Place the dock at (0, 0) in the xy-plane. At 5:00 P.M. boat A is at (0, 0). It's position after 5:00 P.M. is given by (0, -20t) where t is in hours. At 6:00 P.M. boat B is at (0, 0). That's 1 hour after boat A left the point (0, 0) so 1 h x 15 Km/h=15 Km which means at 5:00 P.M. boat B was 15 Km west of the dock at (0, 0) which means it was at (-15, 0) at 5:00 P.M. Boat B's position after 5:00 P.M. is therefore (-15+15t, 0). Use the distance formula to find the distance between the two boats.
<span>d=√((x2-x1)²+(y2-y1)²) </span>
<span>=√((-15+15t-0)²+(0+20t)²) </span>
<span>=√(225-450t+225t²+400t²) </span>
<span>=√(225-450t+625t²) </span>
<span>Find the derivative </span>
<span>d'= (1/2)(225-450t+625t²)^(-1/2)(-450+1250t) </span>
<span>Set equal to zero and you get </span>
<span>-450+1250t=0 </span>
<span>t=450/1250 </span>
<span>=0.36 h=22 minutes </span>
Answer:
a = 2(vt -d)/t^2
Step-by-step explanation:
Add the term containing "a":
d + a(t^2/2) = vt
Subtract d:
a(t^2/2) = vt -d
Multiply by the inverse of the coefficient of "a":
a = 2(vt -d)/t^2
Answer:
A
Step-by-step explanation:
- 3/4 = - 0.75
- 4.5 - 0.75 = - 5.25 = - 5 1/4
Answer:
B. lines BE and CG have to be parallel.
Step-by-step explanation:
Answer:
The coordinates of H’ is (-1,-6)
Step-by-step explanation:
Here, we want go get clockwise rotation about the origin
mathematically, if we had this kind of rotation, it would be that
(x,y) will turn to (y,-x)
So we have it that (6,-1 becomes)
(-1,-6)