Original volume = 2.5 million liters.
After evaporation, the remaining volume is 2 million liters, therefore
Evaporated volume = 2.5 - 2 = 0.5 million liters.
Percent of water evaporated = 100*(0.5/2.5) = 20%
Answer: 20%
Answer:
0.089
Step-by-step explanation:
f(x) = -ln(x) and g(x) = x²
Start by graphing the region. The two curves intersect at about (0.653, 0.426), with g(x) on the left and f(x) on the right. The region is the triangular area between the curves and above the x-axis.
If we were to cut the region horizontally (perpendicular to the y-axis), the resulting line is the width of the square cross section. The thickness of this square is dy. So the volume of the square is:
dV = A dy
dV = s² dy
dV = (x₂ − x₁)² dy
dV = (e⁻ʸ − √y)² dy
The total volume is the sum of all the squares from y=0 to y=0.426.
V = ∫ dV
V = ∫₀⁰'⁴²⁶ (e⁻ʸ − √y)² dy
Evaluate with a calculator:
V ≈ 0.089
Answer:
1 lap in 8/10 (4/5) of a minute
Answer is B, 8/10 minute! or (6a times 5)+(6a times d)
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Step-by-step explanation:
Answer:
- The equation that represents this function is f(x)=x−7.
- The range of this function is increasing as the domain increases.
Step-by-step explanation:
The line has a slope (m) of 1 and a y-intercept (b) of -7. Thus the equation in slope-intercept form is
... y = mx + b
... y = x - 7
For a linear function such as this, the size of the domain is equal to the size of the range (because the slope is 1). Thus when one increases, so does the other.
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<em>Comments on the problem</em>
It appears the first answer choice is a result of mixing up slope and intercept in the equation of the line. A slope (x-coefficient) of -7 is a pretty steep line going downward from left to right. The graph does not have that characteristic.
It is a bit unusual to talk about domain and range increasing or decreasing. The domain is the region over which the function is defined, so is generally fixed. The range is the corresponding set of values of the function, fixed once the domain is determined. Here, since the function is linear, if one were to define it over a larger region, the range of values produced by the function would also get larger.