Answer:
2/3 ≈ 0.6667
Step-by-step explanation:
apply soh cah toa
sine=opposite/hypo
sine=4/6
sine=2/3
Where is the question though
Answer:
elasticity supply of dog food = 2.61
elasticity supply of cat food = 1.71
Step-by-step explanation:
The midpoint formula for elasticity is:
![Elasticity = \frac{(Q2-Q1)/[(Q2+Q1)/2]}{(P2-P1)/[(P2+P1)/2]}](https://tex.z-dn.net/?f=Elasticity%20%3D%20%5Cfrac%7B%28Q2-Q1%29%2F%5B%28Q2%2BQ1%29%2F2%5D%7D%7B%28P2-P1%29%2F%5B%28P2%2BP1%29%2F2%5D%7D)
Point 1: Q = 39.0 and P = 5.50
Point 2: Q = 101.0 and P = 7.75
![Elasticity\ supply\ of\ dog\ food = \frac{(101.0-39.0)/[101.0+39.0)/2]}{(7.75-5.50)/[(7.75+5.50)/2]}=2.61](https://tex.z-dn.net/?f=Elasticity%5C%20supply%5C%20of%5C%20dog%5C%20food%20%3D%20%5Cfrac%7B%28101.0-39.0%29%2F%5B101.0%2B39.0%29%2F2%5D%7D%7B%287.75-5.50%29%2F%5B%287.75%2B5.50%29%2F2%5D%7D%3D2.61)
Doing the same for the cat food:
![Elasticity\ supply\ of\ cat\ food = \frac{(71.0-39.0)/[71.0+39.0)/2]}{(7.75-5.50)/[(7.75+5.50)/2]}=1.71](https://tex.z-dn.net/?f=Elasticity%5C%20supply%5C%20of%5C%20cat%5C%20food%20%3D%20%5Cfrac%7B%2871.0-39.0%29%2F%5B71.0%2B39.0%29%2F2%5D%7D%7B%287.75-5.50%29%2F%5B%287.75%2B5.50%29%2F2%5D%7D%3D1.71)
Answer with explanation:
The equation, y=3 x,
y=Distance traveled
x =Total time
![\frac{\text{Distance traveled}}{\text{Time}}=\frac{y}{x}=3\\\\ \frac{dy}{dx}={\text{Rate of change}}={\text{Velocity}}=3 {\text{unit of time}]](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctext%7BDistance%20traveled%7D%7D%7B%5Ctext%7BTime%7D%7D%3D%5Cfrac%7By%7D%7Bx%7D%3D3%5C%5C%5C%5C%20%5Cfrac%7Bdy%7D%7Bdx%7D%3D%7B%5Ctext%7BRate%20of%20change%7D%7D%3D%7B%5Ctext%7BVelocity%7D%7D%3D3%20%7B%5Ctext%7Bunit%20of%20time%7D%5D)
Also, in terms of straight line
Slope =3= uniform Velocity
Point (3,9) and (5,15) represents Distance traveled in 3 (unit of time) =9 unit ,and 15 unit=Distance traveled in 5 (Unit of time).
→Alonso is moving with uniform speed=3 (unit of time), as velocity remains constant in the entire process.