Asked by joseph
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
Answered by
oobleck
almost.
x^2 + y^2 = 500^2
x^2 + y^2 = 500^2
Answered by
Mohammad
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!!
Answered by
Pearl
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
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.