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:
0.2692307692
Step-by-step explanation:
14/52 = 0.2692307692
Each ream weighs 5 lb . so, reducing 10 lb means removing 2 reams from the 12. so, how do we arrange the 10 reams so that its easy to carry?
i thinks that 3 stacks will not work as it wont be a symmetric arrangement as 1 will be left out. So, 2 stacks of 5 each would be easy to carry. but the stacks should be placed in such a way that the lengths are parallel to each other and not in-line which would increase the length making it comparatively longer. its easier to hold a (2*8.5,11,2*5=17,11,17)compact box.
Answer:
3)x=-9
4)x=-2
5)x=-4
6)x=-5
7)x=-12
8)x=-11
Step-by-step explanation:
