A student has a class with four tests, each worth 100 points, and has earned these
1 answer:
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Answer:
The coordinates of the explosion is (3300, -2750)
Step-by-step explanation:
I have plot the points and attached to thus answer for easy understanding.
Now, from the question, since station A is the first to hear the explosion, we'll make it the foci of the parabola in the graph I attached and it will be horizontal since the distance between station C and A is much more than that between station B and A. Thus, the reason why station C will have to be the other foci with the hyperbola centred at the origin.
Now, sound travels at a speed of 1100 ft/s and station B is located 1100 ft from station A. Thus, the explosion would likely have occurred at a point on the line x = 3300ft . Since station A is 3300ft from centre C = 3300,hence C² = 3300² = 10,890,000. Since it takes 4 seconds longer for the sound to reach station C than A, the sound has traveled 4(1100)= 4400 ft.
Thus, 4400 = d1 = d2 = 2a
So,2a = 4400 and so, a =2200
a² = 2200² = 4,840,000 where d1 is the distance from station C to the explosion and d2 is the distance from station A to the explosion. To find b², let's use the equation ;
c² = a² + b² and so; b² = c² - a² = 10,890,000 - 4,840,000 = 6,050,000
Equation of hyperbola is given as;
(x²/a²) - (y²/b²) = 1
Plugging in the values of a² and b², we obtain ;
(x²/4,840,000) - (y²/6,050,000) = 1
Since we have deduced that the explosion must occur on the line x= 3300, we'll put in 3300 for x to obtain ;
(3300²/4,840,000) - (y²/6,050,000) = 1
2.25 - 1 = (y²/6,050,000)
y² = (1.25 x 6,050,000)
y² = 7562500
y = √7562500
y = ± 2750
Due to the fact that the explosion will occur at a point further from station B than from station A, the explosion will take place in quadrant 4. Thus, we will take the negative value of y which is - 2750.
So explosion will occur at the coordinate (3300, -2750)
Answer:
<h2>AC = 36.01</h2>Step-by-step explanation:
Given ΔABC and ΔADB, since both triangles are right angled triangles then the following are true.
From ΔADB, AB² = AD²+BD²
Given AB = 24 and AD = 16
BD² = AB² - AD²
BD² = 24²-16²
BD² = 576-256
BD² = 320
BD =
BD = 17.9
from ΔABC, AC² = AB²+BC²
SInce AC = AD+DC and BC² = BD² + DC² (from ΔBDC )we will have;
(AD+DC)² = AB²+ (BD² + DC²)
Given AD = 16, AB = 24 and BD = 17.9, on substituting
(16+DC)² = 24²+17.9²+ DC²
256+32DC+DC² = 24²+17.9²+ DC²
256+32DC = 24²+17.9²
32DC = 24²+17.9² - 256
32DC = 640.41
DC =
DC = 20.01
Remember that AC = AD+DC
AC = 16+20.01
AC = 36.01