Can someone help me with this?
2 answers:
Answer:
E. Graph E
Step-by-step explanation:
1: Find critical points of inequality.
- One critical point is 0; we have to solve for the other.
Step 1: Square both sides.
Step 2: Check if solution is correct.
2. Check the intervals between critical points.
- x ≤ 0: doesn't work
- 0 ≤ x ≤ 4: does work
- x ≥ 4: doesn't work
3. Now that we know the interval is 0 ≤ x ≤ 4, we look for the graph that matches this inequality.
- Note: This inequality means that the values include 0, 4, and the values in between them
- Given the information above, the correct answer is E because it includes 0, 4, and the numbers between.
Have a lovely rest of your day/night, and good luck with your assignments! ♡
You might be interested in
First, we need to find the length of the side of the square.
Use phytagorean theorem to find the length of the side. The side acts as hypotenuse, the distance of x and the distance of y acts as the perpendicular side of a right triangle. For clear understanding, see image attached.
General phytagorean theorem
c² = a² + b²
c represents hypotenuse, a and b are the side perpendicular to each other.
In this case, we could write it as
s² = Δx² + Δy²
s represents the length of the side, Δx represents distance of x, Δy represents distance of y
Plug in the numbers, use two of the vertices
I use (4,-1) and (7,3)
s² = Δx² + Δy²
s² = (7-4)² + (3 - (-1))²
s² = 3² + (3 + 1)²
s² = 3² + 4²
s² = 9 + 16
s² = 25
s = √25
s = 5
The length of the side is equal to 5 units length.
Second, find the area of the square
General area to find the area of a square
a = s²
Plug in the numbers
a = s²
a = 5²
a = 25
The area of the square is equal to 25 units area.
Use phytagorean theorem to find the length of the side. The side acts as hypotenuse, the distance of x and the distance of y acts as the perpendicular side of a right triangle. For clear understanding, see image attached.
General phytagorean theorem
c² = a² + b²
c represents hypotenuse, a and b are the side perpendicular to each other.
In this case, we could write it as
s² = Δx² + Δy²
s represents the length of the side, Δx represents distance of x, Δy represents distance of y
Plug in the numbers, use two of the vertices
I use (4,-1) and (7,3)
s² = Δx² + Δy²
s² = (7-4)² + (3 - (-1))²
s² = 3² + (3 + 1)²
s² = 3² + 4²
s² = 9 + 16
s² = 25
s = √25
s = 5
The length of the side is equal to 5 units length.
Second, find the area of the square
General area to find the area of a square
a = s²
Plug in the numbers
a = s²
a = 5²
a = 25
The area of the square is equal to 25 units area.