Let y(t) represent the level of water in inches at time t in hours. Then we are given ...
y'(t) = k√(y(t)) . . . . for some proportionality constant k
y(0) = 30
y(1) = 29
We observe that a function of the form
y(t) = a(t - b)²
will have a derivative that is proportional to y:
y'(t) = 2a(t -b)
We can find the constants "a" and "b" from the given boundary conditions.
At t=0
30 = a(0 -b)²
a = 30/b²
At t=1
29 = a(1 - b)² . . . . . . . . . substitute for t
29 = 30(1 - b)²/b² . . . . . substitute for a
29/30 = (1/b -1)² . . . . . . divide by 30
1 -√(29/30) = 1/b . . . . . . square root, then add 1 (positive root yields extraneous solution)
b = 30 +√870 . . . . . . . . simplify
The value of b is the time it takes for the height of water in the tank to become 0. It is 30+√870 hours ≈ 59 hours 29 minutes 45 seconds
Because it equals to 1 either way. To divide 5/6 by 5/6 you turn a fraction upside down and multiply together. 5/6÷5/6→ 5/6×6/5= 1
In trigonometry laws, there's a equation to solve this problem :
sin (a-b) = sin(a) . cos(b) - sin (b) . cos(a),
so by assuming that a = π, b = θ, so the equation will be like this..
sin (π-θ) = sin(π) . cos(θ) - sin (θ) . cos(π),
= 0 . cos(θ) - sin(θ) . (-1)
= sin(θ) = 0.57
Hope this will help you :)
Answer:
Number 2 is correct
Step-by-step explanation
First you multiply 5 and 0
then simplify and multiply5 and 1
5*1=5
Hope this helps
