The quantity demanded per month, x, of a certain make of personal computer (PC) is related to the average unit price, p (in dollars), of PCs by the equationx = f(p) =1009810,000 − p2It is estimated that t mo from now, the average price of a PC will be given byp(t) =4001 +18t+ 200 (0 ≤ t ≤ 60)dollars. Find the rate at which the quantity demanded per month of the PCs will be changing 16 mo from now. (Round your answer to one decimal place.)

Answers

Answer 1

Answer:

Answer:

The value of dx/dt is $(100p)/(9√((810000-p^2))).(1600)/(√(t)\left(√(t)+8\right)^2)\n$ while the rate of change of quantity demanded per month after 16 months is 18.7.

Step-by-step explanation:

From the given data the equation of quantity demanded for average price is given as is given as

$x=f(p)=(100)/(9)√(810000-p^2)$

The equation of average price p for a given value of t is given as

$p(t)=(400)/(1+(1)/(8)√(t))+200$

Now in order to determine the rate at which the quantity demanded will be changing is given as $(dx)/(dt)$

This is found by using the chain rule as

$(dx)/(dt)=(dx)/(dp).(dp)/(dt)$

Now

$(dx)/(dp)=(d((100)/(9)√(810000-p^2)))/(dp)\n(dx)/(dp)=(100)/(9)(d)/(dp)[(810000-p^2)^(1/2)]\n(dx)/(dp)=(100)/(9)(1)/(2)[(810000-p^2)^(-1/2)](d)/(dp)[(810000-p^2)]\n(dx)/(dp)=(50)/(9)[(810000-p^2)^(-1/2)](-2p)\n(dx)/(dp)=(100p)/(9√((810000-p^2)))$

Now

$\n(dp)/(dt)=(d((400)/(1+(1)/(8)√(t))+200))/(dt)\n(dp)/(dt)=(d)/(dt)\left((400)/(1+(1)/(8)√(t))+200\right)\n(dp)/(dt)=(d)/(dt)\left((400)/(1+(1)/(8)√(t))\right)+0\n(dp)/(dt)=400(d)/(dt)\left((1)/(1+(1)/(8)√(t))\right)\n(dp)/(dt)=400(d)/(du)\left(\left(1+(1)/(8)√(t)\right)^(-1)\right)(d)/(dt)\left(1+(1)/(8)√(t)\right)\n(dp)/(dt)=-(1600)/(√(t)\left(√(t)+8\right)^2)$

So now the value of dx/dt is given as

$(dx)/(dt)=(dx)/(dp).(dp)/(dt)\n(dx)/(dt)=(100p)/(9√((810000-p^2))).(1600)/(√(t)\left(√(t)+8\right)^2)\n$

So the value of dx/dt is $(100p)/(9√((810000-p^2))).(1600)/(√(t)\left(√(t)+8\right)^2)\n$

Now for the time =16 months price is given as

$p(t)=(400)/(1+(1)/(8)√(t))+200\np(16)=(400)/(1+(1)/(8)√(16))+200\np(16)=466.67$

Now the value of x is given as

$(dx)/(dt)=(100p)/(9√((810000-p^2))).(1600)/(√(t)\left(√(t)+8\right)^2)\n\n(dx)/(dt)=(100*466.67)/(9√((810000-466.67^2))).(1600)/(√(16)\left(√(16)+8\right)^2)\n(dx)/(dt)=18.72$

So the rate of change of quantity demanded per month after 16 months is 18.72

Related Questions

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Select the mathematical sentence that is true. A. 3 • (7 – 3) + 5 ≤ 6 + 2 • 1 B. 3 • (7 – 3) + 5 < 6 + 2 • 1 C. 3 • (7 – 3) + 5 > 6 + 2 • 1 D. 3 • (7 – 3) + 5 = 6 + 2 • 1
F(x) = 3x + 9, g(x) = 5x2 Find (f + g)(x).
Find the vertex of the parabola by completing the square x^2-6x+8=y?
What is the inverse f(x)= 2x+1

Arrange the circles (represented by their equations in general form) in ascending order of their radius lengths.x2 + y2 − 2x + 2y − 1 = 0
x2 + y2 − 4x + 4y − 10 = 0
x2 + y2 − 8x − 6y − 20 = 0
4x2 + 4y2 + 16x + 24y − 40 = 0
5x2 + 5y2 − 20x + 30y + 40 = 0
2x2 + 2y2 − 28x − 32y − 8 = 0
x2 + y2 + 12x − 2y − 9 = 0

Answers

The correct answer is:

x²+y²-2x+2y-1 = 0;
x²+y²-4x+4y-10 = 0;
5x²+5y²-20x+30y+40 = 0;
x²+y²-8x-6y-20 = 0;
x²+y²+12x-2y-9 = 0;
4x²+4y²+16x+24y-40 = 0; and
2x²+2y²-28x-32y-8 = 0

Explanation:

For each of these, we want to write the equation in the form
(x+h)²+(y+k)² = r².

To do this, we evaluate the terms 2hx and 2ky in each equation. We will take half of this; this will tell us what h and k are for each equation.

For the first equation:
2hx = -2x and 2ky = 2y.

Half of -2x = -1x and half of 2y = 1y; this means h = -1 and k = 1:
(x-1)² + (y+1)² + ___ - 1 = 0

When we multiply (x-1)², we get
x²-2x+1.
When we multiply (y+1)², we get
y²+2y+1.

This gives us 1+1 = 2 for the constant. We know we must add something to 2 to get -1; 2 + ___ = -1; the missing term is -3. Add that to each side (to have r² on the right side of the equals) and we have
(x-1)² + (y+1)² = 3
This means that r² = 3, and r = √3 = 1.732.

For the second equation, 2hx = -4x and 2ky = 4y; this means h = -4/2 = -2 and k = 4/2 = 2. This gives us
(x-2)² + (y+2)² -10 + ___ = 0.

Multiplying (x-2)² gives us
x²-4x+4.
Multiplying (y+2)² gives us
y²+4x+4.
This gives us 4+4= 8 for our constant so far.

We know 8 + ___ = -10; this means the missing term is -18. Add this to each side of the equation to have
(x-2)²+(y+2)² = 18; r² = 18; r = √18 = 3√2 = 4.243.

For the third equation, 2hx = -8x and 2ky = -6y. This means h = -8/2 = -4 and k = -6/2 = -3. This gives us:
(x-4)²+(y-3)²-20 = 0

Multiplying (x-4)² gives us
x²-8x+16.
Multiplying (y-3)² gives us
y²-6y+9.

This gives us 16+9 = 25 for the constant. We know that 25+___ = -20; the missing term is -45. Add this to each side for r², and we have that
r²=45; r = √45 = 3√5 = 6.708.

For the next equation, we factor 4 out of the entire equation:
4(x²+y²+4x+6y-10)=0.
This means 2hx = 4x and 2ky = 6y; this gives us h = 4/2 = 2 and k = 6/2 = 3. This gives us
4((x+2)²+(y+3)² - 10) = 0.

Multiplying (x+2)² gives us
x²+4x+4.
Multiplying (y+3)² gives us
y²+6y+9.

This gives us a constant of 4+9 = 13. We know 13+__ = -10; this missing value is -23. Since we had factored out a 4, that means we have 4(-23) = -92. Adding this to each side for r², we have
r²=92; r = √92 = 2√23 = 9.59.

For the next equation, we factor out a 5 first:
5(x²+y²-4x+6y+8) = 0. This means that 2hx = -4x and 2ky = 6y; this gives us h = -4/2 = -2 and k = 6/2 = 3:

5((x-2)²+(y+3)²+8) = 0.

Multiplying (x-2)² gives us
x²-4x+4.
Multiplying (y+3)² gives us
y²+6y+9.

This gives us a constant of 4+9 = 13. We know that 13+__ = 8; the missing value is -5. Since we factored a 5 out, we have 5(-5) = -25. Adding this to each side for r² gives us
r²=25; r = √25 = 5.

For the next equation, we first factor a 2 out:
2(x²+y²-14x-16y-4) = 0. This means 2hx = -14x and 2ky = -16y; this gives us h = -14/2 = -7 and k = -16/2 = -8:

2((x-7)²+(y-8)²-4) = 0.

Multiplying (x-7)² gives us
x²-14x+49.
Multiplying (y-8)² gives us
y²-16x+64.

This gives us a constant of 49+64=113. We know that 113+__ = -4; the missing value is -117. Since we first factored out a 2, this gives us 2(-117) = -234. Adding this to each side for r² gives us
r²=234; r = √234 = 3√26 = 15.297.

For the last equation, 2hx = 12x and 2ky = -2; this means h = 12/2 = 6 and k = -2/2 = -1:
(x+6)²+(y-1)²-9 = 0

Multiplying (x+6)² gives us
x²+12x+36.
Multiplying (y-1)² gives us
y²-2y+1.

This gives us a constant of 36+1 = 37. We know that 37+__ = -9; the missing value is -46. Adding this to each side for r² gives us
r² = 46; r=√46 = 6.78.

Find the radius of each equation:

1.
$x^2 + y^2-2x+2y-1 = 0, \n x^2-2x+1-1 + y^2+2y+1-1-1 = 0, \n (x-1)^2+(y+1)^2=3$ , then $r_1= √(3)$ .

2.
$x^2 + y^2-4x + 4y- 10 = 0, \n x^2 -4x+4-4+ y^2 + 4y+4-4- 10 = 0, \n (x-2)^2+(y+2)^2=18$ , then $r_2= √(18)=3 √(2)$ .

3.
$x^2 + y^2-8x- 6y- 20 = 0, \n x^2-8x+16-16+ y^2- 6y+9-9- 20 = 0, \n (x-4)^2+(y-3)^2=45$ , then $r_3= √(45) =3 √(5)$ .

4.
$4x^2 + 4y^2+16x+24y- 40 = 0, \n 4x^2+16x+16-16+ 4y^2+24y+36-36- 40 = 0, \n 4(x+2)^2+4(y+3)^2=92,\n (x+2)^2+(y+3)^2=23$ , then $r_4= √(23)$ .

5.
$5x^2 + 5y^2-20x+30y+ 40 = 0, \n 5x^2-20x+20-20+ 5y^2+30y+45-45- 40 = 0, \n 5(x-2)^2+5(y+3)^2=105,\n (x-2)^2+(y+3)^2=21$ , then $r_5= √(21)$ .

6.
$2x^2 + 2y^2-28x-32y- 8= 0, \n 2x^2-28x+98-98+ 2y^2-32y+128-128- 8= 0, \n 2(x-7)^2+2(y-8)^2=234,\n (x+2)^2+(y+3)^2=117$ , then $r_6= √(117)=3√(13)$ .

7.
$x^2 + y^2+12x-2y-9 = 0, \n x^2+12x+36-36+ y^2-2y+1-1- 9 = 0, \n (x+6)^2+(y-1)^2=46$ , then $r_7= √(46)$ .

Hence
$r_1= √(3)$ , $r_2=3 √(2)$ , $r_3=3 √(5)$ , $r_4= √(23)$ , $r_5= √(21)$ , $r_6= 3√(13)$ , $r_7= √(46)$ and $r_1\ \textless \ r_2\ \textless \ r_5\ \textless \ r_4\ \textless \ r_3\ \textless \ r_7\ \textless \ r_6$ .

Identify the transformation.

Answers

Answer:

C) translation 5 united left, 1 unit up

Step-by-step explanation:

to figure this out you can take one point and translate it into 5 units left and 1 up and it will be in the spot of the new figure, Reflection and rotation are not correct because the figure would be oriented differently, and it is not D because the figure with ' is the new transformation

Andrew stays in shape by running various distances throughout the week. On monday he ran 3 miles on tuesday he ran 4.6 on thrusday 6.76 and on saturday he runs 4.8 miles. What is the mean distance that andrew runs per day

Answers

Answer:

The mean distance that Andrew runs per day is 4. 79 miles.

Step-by-step explanation:

Distance Covered on Monday = 3 miles

Distance Covered on Tuesday = 4.6 miles

Distance Covered on Thursday = 6.76 miles

Distance Covered on Saturday = 4.6 miles

Now,

Average Distance = $\frac{\textrm{Sum of the Ditance Covered}}{\textrm{Toatk number of Days}}$

Average mean distance = $(3 + 4.6 + 6.76 +4.8)/(4) = (19.16)/(4)$ = 4.79 miles

Hence, mean distance that Andrew runs per day is 4. 79 miles.

What two number mutiply to -240 but add up to 8

Answers

Let us assume the two unknown numbers to be "x" and "y".
Then
xy = -240
x = - (240/y)
And
x + y = 8
Now let us put the value of x in the second equation. We get
x + y = 8
-( 240/y) + y = 8
y^2 - 240 = 8y
y^2 - 8y - 240 = 0
y^2 - 20y + 12y - 240 = 0
y(y - 20) + 12(y - 20) = 0
(y - 20) (y + 12) = 0
So we can see that the unknown number y can have two values and hence the unknown number will also have two values.
When
y - 20 = 0
y = 20
Then
xy = - 240
20x = - 240
x = -12
So we can see that when y is equal to 20 then x is equal to -12 and when y is equal to -12 then x is equal to 20. So the two numbers are 20 and -12.

Answer:

20 and 10

Step-by-step explanation:

Which quadratic function has its vertex at (-2,7) and opens down?

Answers

$The \ vertex \ form \ of \ the \ function \ is: \n \ny = a(x - h)^2 + k\n \nvertex = \ is \ the point \n \n(h, k) = (-2,7)\n\nh=-2\nk=7$

$y=a(x-(-2))^(2) +7\n\ny=a(x+2)^(2) +7 \n\n opens \ up \ down \ for : \ a < 0 \n For \ instance, \ letting \ a = -1 \ gives \ this \ parabola: \n\ny= -(x+2)^(2) +7$

$The\ vertex\ form:y=a(x-h)^2+k\n\nvertex(h;\ k)\to(-2;\ 7)\n\no pens\ down\ then\ a < 0\n\nAnswer:y=a(x+2)^2+7\ where\ a < 0.$

Identify the slope and y-intercept for eachUse appropriate values for the x-axis and y-axis and label each axis (see graphs above).
Graph the y-intercept.
Show the lines that indicate the use of slope to graph two more points (as in the quick practice and lesson).
Draw the line representing the graph of the equation (as above).
Save as a file and submit it.

Problems:
John began his job making $20 the first day. After that he was paid $6.00 per hour. Y = 6x + 20.
A membership to Movie Night Movie Club costs $10 plus $2 per movie. Y = 2x +10.
You start out with $20 and then spend money in a store where every item is $3. Y = -3x + 20.
Your weight-lifting class cost you a $10 fee up front and $5 per class after. Y = 5x + 10.
You join a paperback book club for $6, and then spend $3 per book. Y = 3x + 6.

Answers

Answer:

Graph attached of all the problems.

Problem 1- Slope is m=6 and y-intercept is c=20

Problem 2-Slope is m=2 and y-intercept is c=10

Problem 3-Slope is m=-3 and y-intercept is c=20

Problem 4-Slope is m=5 and y-intercept is c=10

Problem 5-Slope is m=3 and y-intercept is c=6

Step-by-step explanation:

Given : 5 problems

To identify : The slope and y-intercept for each problem and also show the graph.

Solution :

Slope intercept form is y=mx+c

where m is the slope and c is the y-intercept

Problems:

Problem 1 - John began his job making $20 the first day. After that he was paid $6.00 per hour.

Y = 6x + 20