6(2(5) - 4)
6(10 - 4)
6(6)
36
Answer:
3 sqrt(3) =x
Step-by-step explanation:
Since this is a right triangle, we can use trig functions
cos theta = adj / hyp
cos 30 = x/6
6 cos 30 = x
6 ( sqrt(3)/2) = x
3 sqrt(3) =x
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
3 and -3
They are both the same distance from zero (definition of absolution value)
K = 70 i did 6 x -5 which was -30 then -30+100 = 70
