Saturday

March 28, 2015

March 28, 2015

Posted by **CRYSTAL** on Monday, October 24, 2011 at 10:07am.

- CALCULUS -
**Steve**, Monday, October 24, 2011 at 11:40amAfter t hours, distance d is

d^2 = (40+23t)^2 + (19t)^2

2d dd/dt = 2(40+23t)(23) + 2(19t)(19)

When t=5, d^2 = 155^2 + 95^2, d=181.8

2(181.8) dd/dt = 2(155)(23) + 2(95)(19)

363.6 dd/dt = 10740

dd/dt = 29.5 knots

check my math . . .

- CALCULUS -
**John Lee**, Thursday, March 8, 2012 at 4:36pmYour math be wrong.

**Answer this Question**

**Related Questions**

Calculus - At noon, ship A is 50 nautical miles due west of ship B. Ship A is ...

calculus - At noon, ship A is 10 nautical miles due west of ship B. Ship A is ...

calculus - At noon, ship A is 10 nautical miles due west of ship B. Ship A is ...

calculus - At noon, ship A is 10 nautical miles due west of ship B. Ship A is ...

calculus - At noon, ship A is 30 nautical miles due west of ship B. Ship A is ...

Calculus - At noon, ship A is 20 nautical miles due west of ship B. Ship A is ...

Calculus - At noon, ship A is 10 nautical miles due west of ship B. Ship A is ...

Calculus - At noon, ship A is 10 nautical miles due west of ship B. Ship A is ...

calculus 1 - At noon, ship A is 20 nautical miles due west of ship B. Ship A is ...

calculus 1 - At noon, ship A is 30 nautical miles due west of ship B. Ship A is ...