Ann wants to choose from two telephone plans. Plan A involves a fixed charge of $10 per month and call charges at $0.10 per minute. Plan B involves a fixed charge of $15 per month and call charges at $0.08 per minute.
Plan A $10 + .10/minute
Plan B $15 + .08/minute
If 250 minutes are used:
Plan A: $10+$25=$35
Plan B: $15+$20=$35
If 400 minutes are used:
Plan A: $10+$40=$50
Plan B: $15+$32=$47
B is the correct answer. How to test it:
Plan A: $10+(.10*249 minutes)
$10+$24.9=$34.9
Plan B: $15+(.08*249 minutes)
$15+$19.92=$34.92
Plan A < Plan B if less than 250 minutes are used.
Given:
The first two terms in an arithmetic progression are -2 and 5.
The last term in the progression is the only number in the progression that is greater than 200.
To find:
The sum of all the terms in the progression.
Solution:
We have,
First term : 
Common difference : 


nth term of an A.P. is

where, a is first term and d is common difference.

According to the equation,
.



Divide both sides by 7.

Add 1 on both sides.

So, least possible integer value is 30. It means, A.P. has 30 term.
Sum of n terms of an A.P. is
![S_n=\dfrac{n}{2}[2a+(n-1)d]](https://tex.z-dn.net/?f=S_n%3D%5Cdfrac%7Bn%7D%7B2%7D%5B2a%2B%28n-1%29d%5D)
Substituting n=30, a=-2 and d=7, we get
![S_{30}=\dfrac{30}{2}[2(-2)+(30-1)7]](https://tex.z-dn.net/?f=S_%7B30%7D%3D%5Cdfrac%7B30%7D%7B2%7D%5B2%28-2%29%2B%2830-1%297%5D)
![S_{30}=15[-4+(29)7]](https://tex.z-dn.net/?f=S_%7B30%7D%3D15%5B-4%2B%2829%297%5D)
![S_{30}=15[-4+203]](https://tex.z-dn.net/?f=S_%7B30%7D%3D15%5B-4%2B203%5D)


Therefore, the sum of all the terms in the progression is 2985.
Answer:
I think it is add 11 both sides? If it is wrong i am sooooo sorry
Step-by-step explanation:
71/100 because 71 cant be divided by anything
Let me know if you can see this photo