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klio [65]
3 years ago
11

Find the largest possible revenue from the demand equation "Q= -2p + 1000

Mathematics
1 answer:
Mariana [72]3 years ago
3 0
Revenue=quantity x price= (-2p+1000) (p)= -2p^2+1000p
The maximum revenue will occur when the first derivative is zero so when 2(-2p)+1000=0;p=250
Which generates 125,000 in revenue
Try prices of 245 and 255 and you will see they both are less than 250 thereby proving the max revenue is 250
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If f(x)= 1/3x and g(x)=3x, then what are g * f (x) and g*f(1/2)
MrMuchimi
The correct answer is:  3)  " x ;  1/2" .
_____________________________________________________
Explanation:
_____________________________________________________
Question 1) 

g [ f(x) ] = ? ;

→ Given:  " f(x) = 1/3 x " ;

→ Given:  " g(x) = 3x " ;  
__________________________________
g [ f(x) ] = g(1/3 x) = 3(1/3 x) = 1x = "x" .
__________________________________
Question 2)

g [ f(1/2) ] = ? ; 
__________________________________
→  Given:  " f(x) = 1/3 x " ;

→  f(1/2) = (1/3) * (1/2) = (1*1) / (3*2) = (1/6) ;

→  g [ f(x) ] =  

          g(1/6) = 3* (1/6) = (3/1) * (1/6) = (3*1) / (1*6) = 3/6 = (3÷3) / (6÷3) = " 1/2 "  .
___________________________________________________________ 


3 0
3 years ago
What is the volume of the prism
satela [25.4K]

Answer: 336cm^3

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6 0
3 years ago
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anyanavicka [17]

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4 0
3 years ago
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Answer

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4 0
3 years ago
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