Here's some help on #1. Please post one question per post. You will get quicker response that way.
1. To get the profit, you need to know the dealer's cost of each cassette as well as the revenue (the amount of money taken in).
Let N = number of cassettes sold
N = 150 - 15*[(P - 10)/0.50]
= 450 - 30 P
P is the price in dollars. If P = 15, none are sold.
Revenue = R = NP = 450 P - 30 P^2
For maximum revenue, dR/dP = 0
450 - 60 P = 0
That results in P = $7.50 for maximum revenue, but you'll probably be losing money at that price. What you need is an equation for PROFIT vs. number sold, not total revenue. For that, you need the wholesale cost per cassette.
I don't get either of these equations. Please help me out!
1. Music Mart sells an average of 150 cassette tapes daily at $10.00 a tape. If 15 sales are lost for each $0.50 increase in unit price, what price will maximize profits? What will be the resulting daily profit?
2. S. Piker served a volleyball from 1 meter off the ground with an initial upward velocity of 8 meters per second. Write an equation for the height of the ball "h" in meters after "t" seconds. What will be the height of the ball after .5 seconds? What is the maximum height of the ball? If no one touches the ball, when will it hit the floor?
3 answers
1. Music Mart sells an average of 150 cassette tapes daily at $10.00 a tape. If 15 sales are lost for each $0.50 increase in unit price, what price will maximize profits? What will be the resulting daily profit?
A few punches on the calculator will quickly expose the answer.
Alternatively,
Profit, P = (150 - 15x)(10 + .5x)
Multiplying out yields 7.5x^1 75x -1500.
Solve for the unexpected result using the quadratic formula.
2. S. Piker served a volleyball from 1 meter off the ground with an initial upward velocity of 8 meters per second. Write an equation for the height of the ball "h" in meters after "t" seconds. What will be the height of the ball after .5 seconds? What is the maximum height of the ball? If no one touches the ball, when will it hit the floor?
Judicious use of the equations of motion will take you to the answers.
The relationships between initial velocity, final velocity, distance covered and time, in uniformlay accelerated motion are defined mathematically by the following three equations:
1--From a = (Vo - Vf)/t, the final velocity of a body under constant acceleration is given by Vf = Vo + at.
2--The second equation regarding accelerated motion defines the distance traveled by a body under uniform acceleration.The average velocity of a moving body during a time interval "t" is expressed by Vav = (Vo + Vf)/2. The distance traveled, "s", during this time interval "t" is the product of the average velocity and the duraion of the time interval or s = (Vo + Vf)t/2. Substituting Vf = Vo + at into this expression yields s = [Vo + (Vo + at)]t/2 or s = Vot + at^2/2.
3--The third equation of uniformly accelerated motion is derived from the first two by eliminating the time interval "t". Multiplying the two expressions results in as = (Vf - Vo)/2 x (Vo + Vf)t/2 or Vf^2 = Vo^2 + 2as.
In summary,
Vf = Vo + at
s = Vot + at^2/2
Vf^2 = Vo^2 + 2as
These same equations apply to rising and falling bodies with the exception that a is replaced by g, the acceleration due to gravity.
For rising bodies,
Vf = Vo - gt
s = Vot - gt^2/2
Vf^2 = Vo^2 - 2gs.
For falling bodies,
Vf = Vo + gt
s = Vot + gt^2/2
Vf^2 = Vo^2 + 2gs
A few punches on the calculator will quickly expose the answer.
Alternatively,
Profit, P = (150 - 15x)(10 + .5x)
Multiplying out yields 7.5x^1 75x -1500.
Solve for the unexpected result using the quadratic formula.
2. S. Piker served a volleyball from 1 meter off the ground with an initial upward velocity of 8 meters per second. Write an equation for the height of the ball "h" in meters after "t" seconds. What will be the height of the ball after .5 seconds? What is the maximum height of the ball? If no one touches the ball, when will it hit the floor?
Judicious use of the equations of motion will take you to the answers.
The relationships between initial velocity, final velocity, distance covered and time, in uniformlay accelerated motion are defined mathematically by the following three equations:
1--From a = (Vo - Vf)/t, the final velocity of a body under constant acceleration is given by Vf = Vo + at.
2--The second equation regarding accelerated motion defines the distance traveled by a body under uniform acceleration.The average velocity of a moving body during a time interval "t" is expressed by Vav = (Vo + Vf)/2. The distance traveled, "s", during this time interval "t" is the product of the average velocity and the duraion of the time interval or s = (Vo + Vf)t/2. Substituting Vf = Vo + at into this expression yields s = [Vo + (Vo + at)]t/2 or s = Vot + at^2/2.
3--The third equation of uniformly accelerated motion is derived from the first two by eliminating the time interval "t". Multiplying the two expressions results in as = (Vf - Vo)/2 x (Vo + Vf)t/2 or Vf^2 = Vo^2 + 2as.
In summary,
Vf = Vo + at
s = Vot + at^2/2
Vf^2 = Vo^2 + 2as
These same equations apply to rising and falling bodies with the exception that a is replaced by g, the acceleration due to gravity.
For rising bodies,
Vf = Vo - gt
s = Vot - gt^2/2
Vf^2 = Vo^2 - 2gs.
For falling bodies,
Vf = Vo + gt
s = Vot + gt^2/2
Vf^2 = Vo^2 + 2gs
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13 answers