THIS IS THE COMPLETE QUESTION BELOW
The demand equation for a product is p=90000/400+3x where p is the price (in dollars) and x is the number of units (in thousands). Find the average price p on the interval 40 ≤ x ≤ 50.
Answer
$168.27
Step by step Explanation
Given p=90000/400+3x
With the limits of 40 to 50
Then we need the integral in the form below to find the average price
1/(g-d)∫ⁿₐf(x)dx
Where n= 40 and a= 50, then if we substitute p and the limits then we integrate
1/(50-40)∫⁵⁰₄₀(90000/400+3x)
1/10∫⁵⁰₄₀(90000/400+3x)
If we perform some factorization we have
90000/(10)(3)∫3dx/(400+3x)
3000[ln400+3x]₄₀⁵⁰
Then let substitute the upper and lower limits we have
3000[ln400+3(50)]-ln[400+3(40]
30000[ln550-ln520]
3000[6.3099×6.254]
3000[0.056]
=168.27
the average price p on the interval 40 ≤ x ≤ 50 is
=$168.27
Answer:
A
Step-by-step explanation:
Answer: i took the test and got it right
D (0.55,0.75)
Step-by-step explanation:
Answer:
dddadad
Step-by-step explanation:
Answer:
x = 5/39
, y = 539/39
Step-by-step explanation:
Solve the following system:
{y - 2.5 x = 13.5
12.25 x - y = -12.25
In the first equation, look to solve for y:
{y - 2.5 x = 13.5
12.25 x - y = -12.25
y - 2.5 x = y - (5 x)/2 and 13.5 = 27/2:
y - (5 x)/2 = 27/2
Add (5 x)/2 to both sides:
{y = 1/2 (5 x + 27)
12.25 x - y = -12.25
Substitute y = 1/2 (5 x + 27) into the second equation:
{y = 1/2 (5 x + 27)
1/2 (-5 x - 27) + 12.25 x = -12.25
(-5 x - 27)/2 + 12.25 x = 12.25 x + (-(5 x)/2 - 27/2) = 9.75 x - 27/2:
{y = 1/2 (5 x + 27)
9.75 x - 27/2 = -12.25
In the second equation, look to solve for x:
{y = 1/2 (5 x + 27)
9.75 x - 27/2 = -12.25
9.75 x - 27/2 = (39 x)/4 - 27/2 and -12.25 = -49/4:
(39 x)/4 - 27/2 = -49/4
Add 27/2 to both sides:
{y = 1/2 (5 x + 27)
(39 x)/4 = 5/4
Multiply both sides by 4/39:
{y = 1/2 (5 x + 27)
x = 5/39
Substitute x = 5/39 into the first equation:
{y = 539/39
x = 5/39
Collect results in alphabetical order:
Answer: {x = 5/39
, y = 539/39
