 0 -2(-2) =22
18 + 0 +4= 22
22=22
x+5y+z =4
6 +5(0)+ -2 =4
6 + 0 +-2 = 4
4=4
x+3z=0
6 + 3(-2) =0
6 -6 = 0
0=0
Answer:
hi thanks
Step-by-step explanation:
Answer:
10
Step-by-step explanation:
(2/3) x (15)
is the same as
(2/3) x (15/1)
which means after you multiply them you get:
30/3
which simplifies to:
10
