Answer:
The concentration of salt in the tank approaches
Step-by-step explanation:
Data provide in the question:
Water contained in the tank = 8000 L
Salt per litre contained in Brine = 35 g/L
Rate of pumping water into the tank = 25 L/min
Concentration of salt 
Now,
Dividing both numerator and denominator by
, we have

Here,
The concentration of salt in the tank approaches
Let the distance of the first part of the race be x, and that of the second part, 15 - x, then
x/8 + (15 - x)/20 = 1.125
5x + 2(15 - x) = 40 x 1.125
5x + 30 - 2x = 45
3x = 45 - 30 = 15
x = 15/3 = 5
Therefore, the distance of the first part of the race is 5 miles and the time is 5/8 = 0.625 hours or 37.5 minutes
The distance of the second part of the race is 15 - 5 = 10 miles and the time is 1.125 - 0.625 = 0.5 hours or 30 minutes.
Answer: The correct option is (A) reduction.
Step-by-step explanation: Given that the quadrilateral A'B'C'D' is a dilation of the quadrilateral ABCD.
As shown in the given figure, the lengths of the sides of quadrilateral ABCD are as follows:
AB = 5 units, BC = 4 units, CD = 10 units and DA = 6 units.
And, the lengths of the sides of quadrilateral A'B'C'D' are as follows:

We know that the dilation will be an enlargement if the scale factor is greater than 1 and it will be a reduction if the scale factor is less than 1.
Now, the scale factor is given by

Since the scale factor is less than 1, so the dilation will be a reduction.