A] David's speed during the period:
i] AB
speed=distance/time
time taken between AB=0.5 hours
distance=20 km
thus
speed=20/0.5=40 km/h
ii]BC
time taken between BC=1.5 hours
distance=20
speed=20/1.5=13 1/3 km/h
b] Average speed for the entire time will be:
average speed=(total distance)/(total time)
total distance=(20+20+20)=60 km
total time=3 hours
thus
average speed=60/3=20 km/h
c] David's distance 2 hours and 15 minutes from the starting point was:
distance=20+20+10=50 km
<span>We want to optimize f(x,y,z)=x^2 y^2 z^2, subject to g(x,y,z) = x^2 + y^2 + z^2 = 289.
Then, ∇f = λ∇g ==> <2xy^2 z^2, 2x^2 yz^2, 2x^2 y^2 z> = λ<2x, 2y, 2z>.
Equating like entries:
xy^2 z^2 = λx
x^2 yz^2 = λy
x^2 y^2 z = λz.
Hence, x^2 y^2 z^2 = λx^2 = λy^2 = λz^2.
(i) If λ = 0, then at least one of x, y, z is 0, and thus f(x,y,z) = 0 <---Minimum
(Note that there are infinitely many such points.)
(f being a perfect square implies that this has to be the minimum.)
(ii) Otherwise, we have x^2 = y^2 = z^2.
Substituting this into g yields 3x^2 = 289 ==> x = ±17/√3.
This yields eight critical points (all signage possibilities)
(x, y, z) = (±17/√3, ±17/√3, ±17/√3), and
f(±17/√3, ±17/√3, ±17/√3) = (289/3)^3 <----Maximum
I hope this helps! </span><span>
</span>
It allows you to do simple multiplicative and division problems faster
Answer:
We need more info.
Step-by-step explanation:
Answer:
21/50, 42%
Step-by-step explanation:
The total times the coin was flipped is 50 and the number of times it landed on tails is 50-29=21. So we need to divide 21 by 50, which gives us 42%.
