3
x
+
2
y
>
24
3
x
+
2
y
>
24
Solve for
y
y
.
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y
>
−
3
x
2
+
12
y
>
-
3
x
2
+
12
Use the slope-intercept form to find the slope and y-intercept.
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Slope:
−
3
2
-
3
2
Y-Intercept:
12
12
Graph a dashed line, then shade the area above the boundary line since
y
y
is greater than
−
3
x
2
+
12
-
3
x
2
+
12
.
y
>
−
3
x
2
+
12
y
>
-
3
x
2
+
12
Answer:
The weight of the water in the pool is approximately 60,000 lb·f
Step-by-step explanation:
The details of the swimming pool are;
The dimensions of the rectangular cross-section of the swimming pool = 10 feet × 20 feet
The depth of the pool = 5 feet
The density of the water in the pool = 60 pounds per cubic foot
From the question, we have;
The weight of the water in Pound force = W = The volume of water in the pool given in ft.³ × The density of water in the pool given in lb/ft.³ × Acceleration due to gravity, g
The volume of water in the pool = Cross-sectional area × Depth
∴ The volume of water in the pool = 10 ft. × 20 ft. × 5 ft. = 1,000 ft.³
Acceleration due to gravity, g ≈ 32.09 ft./s²
∴ W = 1,000 ft.³ × 60 lb/ft.³ × 32.09 ft./s² = 266,196.089 N
266,196.089 N ≈ 60,000 lb·f
The weight of the water in the pool ≈ 60,000 lb·f
By exterior angle of triangle:
17x+4+8x+1=80
x=3
I converted the four months into a fraction, and got 1/3. There's 12 months in a year, and divided by 4, I got 3.