Answer:
D. 5 +6k/n
Step-by-step explanation:
The width of the interval is (5 -2) = 3. The width of one of n parts of it will be ...
3/n
Then the difference between the left end point of the interval and the value of x at the right end of the k-th rectangle will be ...
k·(3/n) = 3k/n
So, the value of x at that point is that difference added to the interval's left end:
2 + 3k/n
The value of the function for this value of x is ...
f(2 +3k/n) = 2(2 +3k/n) +1 = (4 +6k/n) +1
= 5 +6k/n
Answer:
CACULATOR: shift solve= ---> x= π
Step-by-step explanation:
What multiplied by (x-4) would make x2+5x-36
They are all the x points on the graph like the range is the y
For this case we find the slopes of each of the lines:
The g line passes through the following points:

So, the slope is:

Line h passes through the following points:

So, the slope is:

By definition, if two lines are parallel then their slopes are equal. If the lines are perpendicular then the product of their slopes is -1.
It is observed that lines g and h are not parallel. We verify if they are perpendicular:

Thus, the lines are perpendicular.
Answer:
The lines are perpendicular.