Answer:
n = 133
Step-by-step explanation:
The sum of n terms of an arithmetic sequence is ...
Sn = n(2·a1 +d(n -1))/2
You want ...
n(10 +1/2(n -1))/2 > 5000 . . . . . filling in a1=5, d=1/2
(n^2 +19n)/4 > 5000 . . . . . . . . .simplify
n^2 +19n -20000 > 0 . . . . . . . write in standard form
We can solve this quadratic using the quadratic formula ...
n > (-19 +√(19^2 -4(1)(-20000)))/2 = (-19 +√80361)/2 ≈ 132.24
The least value of n such that the sum of n terms exceeds 5000 is ...
n = 133.
Answer: use google calculator It’s reliable
Step-by-step explanation:
Answer: 1.8kg
55.5-53.8= 1.7 in the first weeks
3.5- 1.7= 1.8 for final week
<u> Solution-</u>
The given function is,
Therefore, at x=0, -1, 1 , f(x) will be 0 . Hence, 0, -1 ,1 are the x-intercepts.
Plotting the graph on desmos, the graph will be as in the attachment.
