40 POINTS!! PLEASE!! I AM SOOO LOST!!
1 answer:
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First, lets create a equation for our situation. Let 
be the months. We know four our problem that <span>Eliza started her savings account with $100, and each month she deposits $25 into her account. We can use that information to create a model as follows:
</span>
<span>
We want to find the average value of that function </span>from the 2nd month to the 10th month, so its average value in the interval [2,10]. Remember that the formula for finding the average of a function over an interval is:
. So lets replace the values in our formula to find the average of our function:
![\frac{25(10)+100-[25(2)+100]}{10-2}](https://tex.z-dn.net/?f=%20%5Cfrac%7B25%2810%29%2B100-%5B25%282%29%2B100%5D%7D%7B10-2%7D%20)
We can conclude that <span>the average rate of change in Eliza's account from the 2nd month to the 10th month is $25.</span>
</span>
<span>
We want to find the average value of that function </span>from the 2nd month to the 10th month, so its average value in the interval [2,10]. Remember that the formula for finding the average of a function over an interval is:
We can conclude that <span>the average rate of change in Eliza's account from the 2nd month to the 10th month is $25.</span>
Answer:
first option
Step-by-step explanation:
Given
f(x) = ← factorise the numerator
= ← cancel (x + 4) on numerator/ denominator
= 2x - 3
Cancelling (x + 4) creates a discontinuity ( a hole ) at x + 4 = 0, that is
x = - 4
Substitute x = - 4 into the simplified f(x) for y- coordinate
f(- 4) = 2(- 4) - 3 = - 8 - 3 = - 11
The discontinuity occurs at (- 4, - 11 )
To obtain the zero let f(x) = 0, that is
2x - 3 = 0 ⇒ 2x = 3 ⇒ x =
There is a zero at ( , 0 )
Thus
discontinuity at (- 4, - 11 ), zero at ( , 0 )