Answer: B or the Second graph right next to the first one.
Step-by-step explanation: Graph by using a few selected points... Horizontal Asymptote - y=7.
x----------y
0 -118
I believe the answer is 17.
<span>x^2/8 - x/4 + 1/8 = 0
A parabola is defined as the set of all points such that each point has the same distance from the focus and the directrix. Also the parabola's equation will be a quadratic equation of the form ax^2 + bx + c. So if we can determine 3 points on the parabola, we can use those points to calculate the desired equation.
First, let's draw the shortest possible line from the focus to the directrix. The midpoint of that line will be a point on the desired parabola. Since the slope of the directrix is 0, the line will have the equation of x=1. This line segment will be from (1,2) to (1,-2) and the midpoint will be ((1+1)/2, (2 + -2)/2) = (2/2, 0/2) = (1,0).
Now for the 2nd point, let's draw a line that's parallel to the directrix and passing through the focus. The equation of that line will be y=2. Any point on that line will have a distance of 4 from the directrix. So let's give it an x-coordinate value of (1+4) = 5. So another point for the parabola is (5,2). And finally, if we subtract 4 instead of adding 4 to the x coordinate, we can get a third point of 1-4 = -3. So that 3rd point is (-3,2).
So we now have 3 points on the parabola. They are (1,0), (5,2), and (-3,2). Let's create some equations of the form ax^2 + bx + c = y and then substitute the known values into those equations. SO
ax^2 + bx + c = y
(1) a*1^2 + b*1 + c = 0
(2) a*5^2 + b*5 + c = 2
(3) a*(-3)^2 + b*(-3) + c = 2
Let's do the multiplication for those expressions. So
(4) a + b + c = 0
(5) 25a + 5b + c = 2
(6) 9a - 3b + c = 2
Equations (5) and (6) above look interesting. Let's subtract (6) from (5). So
25a + 5b + c = 2
- 9a - 3b + c = 2
= 16a + 8b = 0
Now let's express a in terms of b.
16a + 8b = 0
16a = -8b
a = -8b/16
(7) a = -b/2
Now let's substitute the value (-b/2) for a in expression (4) above. So
a + b + c = 0
-b/2 + b + c = 0
And solve for c
-b/2 + b + c = 0
b/2 + c = 0
(8) c = -b/2
So we know that a = -b/2 and c = -b/2. Let's substitute those values for a and c in equation (5) above and solve for b.
25a + 5b + c = 2
25(-b/2) + 5b - b/2 = 2
-25b/2 + 5b - b/2 = 2
2(-25b/2 + 5b - b/2) = 2*2
-25b + 10b - b = 4
-16b = 4
b = -4/16
b = -1/4
So we now know that b = -1/4. Using equations (7) and (8) above, let's calculate a and c.
a = -b/2 = -(-1/4)/2 = 1/4 * 1/2 = 1/8
c = -b/2 = -(-1/4)/2 = 1/4 * 1/2 = 1/8
So both a and c are 1/8. So the equation for the parabola is
x^2/8 - x/4 + 1/8 = 0
Let's test to make sure it works. First, let's use an x of 1.
x^2/8 - x/4 + 1/8 = y
1^2/8 - 1/4 + 1/8 = y
1/8 - 1/4 + 1/8 = y
1/8 - 2/8 + 1/8 = y
0 = y
And we get 0 as expected. Let's try x = 2
x^2/8 - x/4 + 1/8 = y
2^2/8 - 2/4 + 1/8 = y
4/8 - 1/2 + 1/8 = y
4/8 - 1/2 + 1/8 = y
1/2 - 1/2 + 1/8 = y
1/8 = y.
Let's test if (2,1/8) is the same distance from both the focus and the directrix. The distance from the directrix is 1/8 - (-2) = 1/8 + 2 = 1/8 + 16/8 = 17/8
The distance from the focus is
d = sqrt((2-1)^2 + (1/8-2)^2)
d = sqrt(1^2 + -15/8^2)
d = sqrt(1 + 225/64)
d = sqrt(289/64)
d = 17/8
And the distances match again. So we do have the correct equation of:
x^2/8 - x/4 + 1/8 = 0</span>
So first calculate the time between the plane flied over the Air Force Bases (AFB) and Navy Base Station (NBS):
(time passed Air Force Bases) - (time passed Air Force Bases) = 10:32 - 10:20 = 12 mins
Since we know the distance that plane travel from AFB to NBS is 120 miles and the plane traveled that distance within 12 minutes.
We have
120 miles / 12 minutes = 10 miles / minute
Calculate it into miles per hour (60 minutes):
10 miles/minute x 60 minutes = 600 miles per hour.
ANSWER : 600 mph
