Ma
17.
:
A new company, Twimota, has been
Regular
steadily increasing its customer base for
Company
Customers
several months. After 2 months, it had
120
50 regular customers and after 5 months,
115
it had 125 regular customers. The table
2
shows the growth rate of several other
3 :
198
new companies. 7.RP.2
4 .
26
.245
Part A: Sort the companies into the
appropriate bins based on whether they
are gaining customers at a higher or lower rate compared to Twimota."
Assume the growth rate of each company is constant.
Months
5
4.
.8
1
T 10
ILUISW-Hill Education. Permission is granted to reproduce for classroom use.
Company 1
Company 2. Company 3
Company 4
Company 5
Lower Rate than Twimota
Higher Rate than Twimota
Part B: How many customers can Twimota reasonably expect to have after
one year assuming this trend continues?
How do I figure this put

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation:

let's cover over the formula of the vertex:

y = a(x – h)^2 + k

in this formula, (h,k) are the vertex, always remember that the positive or negative value of the vertex is based on the sign it has.

for example, if the vertex is -3,-6 then both the x and y value would have the positive sign attached to it and if it were a positive x and y value, then a negative sign would be attached to it.

with that being said, for our problem the x value is a positive one so we are looking for a number with a negative sign next to it, in this case the first and second choice.

however, for the first answer choice it isn't quite part of the formula so we know that the answer choice correctly displays the x value.

to confirm, we can see if the -1 fits correctly with the second answer choice, and as mentioned above a negative number will have a positive sign attached to it. so, for our problem we can see that -1 has been changed to +1, so we can confirm that the second answer choice is correct.

hence, our final answer is y=(x - 1)^2 + 1

i would be happy to . most people on would also love to i'm sure. if you just want someone who has done algebra before and want a little more personal relationship than just begging strangers for , i can certainly be of service. i do have college myself so i might not be 24/7.