To determine the rates at which the inlet and outlet pipes fill and empty the reservoir, we remember that work done equals rate multiplied by time. Let’s denote the inlet rate as i and for the outlet pipe as 0. Therefore,
i(24) = 1
o(28) = 1
In this context, the '1' represents the total number of reservoirs, since the problem states the time needed for each pipe to either fill or empty a singular reservoir. Solving for rates yields:
i = 1/24 reservoirs/hour
o = 1/28 reservoirs/hour
Over the first six hours, the inlet pipe fills (1/24)(6) = 1/4 reservoirs and during the same period, the outlet pipe empties (1/28)(6) = 3/14 reservoirs. To calculate the net volume of the reservoir filled, we subtract the emptying total from the filling total:
1/4 - 3/14 = 1/28 reservoirs (note that if emptying exceeds filling, a negative value results. In such cases, treat that negative value as zero, indicating that the outlet rate surpasses the inlet rate, leading to an empty reservoir).
Now we need to find out how long it will take to fill up one reservoir since we’ve already partially filled 1/28 of it, after closing the outlet pipe. In simpler terms, we need to determine the time required for the inlet pipe to finish filling the remaining 27/28 of the reservoir. Fortunately, we have already established the filling rate for the inlet pipe, leading to the equation:
(1/24)t = 27/28
Solving for t gives us 23.14 hours. Remember to add the initial 6 hours to this result since the question seeks the total time. Thus, the final total is 29.14 hours.
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Answer:
D
Step-by-step explanation:
When you multiply 3/4 by A, B, or C, it results in a decimal value; however, multiplying 3/4 by 8 yields 6, which is a whole number.
The slope equals $0.10 (since $1.00 per 10 tokens translates to $0.10 per token)
The y-intercept is $60 (the fixed yearly membership fee)
The linear equation is y = 0.10x + 60 (following y = mx + b)
The domain consists of all x values where x ≥ 0 (negative token quantities are impossible)
The range includes all y values with y ≥ 60 (plugging the domain values into the function)
The y-intercept of this function stands at $60
Since this parabola intersects the center, its formula is:
y = ax². Given that it opens downward, the coefficient a must be negative.
Thus, the equation can be expressed as:
y = - ax², with the axis of symmetry located at x = 0.
The height measures 84 ft when the parabola's opening is 42 ft wide.
This indicates that for the height y, the corresponding x-values are +21 and -21 (due to symmetry).
To find a, let's substitute y and x with their respective values:
y = - ax²
84 = - a(21)²
84 = - a(441), leading to a = - 84/441 ↔ a = - 4/21.
Therefore, the final equation is: y = -4/21 x².
A dilation represents a transformation
, centered at point O with a scale factor of k, which cannot be zero. This transformation keeps O fixed while transforming any other point P into its image P'. Points O, P, and P' are collinear.
In a dilation of
, the scale factor
maps the original figure to its transformed image in such a way that the distances from O to points of the image are half the distances from O to the original figure. Consequently, the image's size is also half that of the original figure.
Thus, <span>If
represents the dilation of △ABC, then the properties of the image △A'B'C'</span> are:
<span>AB is parallel to A'B'.
The distance from A' to the origin is half that from A to the origin.</span>
