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:127
all you do is subtract -70 by 57 and it will give you a negative but get rid of it and you get 127 or just take the - away from -70 and add 57
Answer: 11 year
P(1) = 37,100
P(4) = 58,400
The linear equation (for x ≥ 1)
P(x) = 37,100 + a(x-1)
For x = 4
58,400 = 37,100 + a(4-1)
58,400 - 37,100 = 3a
21300 = 3a
a = 7100
So, the linear equation:
P(x) = 37100 + 7100*(x-1)
P(x) = 37100 + 7100x - 7,100
P(x) = 7100x + 30000
To find when the profit should reach 108100, we can substitute P(x) by 108100.
108100 = 7100x + 30000
108100 - 30000 = 7100x
78100 = 7100x
x = 78100/7100
x = 11
Answer: 11 year
Answer: 0.7, 0.251, 0.18, 0.171
Step-by-step explanation:
Suppose we choose
and
. Then
Now suppose we choose
such that
where we pick the solution for this system such that
. Then we find
Note that you can always find a solution to the system above that satisfies
as long as
. What this means is that you can always find the value of
as a (constant) function of
.
