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
Y intercept - 0.4
x intercept- 0.3
I would look through the numbers to determine their range. It appears to me to be 42 to 69. This suggests a frequency table with numbers 40 to 70, a range easily divided into 6 parts.
Since the width of each group will be (70 -40)/6 = 5, you have to decide on the group boundaries. It might work well to subtract 1/2 from the numbers 40, 45, 50, 55, 60, 65 to define the boundary. That way, for the integers given, you have
.. group 1 = 40 ≤ age < 45
.. group 2 = 45 ≤ age < 50
.. group 3 = 50 ≤ age < 55
.. group 4 = 55 ≤ age < 60
.. group 5 = 60 ≤ age < 65
.. group 6 = 65 ≤ age < 70
Answer:
y = 2x - 3
Step-by-step explanation:
the formula for finding equation of a line is m = (y2 - y1 )/(x2 - x1). that is (7-3)/(5-3) which gives you 4/2 = 2.
you then chose one of the point and fill in the equation y - y1 = m (x -x1)
So let's chose (3,3) and fill in.
y - 3 = 2 (x - 3)
y - 3 = 2x - 6
y - 2x = -6 + 3
y - 2x = -3
Hence, the equation of the line is y = 2x - 3