The quick and easy answer is 3/100
Answer:
The rate at which the distance between them is changing at 2:00 p.m. is approximately 1.92 km/h
Step-by-step explanation:
At noon the location of Lan = 300 km north of Makenna
Lan's direction = South
Lan's speed = 60 km/h
Makenna's direction and speed = West at 75 km/h
The distance Lan has traveled at 2:00 PM = 2 h × 60 km/h = 120 km
The distance north between Lan and Makenna at 2:00 p.m = 300 km - 120 km = 180 km
The distance West Makenna has traveled at 2:00 p.m. = 2 h × 75 km/h = 150 km
Let 's' represent the distance between them, let 'y' represent the Lan's position north of Makenna at 2:00 p.m., and let 'x' represent Makenna's position west from Lan at 2:00 p.m.
By Pythagoras' theorem, we have;
s² = x² + y²
The distance between them at 2:00 p.m. s = √(180² + 150²) = 30·√61
ds²/dt = dx²/dt + dy²/dt
2·s·ds/dt = 2·x·dx/dt + 2·y·dy/dt
2×30·√61 × ds/dt = 2×150×75 + 2×180×(-60) = 900
ds/dt = 900/(2×30·√61) ≈ 1.92
The rate at which the distance between them is changing at 2:00 p.m. ds/dt ≈ 1.92 km/h
Answer:
Enlargement.
Scale Factor: 3
Step-by-step explanation:
Use points to find the enlargement. Typically, you will use all the points.
A(1 , 1) ⇒ A'(3 , 3)
B(2 , 1) ⇒ B'(6 , 3)
C(1 , 2) ⇒ C'(3 , 6)
D(2 , 2) ⇒ D'(6 , 6)
To find the scale factor, simply divide the Point' with the original Point. Use any number.
A'(3 , 3)/(A(1 , 1)) = 3
B'(6 , 3)/(B(2 , 1)) = 3
C'(3 , 6)/(C(1 , 2)) = 3
D'(6 , 6)/(D(2 , 2)) = 3
Your scale factor is 3.
3r-6=-21
3r-6+6=-21+6 (Add 6 to both sides...)
3r=-15
r=-5
3r-6=21
3r-6+6=21+6 (Add 6 to both sides...)
3r=27
r=9
r=-5 or r=9
