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.
Answer:
4th option - over the interval (4,7) the local minimum is -7
Step-by-step explanation:
There's only one local minimum in this graph and it's the one between (4,7), so this is the only plausible answer.
That's 6! (factorial). It's 720.
Answer:
5- 2h = -7 ?
Step-by-step explanation:Im not really sure if im right :(
Answer:
10 to the power of 2
Step-by-step explanation:
