A piecewise function f(x) is defined as shown. f(x) = StartLayout enlarged left-brace 1st Row 1st column negative five-fourths x
+ 90, 2nd column 0 less-than-or-equal-to x less-than 40 2nd row 1st column negative three-eighths x + 75, 2nd column 40 less-than-or-equal-to x less-than-or-equal-to 200 EndLayout Which table could be used to graph a piece of the function? A 2-column table has 3 rows. The first column is labeled x with entries 0, 16, 40. The second column is labeled y with entries 90, 85, 75. A 2-column table has 3 rows. The first column is labeled x with entries 0, 40, 200. The second column is labeled y with entries 90, 40, 0. A 2-column table has 3 rows. The first column is labeled x with entries 40, 120, 200. The second column is labeled y with entries 75, 30, 0. A 2-column table has 3 rows. The first column is labeled x with entries 40, 160, 200. The second column is labeled y with entries 60, 15, 0.
2 answers:
Answer:
A 2-column table has 3 rows. The first column is labeled x with entries 40, 160, 200. The second column is labeled y with entries 60, 15, 0.
Step-by-step explanation:
All of the tables have x=40 in the first column. The function value there is ...
f(x) = (-3/8)(40) +75 = -15 +75 = 60
Only the last table has y = 60 for x = 40.
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A check of the other points in that table shows they are all on the same (second) piece of the graph, as required by the problem statement. Those points are marked with a purple x in the attachment.
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Answer:
The answer is below
Step-by-step explanation:
a) The maximum capacity of he tank is 6 L and initially it contains 11 mg of salt dissolved in 3 L of water. Solution enters the tank at a rate of 3 L/hr, therefore in x hours, the amount of water that have entered the tank = 3x.
Solution also leaves the tank at a rate of 2L/hr, therefore in x hours, the amount of water that have left the tank = 2x
Hence the amount of water present in the tank at x hours is given as:
3 + 3x - 2x = 3 + x
The time taken to full the tank can be gotten from:
3 + x = 6
x = 6 - 3
x = 3 hr
b)
8 mg/ 6 L = 4/3 mg/L