Download to read offline
Uploaded byapexjaipur
4 views
x = 10 + (1p^2) Elasticity , E = (dxdp)(px) .pdf
The document presents a mathematical expression for elasticity as e = (dx/dp)*(p/x) with x defined as 10 + (1/p^2). It calculates the elasticity at p = 0.3, resulting in a value of e = -1.0526. The process is repeated for clarity, leading to the same conclusion.
More Related Content
![[PDF]Solution [PDF].pdf](https://cdn.slidesharecdn.com/ss_thumbnails/pdfsolutionpdf-230412010219-eebd3c96-thumbnail.jpg?width=640&height=640&fit=bounds)
![pOH = -log[OH-] = -log(0.061) = 1.21 pH = 14 - pO.pdf](https://cdn.slidesharecdn.com/ss_thumbnails/poh-logoh-log0-230412010218-75361cc2-thumbnail.jpg?width=640&height=640&fit=bounds)
More from apexjaipur
Recently uploaded
x = 10 + (1p^2) Elasticity , E = (dxdp)(px) .pdf
- 1. x = 10+ (1/p^2) Elasticity , E = (dx/dp)*(p/x) = (-2/p^3)*(p/(10 + (1/p^2) ) Put p = 0.3,we get E = (-2/0.3^3)*(0.3/(10+(1/0.3^2)) E = - 1.0526 Solution x = 10 + (1/p^2) Elasticity , E = (dx/dp)*(p/x) = (-2/p^3)*(p/(10 + (1/p^2) ) Put p = 0.3,we get E = (-2/0.3^3)*(0.3/(10+(1/0.3^2)) E = - 1.0526