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:
23.1%
Step-by-step explanation:
11/48=0.23125
Round up for percent.
0.231x100=23.1
23.1%
Answer:
D. 5x^2 + 2y^2 + 12x + 5y
Step-by-step explanation:
6x + 2y + y^2 + 6x + 2y + y^2 + 4x^2 + x^2 + y
6x + 6x = 12x
12x + 2y + y^2 + 2y + y^2 + 4x^2 + x^2 + y
2y + 2y + y = 5y
12x + 5y + y^2 + y^2 + 4x^2 + x^2
y^2 + y^2 = 2y^2
2y^2 + 4x^2 + x^2 + 12x + 5y
4x^2 + x^2 = 5x^2
2y^2 + 5x^2 + 12x + 5y
5x^2 + 2y^2 + 12x + 5y
Answer:
70/100
Step-by-step explanation:
Surface of the earth covered with water = 7/10
Equivalent fraction of 7/10 with 100 as the denominator = 70/100
This means 70% of the earth surface is covered with water
Percentage covered with land = Total percentage - percentage covered with water
= 100/100 - 70/100
= 100-70/100
= 30/100
Percentage covered with land = 30%
Answer:
y= 3x^{3} + 10
Step-by-step explanation:
* 3(x) = 3x
* 3(2) = 6
* x^{2} + 3x = 3x^{3} because you add the x's
* 6+4 = 10
hope that helps if not i can try to go more in depth :)