Answer:
-816.8 ft²/min
Step-by-step explanation:
The rate of change of area is the product of the rate of change of radius and the circumference of the circle. At the time of interest, ...
dA/dt = C·dr/dt = 2πr·dr/dt
dA/dt = 2π·(26 ft)(-5 ft/min)
dA/dt = -260π ft²/min ≈ -816.8 ft²/min
$310 dollars because after it res and decreased it gained $.10 and 100 x 3 is 300 so thats before you calculate how much it rose so .10 x 100 is 10 so add that to 300 and you get 310
Answer:
3
Step-by-step explanation:
Since it's at the same time of the day, the ratio between the height of the person and the shadow they cast will stay the same. So the man's height to shadow ratio is 6:8 = 3:4. The son's height to shadow ratio would be the same so x:4 = 3:4 therefore his height is 3 feet.
60 = a * (-30)^2
a = 1/15
So y = (1/15)x^2
abc)
The derivative of this function is 2x/15. This is the slope of a tangent at that point.
If Linda lets go at some point along the parabola with coordinates (t, t^2 / 15), then she will travel along a line that was TANGENT to the parabola at that point.
Since that line has slope 2t/15, we can determine equation of line using point-slope formula:
y = m(x-x0) + y0
y = 2t/15 * (x - t) + (1/15)t^2
Plug in the x-coordinate "t" that was given for any point.
d)
We are looking for some x-coordinate "t" of a point on the parabola that holds the tangent line that passes through the dock at point (30, 30).
So, use our equation for a general tangent picked at point (t, t^2 / 15):
y = 2t/15 * (x - t) + (1/15)t^2
And plug in the condition that it must satisfy x=30, y=30.
30 = 2t/15 * (30 - t) + (1/15)t^2
t = 30 ± 2√15 = 8.79 or 51.21
The larger solution does in fact work for a tangent that passes through the dock, but it's not important for us because she would have to travel in reverse to get to the dock from that point.
So the only solution is she needs to let go x = 8.79 m east and y = 5.15 m north of the vertex.
