The easist way i can think of doing this would be to multipy the 22.5 miles by three so you can see how far he would go if he had one hour to travel at that average speed, the answer after multipying by 3 is 67.5 miles per hour, which in most plavces he would b considered speeding.
Answer:
8=8
Step-by-step explanation:
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Question 1
f(x) represents the distance of the cannon from the ground
Question 2
When f(x) = 0, there will two values of 'x'. The point on the positive x-axis where the graph crosses represent the point where the cannon hits the ground.
Question 3
Yes, it would. By knowing the spot where the cannon will hit the ground, we can set the net at the spot.
Question 4
f(x) = 0
-0.05 (x² - 26x -120) = 0
x² - 26x - 120 = 0
Question 5
Please refer to the table attached below
Question 6
The value of p and q that gives the correct factors are -30 and 4 since it gives p+q = -26
Question 7
Factorising the equation completely
-0.05 (x² - 26x - 120) = 0
-0.05 (x+4) (x-30) = 0
(x+4) (x-30) = 0
Question 8
Solving the equation
x+4 = 0 and x-30=0
x=-4 and x=30
Question 9
The roots of the equation is x = -4 and x = 30
Question 10
Please refer to the third graph
Question 11
Yes
Question 12
The negative zero means the initial distance of the cannon, where it was fired from
Question 13
The distance between x = -4 and x = 30 is 34 units. If the cannon was fired from the point when x = -4, the cannon will hit the ground again 34 units from the point it was fired from. If Nik put a net 30 units from the firing point, the cannon will fly pass it.
First, 5 students and 2 teachers is 7 different people. That means there are 7! Ways they can stand in a line. (7! Means 7x6x5x4x3x2x1)
That is 5040 ways.
Having the teachers on either side with the five students in the middle splits the question. There are 2 ways the teachers could stand (TA on left, TB on right or vice versa). There are 5! Ways to arrange the students in the middle. That is, 120. So combined, there are 2x120 ways to have them lined up with the teachers on either side. 240 ways.
Out of the 5040 ways for all of them to line up, one teacher will be in the middle 3 spaces 3/7 of the time. The second teacher will be on one of the remaining 2 central spaces 2/6 (1/3) of the time. 3/7x1/3 = 1/7. That means 1/7 or 14.29% of the time, the two teachers will occupy two of the three middle slots.