9514 1404 393
Answer:
(c) dZ/dt = 11/√544, so moving away at about 0.47 m/s
Step-by-step explanation:
The (x, y) coordinates of the car are related by the fact that they are on a circle centered at (12, 7) with a radius of 5. Then their rates of change are related by the derivative of the circle equation with respect to time.
(x -12)^2 + (y -7)^2 = 25
2(x -12)x' +2(y -7)y' = 0
y' = -(x -12)/(y -7)x'
At the time and point of interest, we have ...
y' = -(15 -12)/(11 -7)(1) = -3/4
__
The distance (z) to the observer is given by the Pythagorean theorem:
z^2 = (x -0)^2 +(y -2)^2
and its rate of change with time is ...
2z·z' = 2(x-0)x' +2(y -2)y'
The distance d at the point of interest is ...
z = √((15 -0)^2 +(11 -2)^2) = √(225 +81) = √306 = 3√34
So, the rate of change of distance to the observer at the time and point of interest is ...
z' = (x(x') +(y -2)y')/z
z' = ((15)(1) +(11-2)(-3/4))/(3√34) = (33/4)/(3√34)
z' = 11/√544 ≈ 0.47 . . . m/s
Vertical line is x=something
we see
(x,y)
(-2,3)
x=-2 is the equation
3364
= 2^2 * 29^2
So
√3364 = √(2^2 * 29^2) = 2 * 29 = 58
Answer:
m=125*n
Step-by-step explanation:
for every minute that passes 125 meals are packed so you multiply 125 by the amount of minutes(n) to get m