Answer/step-by-step explanation
The soldier at point P lie on a parabola because he determined his position and distances from towns A and B through measurement of the difference in timing (phase) of radio signals received from the two towns.
This analysis of the signal time difference gives the difference in distance of the soldier at P, from the towns.
This process is known as hyperbolic navigation.
These distances of point P from towns A and B is estimated by the soldier at point P, by measuring the delay localizes the receiver to a hyperbolic line on a chart.
Two hyperbolic lines will be drawn by taking timing measurements from the
towns A and B .
Point P will be at the intersection of the lines.
These distances of point P(The soldier's positions) from town A and town B were determined using the timing of the signals received from the two towns, due to the fact that point P was on a certain hyperbola.
Answer:
y = |x-3| -1
Step-by-step explanation:
The parent function has been shifted right 3 units and down 1 unit. That is f(x)=|x| has been moved to ...
y = f(x -3) -1
y = |x -3| -1 . . . . matches the last choice
Answer:
1250 m²
Step-by-step explanation:
Let x and y denote the sides of the rectangular research plot.
Thus, area is;
A = xy
Now, we are told that end of the plot already has an erected wall. This means we are left with 3 sides to work with.
Thus, if y is the erected wall, and we are using 100m wire for the remaining sides, it means;
2x + y = 100
Thus, y = 100 - 2x
Since A = xy
We have; A = x(100 - 2x)
A = 100x - 2x²
At maximum area, dA/dx = 0.thus;
dA/dx = 100 - 4x
-4x + 100 = 0
4x = 100
x = 100/4
x = 25
Let's confirm if it is maximum from d²A/dx²
d²A/dx² = -4. This is less than 0 and thus it's maximum.
Let's plug in 25 for x in the area equation;
A_max = 25(100 - 2(25))
A_max = 1250 m²
300 is greater than 3 by 297.
297+3=300
