That is the slope
rise/run
we see we have (2/5,1/2)
rise is y, run is x
(x,y)
((1/2)/(2/5)=(1/2)(2/5)=2/10=1/5
unit rate is y=(1/5)x
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
H= ut -16 t^2
h = 144(2) - 16 (4)
= 288 - 64 = 224
Answer:
The answer is y=8/9x
Step-by-step explanation:
Answer:
The class average is 83.6
Step-by-step explanation:
:D
