Answer:
Step-by-step explanation:
f(x) = x⁵ – 8x⁴ + 16x³
As x approaches +∞, the highest term, x⁵, approaches +∞.
As x approaches -∞, x⁵ approaches -∞ (a negative number raised to an odd exponent is also negative).
Now let's factor:
f(x) = x³ (x² – 8x + 16)
f(x) = x³ (x – 4)²
f(x) has roots at x=0 and x=4. x=4 is a repeated root (because it's squared), so the graph touches the x-axis but does not cross at x=4.
The graph crosses the x-axis at x=0.
14/15=28/30
1/2=15/30
28/30-15/30=13/30
the answer is 13/30 because you need to find the LCD(least common denominator) in order to minus，the LCD of 15 and 2 is 30 so you get 28/30 minus 15/30 and you get 13/30
Answer:
y = -2.4
Step-by-step explanation:
y = 1.8 - 4.2 = -2.4
Answer: They all go into (3/2)/(4/5) = n
Step-by-step explanation:
First Blue Box n = 4/5 / 2/3 = 4/5 * 3/2 = 3/5
Second Blue Box n = 3/2 / 4/5 = 3/2 * 5/4 = 15/8
Plug in n for the 3 smaller boxed equations
Answer:
a) Let r be a random number generated in the interval (0, 1). If we multiply r by n we will obtain a random number in the interval (0, n). We take the integer part of r*n+1 to generate a random integer between 1 and n.
Suppose that the tickets are numered and we order them ascendently. We generate a random integer K from 1 to 150 and we take the Kth ticket in the ordered list of tickets to design the first passanger to go to first class. We put the remaining tickets in order and pick another one by selecting, this time, a number from 1 to 149 (because we removed one ticket). Then we do this a third time to pick a passanger to the business class by picking a number from 1 to 148 and selecting the respective ticket.
After we select the 3 affortunate passangers, we select other 6 more using this same method to rebook them to a later flight.
b) You may move randomly two passangers from business class to first class and then move 3 passangers from economy class to business class, thus, a passanger that pay for a more economic ticket wont be in a more expensive site. Also, if the tickets are numered in order of purchase, then you can, instead of making a random selection, move only passangers with the lowest numbers.
