Answer: 504
Step-by-step explanation:
Answer:
We conclude that If Tawnee increases the length and width of the playground by a scale factor of 2, the perimeter of the new playground will be twice the perimeter of the original playground.
Step-by-step explanation:
We know that the perimeter of a rectangle = 2(l+w)
i.e.
P = 2(l+w)
Here
Given that the length and width of the playground by a scale factor of 2
A scale factor of 2 means we need to multiply both length and width by 2.
i.e
P = 2× 2(l+w)
P' = 2 (2(l+w))
= 2P ∵ P = 2(l+w)
Therefore, we conclude that If Tawnee increases the length and width of the playground by a scale factor of 2, the perimeter of the new playground will be twice the perimeter of the original playground.
Answer:
y = -6x + 40
kinda confused by the question but lmk if you have any questions!
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
