Answer:
y = -14/3x + 43/3
Step-by-step explanation:
y2 - y1 / x2 - x1
5 - (-9) / 2 - 5
14 / -3
- 14/3
y = -14/3x + b
5 = -14/3(2) + b
5 = -28/3 + b
43/3 = b
y = -14/3x + 43/3
2(4 - 2r) = -2(r + 5)
First, divide both sides by -2. / Your problem should look like:
Second, regroup your terms. / Your problem should look like:
Third, add 4 to both sides. / Your problem should look like:
Fourth, add r + 5 + 4 to get r + 9. / Your problem should look like:
Fifth, subtract r from both sides. / Your problem should look like:
Sixth, subtract 2r - r to get r. / Your problem should look like:

Answer:
r = 9
Answer:
Step-by-step explanation:
i dont know the exact answer ..but i know that the hours would be a larger number and first you need to find out how many ours 1 pump takes so you do 3 divided by 7 which on the calculator its 0.4...
after this you times it but i dont know by what
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