Answer:
13 ft/s
Step-by-step explanation:
t seconds after the boy passes under the balloon the distance between them is ...
d = √((15t)² +(45+5t)²) = √(250t² +450t +2025)
The rate of change of d with respect to t is ...
dd/dt = (500t +450)/(2√(250t² +450t +2025)) = (50t +45)/√(10t² +18t +81)
At t=3, this derivative evaluates to ...
dd/dt = (50·3 +45)/√(90+54+81) = 195/15 = 13
The distance between the boy and the balloon is increasing at the rate of 13 ft per second.
_____
The boy is moving horizontally at 15 ft/s, so his position relative to the spot under the balloon is 15t feet after t seconds.
The balloon starts at 45 feet above the boy and is moving upward at 5 ft/s, so its vertical distance from the spot under the balloon is 45+5t feet after t seconds.
The straight-line distance between the boy and the balloon is found as the hypotenuse of a right triangle with legs 15t and (45+5t). Using the Pythagorean theorem, that distance is ...
d = √((15t)² + (45+5t)²)
Answer:
D = 4, Z = 24
Step-by-step explanation:
6D+3Z = 96
5D+4Z = 116
multiply top by 5 and bottom by -6
top equation = 30D+15Z = 480
bottom equation = -30D-24Z = -696
Cancel 30D and -30 D
top equation = 15Z = 480
Bottom equation = -24Z = -696
15Z - 24 Z = 480 - 696
-9Z = 216
Z = 216/9
Z = 24
Now that we have the value of Z, we can substitute it in any equation to find D
6D + 3(24) = 96
6D + 72 = 96
6D = 96 - 72
6D = 24
D = 24/6
D = 4
Given:
The graph of a line segment.
The line segment AB translated by the following rule:
To find:
The coordinates of the end points of the line segment A'B'.
Solution:
From the given figure, it is clear that the end points of the line segment AB are A(-2,-3) and B(4,-1).
We have,
Using this rule, we get
Similarly,
Therefore, the endpoint of the line segment A'B' are A'(2,-6) and B'(8,-4).
