Answer:
$7812.5million
$7812.5million
Step-by-step explanation:
From the question, the demand quantity <em>D(x) is 410-x </em> and the supply quantity <em>S(x)=160+x.</em> We determine the equilibrium quantity by equating the Demand quantity and the Supply quantity i.e
<em>D(x)=S(x). </em>
<em>
</em>
Hence the equilibrium quantity is 125.
Next we determine the equilibrium price. This can be obtain by just substituting the equilibrium quantity into either the demand quantity or the supply quantity. I prefer using the Demand quantity.
Equilibrium price=
.
Next we write the expression for the Consumer Surplus
![CS=\int\limits^q_0 {D(x)} \, dx -pq](https://tex.z-dn.net/?f=CS%3D%5Cint%5Climits%5Eq_0%20%7BD%28x%29%7D%20%5C%2C%20dx%20-pq)
where <em>p</em> and <em>q </em> are the equilibrium price and equilibrium quantity respectively.
By substituting values we have
![CS=\int\limits^q_0 {(410-x)} \, dx (285)(125)\\](https://tex.z-dn.net/?f=CS%3D%5Cint%5Climits%5Eq_0%20%7B%28410-x%29%7D%20%5C%2C%20dx%20%28285%29%28125%29%5C%5C)
![CS=/410x-\frac{x^{2} }{2} /^{125}_{0} \\](https://tex.z-dn.net/?f=CS%3D%2F410x-%5Cfrac%7Bx%5E%7B2%7D%20%7D%7B2%7D%20%2F%5E%7B125%7D_%7B0%7D%20%5C%5C)
By carrying out simple arithmetic we arrive at
.
To determine the producer surplus, we use the expression below
![PS=pq-\int\limits^q_0 {S(x)} \,dx\\](https://tex.z-dn.net/?f=PS%3Dpq-%5Cint%5Climits%5Eq_0%20%7BS%28x%29%7D%20%5C%2Cdx%5C%5C)
Hence if we substitute values we arrive at
.
By simply simplification we arrive at
![PS=(285)(125)-(20000+7812.5)\\PS=35625-27812.5\\PS=$7812.5million](https://tex.z-dn.net/?f=PS%3D%28285%29%28125%29-%2820000%2B7812.5%29%5C%5CPS%3D35625-27812.5%5C%5CPS%3D%247812.5million)
Answer:
It is a fraction
. f (x) = x +12×2-x-1 is an example of rational function.
It has numerator and denominator properties.
vertical asymptotes is the another property of rational function.
Numerater and denominator both are polynomials.
function expressed by numerator and denominator with ratio.
Yoo kid for these type of problems your better o go with an app called Photomath
Answer:
![Area\ of\ a\ sector = 11\pi\ inches^{2}](https://tex.z-dn.net/?f=Area%5C%20of%5C%20a%5C%20sector%20%3D%2011%5Cpi%5C%20inches%5E%7B2%7D)
Step-by-step explanation:
Given:
The radius of the circle is 6 inches.
And the central angle is 110 degree.
The area of the sector =
-------(1)
Where ![x= central\ angle](https://tex.z-dn.net/?f=x%3D%20central%5C%20angle)
![r = radius](https://tex.z-dn.net/?f=r%20%3D%20radius)
Put radius and central angle value in equation 1.
area of the sector = ![\frac{110}{360}\pi (6)^{2}](https://tex.z-dn.net/?f=%5Cfrac%7B110%7D%7B360%7D%5Cpi%20%286%29%5E%7B2%7D)
area of the sector = ![\frac{11}{36}\pi\times 36](https://tex.z-dn.net/?f=%5Cfrac%7B11%7D%7B36%7D%5Cpi%5Ctimes%2036)
![Area\ of\ a\ sector = 11\pi\ inches^{2}](https://tex.z-dn.net/?f=Area%5C%20of%5C%20a%5C%20sector%20%3D%2011%5Cpi%5C%20inches%5E%7B2%7D)
Therefore, the area of the sector is ![11\pi \ inches^{2}](https://tex.z-dn.net/?f=11%5Cpi%20%5C%20inches%5E%7B2%7D)