Answer:
the answer is 5.1%
Step-by-step explanation:
first you divide 37 by 39 to get .94871795
next you would subtract 1 by that number and you get
.05128205 which can be rounded to .051 or 5.1%
I think it is 4 dollars per bag of popcorn
Answer:
Equation of movement y(t)

Amplitude: 1 inch
Period: 0.628 seconds
Step-by-step explanation:
If there is no friction, the amplitude will be the length it was streched from the equilibrium. In this case, this value is 1 inch

The period depends on the mass and the spring constant.
The formula for the period is:

The model y(t) for the movement of the mass-spring system is

Answer:
Square root of 3
Step-by-step explanation:
Answer:
Part 4) 
Part 10) The angle of elevation is 
Part 11) The angle of depression is 
Part 12)
or 
Part 13)
or 
Step-by-step explanation:
Part 4) we have that

The angle theta lies in Quadrant II
so
The sine of angle theta is positive
Remember that

substitute the given value




Part 10)
Let
----> angle of elevation
we know that
----> opposite side angle theta divided by adjacent side angle theta

Part 11)
Let
----> angle of depression
we know that
----> opposite side angle theta divided by hypotenuse


Part 12) What is the exact value of arcsin(0.5)?
Remember that

therefore
-----> has two solutions
----> I Quadrant
or
----> II Quadrant
Part 13) What is the exact value of 
The sine is negative
so
The angle lies in Quadrant III or Quadrant IV
Remember that

therefore
----> has two solutions
----> IV Quadrant
or
----> III Quadrant