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>
Answer: The correct option is A.
Step-by-step explanation: We are given a polynomial which is a sum of other 2 polynomials.
We are given the resultant polynomial which is : 
One of the polynomial which are added up is : 
Let the other polynomial be 'x'
According to the question:


Solving the like terms in above equation we get:


Hence, the correct option is A.
Answer with Step-by-step explanation:
We are given that


We have to find the speed of river's current and the direction of current flowing.

Speed of river's current=
Speed of river's current=
Speed of river's current=
By using the formula

Direction of current flowing,

The angle lies in third quadrant therefore
