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:
1/6
Step-by-step explanation:
If you divide the fractions to decimals 1/6 equals .16 which is closest to 0
Answer:248
Step-by-step explanation:
First you need to find the unit rate by dividing 93 by 3 which would be 31 then take that number (which is how many miles for one gallon) and multiply it by 8 which would get you 248( if you need this as an equation here 93\3= 31 31*8 = 248). If they gave you the distance instead you would have had to divide by 31 instead of using multiplication. Hope this helps!
How the question has been worded makes me think the answer is (2,5)
Answer:
9.56 ft/sec
Step-by-step explanation:
We are told that a 5.8-ft-tall person walks away from a 9-ft lamppost at a constant rate of 3.4 ft/sec.
I've attached an image showing triangle that depicts this;
Thus; dx/dt = 3.4 ft/sec
From the attached image and using principle of similar triangles, we can say that;
9/y = 5.8/(y - x)
9(y - x) = 5.8y
9y - 9x = 5.8y
9y - 5.8y = 9x
3.2y = 9x
y = 9x/3.2
dy/dx = 9/3.2
Now, to find how fast the tip of the shadow is moving away from the lamp post, it is;
dy/dt = dy/dx × dx/dt
dy/dt = (9/3.2) × 3.4
dy/dt = 9.5625 ft/s ≈ 9.56 ft/sec