Answer:
Answer D is correct
Step-by-step explanation:
1: (1, 1.25) (2, 2)
So you have your first problem, your first number is x1 while your next number in the first set of parenthesis is y1. Your next set of parenthesis will be x2 and y2 like this:
x1 y1 x2 y2
(1, 1.25) (2, 2)
Then you set up a equation like this!
x2-x1
-------Divided
y2-y1
so we now plug in the numbers and get this
2-1.25 = 0.75
--------- ---- or 0.75 BUT not 1.25 like we need!
2-1 = 1
X-3y=6
x+y=2
this is an substitution problem
so first you can do is rewrite the problem by subjection one variable
x=3y+6
then substitute this in the other proble
x+y=2
(3y+6)+y=2
4y+6=2
4y=2-6
4y=-4
y=-1
then substitute the no. in the original equation.
x=3y+6
x=3(-1)+6
x=-3+6
x=3
now you got the intercepts and you draw the line and check.
it's in the IV quadrant
Using function concepts, it is found that it is increasing on the interval:
(–∞, –5] ∪ [3, ∞)
---------------
The function is given by:

The graph is given at the end of this question.
- If the function is pointing upwards, it is increasing. Otherwise, it is decreasing.
- In the graph, it can be seen that it is pointing upwards for x of -5 and less, or 3 and higher, thus, the interval is:
(–∞, –5] ∪ [3, ∞)
A similar problem is given at brainly.com/question/13539822
Hi there
If the amount deposited at (end) of each year, use the formula of the (future/present) value of annuity ordinary
If the amount deposited at the (beginning) of each year use the formula of the (future/present) value of annuity due
So
FvAo=5,000×(((1+0.0245)^(5)−1)
÷(0.0245))
=26,255.38...answer
Hope it helps