Answer:
7.5 ft³/min
Step-by-step explanation:
Let x be the depth below the surface of the water. The height, h of the water is thus h = 10 - x.
Now, the volume of water V = Ah where A = area of isosceles base of trough = 1/2bh' where b = base of triangle = 4 ft and h' = height of triangle = 1 ft. So, A = 1/2 × 4 ft × 1 ft = 2 ft²
So, V = Ah = 2h = 2(10 - x)
The rate of change of volume is thus
dV/dt = d[2(10 - x)]/dt = -2dx/dt
Since dV/dt = 15 ft³/min,
dx/dt = -(dV/dt)/2 = -15 ft³/min ÷ 2 = -7.5 ft³/min
Since the height of the water is h = 10 - x, the rate at which the water level is rising is dh/dt = d[10 - x]/dt
= -dx/dt
= -(-7.5 ft³/min)
= 7.5 ft³/min
And the height at this point when x = 8 inches = 8 in × 1 ft/12 in = 0.67 ft is h = (10 - 0.67) ft = 9.33 ft
Answer:
The answer is x=57
Step-by-step explanation:
First, we would set the number to x.
Then, we turn the words into an equation.
x+75 = 3x-39
Isolate the x.
-2x = -114
So...
x = 57
<span>There are 8 pints in 1 gallon, so 3 pints is 3/8 of a gallon </span>
Answer:
5x+3
Step-by-step explanation:
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Answer:
there is no greatest load
Step-by-step explanation:
Let x and y represent the load capacities of my truck and my neighbor's truck, respectively. We are given two relations:
x ≥ y +600 . . . . . my truck can carry at least 600 pounds more
x ≤ (1/3)(4y) . . . . . my truck carries no more than all 4 of hers
Combining these two inequalities, we have ...
4/3y ≥ x ≥ y +600
1/3y ≥ 600 . . . . . . . subtract y
y ≥ 1800 . . . . . . . . multiply by 3
My truck's capacity is greater than 1800 +600 = 2400 pounds. This is a lower limit. The question asks for an <em>upper limit</em>. The given conditions do not place any upper limit on truck capacity.
