Question
a ship drops its anchor into the water and creates a circular ripple. the radius of this ripple increases at a rate of 50m/s
a) find an equation for the circle 10 seconds after the anchor is dropped
i got x^2+y^2=500? is that correct
a) find an equation for the circle 10 seconds after the anchor is dropped
i got x^2+y^2=500? is that correct
Answers
almost.
x^2 + y^2 = 500^2
x^2 + y^2 = 500^2
a) The circle's radius would be:
50 x 10 = 500m,
However, if you want that 500m in the equation form (x)2 + (y)2 = r2, than you have to plug the 500m into the formula like this:
(x)2 + (y)2 = r2
(x)2 + (y)2 = 5002 (the 2 means squared, idk how to make it in superscript)
(x)2 + (y)2 = 250,000
There ya go pal!!
50 x 10 = 500m,
However, if you want that 500m in the equation form (x)2 + (y)2 = r2, than you have to plug the 500m into the formula like this:
(x)2 + (y)2 = r2
(x)2 + (y)2 = 5002 (the 2 means squared, idk how to make it in superscript)
(x)2 + (y)2 = 250,000
There ya go pal!!
Simply saying
50×10=500m
X^2+y^2=500^2
X^2+y^2=250000
50×10=500m
X^2+y^2=500^2
X^2+y^2=250000
Related Questions
A stone dropped into water produces a circular ripple which increases in radius at a rate of 1.50 m/...
A stone dropped into a still pond sends out a circular ripple whose radius increases at a constant...
A stone is dropped into a still pond sends out a circular ripple whose radius increases at a constan...
A big ship drops its anchor.
The variable l models the anchor's elevation (in meters) relative to t...