The interval where the function is nonlinear and decreasing is 0 < x < 4
<h3>How to determine the interval where the function is nonlinear and decreasing?</h3>
The straight lines on the graph are the intervals where the graph is linear
This means that the straight lines on the graph will not be considered
Considering the curve, the graph decrease from x = 0 to x = 4
This can be rewritten as:
0 < x < 4
Hence, the interval where the function is nonlinear and decreasing is 0 < x < 4
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Answer:
In 2003, the population was 59000 and the population has been growing by 1,700 people each year.
A.
The equation will be:
59000+1700x = (population 'x' years after 2003)
For x, you plug in the amount of years after 2003.
Like if it is the year 2003, the population is 
= 59000
when it is year 2005, the population is 
= 62400
B.
The town's population in 2007 will be :
Population = 65800
C.
=> 
x = 11
Means 
Hence, by year 2014 the population will be 77700.
Answer:Irrational
Step-by-step explanation: There is not a number you can square to get 23. The answer would be a repeating number, being irrational.
Answer:
8+pi/2
Step-by-step explanation:
The area of the rectangle on the bottom is 8, or 2*4. The area of the top is half (because it's a semi circle) of pi*r^2, or just 1/2 pi
Answer:
A
Step-by-step explanation:
From what we have , it is expected that the parent function is a v-shaped graph that starts from the origin
The expected equation parent function should be;
y = |x|
Now, moving on, the transformed function is 1/2 of the parent function
This shows a direct compression of the initial partner function
So, it is expected that the transformed function is more compressed compared to the initial parent function
so, the correct choice of answer here is first option
