Answer:
5/7
Step-by-step explanation:
Let x be the distance it runs on the first day, then:-
x + x + 2 + x + 4 = 54
3x + 6 = 54
3x = 54 - 6 = 48
x = 48/3 = 16 km
So for the first 2 days it runs 16 + 18 = 34 km. Answer
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:
7x²+6x-6
Step-by-step explanation:
1 Combine similar terms
x²+5x+6x²+x-6
7x²+5x+x-6
2 Combine similar terms
7x²+5x+x-6
7x²+6x-6