First find the domain(the set of all first coordinates in a function) by finding where the equation(a mathematical statement that says two expressions have the same value; any number sentence with an = sign) is defined.
After that you should know what to do. I don't know how to work this problem cuz i'm barely in 9 grade. But somehow, I know this.
(−∞,∞)
{x|x∈R}
If this helped, please let me know and give me brainliest, please!!
If this is incorrect, please let me know so I can fix it!
350 divided by 7 is 50 so the man can run 50 feet in one minute
The answer is (3600 - 900π) ft²
Step 1. Find the radius r of circles.
Step 2. Find the area of the portion of the field that will be watered by the sprinklers (A1)
Step 3. Find the total area of the field (A2)
Step 4. Find the area of the portion of the field that will not be watered by the sprinklers (A)
Step 1. Find the radius r of circles
r = ?
According to the image, radius of a square is one fourth of the field side length:
r = s/4
s = 60 ft
r = 60/4 = 15 ft
Step 2. Find the area of the portion of the field that will be watered by the sprinklers.
The area of the field that will be watered by the sprinklers (A1) is actually total area of 4 circles with radius 15 ft.
Since the area of a circle is π r², then A1 is:
A1 = 4 * π r² = 4 * π * 15² = 900π ft²
Step 3. Find the total area of the field (A2)
The field is actually a square with side s = 60 ft.
A2 = s² = 60² = 3600 ft²
Step 4. Find the area of the portion of the field that will not be watered by the sprinklers (A).
To get the area of the portion of the field that will not be watered by the sprinklers (A) we need to subtract the area of 4 circles from the total area:
A = A2 - A1
A = (3600 - 900π) ft²
Answer:
Step-by-step explanation:
You should use a ^ to indicate an exponent. x^2 is the same as x².
x² + 9x + 20
Replace the x with (-5).
(-5)² + 9(-5) + 20 = 25-45+20 = 0
Then do the same with x = (-3):
(-3)² + 9(-3) + 20 = 9 - 27 + 20 = 2
You can do x = (-6)
B; make sure to make the denominators equal before adding
