Given information:
Total animals = 13
Total legs = 40
Assumed all animals have their legs intact ...
Pig = 4 legs
Chicken = 2legs
Assumed all chickens:
Total legs = 2 x 13 = 26
Extra legs = 40 - 26 = 14
Extra legs belongs to the pig (each pig has extra 2 legs)
14 ÷ 2 = 7
There are 7 pigs
13 - 7 = 6
There are 6 chickens
---------- OR ALGEBRAICALLY --------
Let the number of pigs be x and number of chickens be y
pigs = x
chickens = y
x + y = 13 -------------- (1)
4x + 2y = 40 ---------- (2)
From (1):
x + y = 13
x = 13 - y -------------- sub into (2)
4(13 - y) + 2y = 40
52 - 4y + 2y = 40
52 - 2y = 40
2y = 52 - 40
2y = 12
y = 6 ----------- sub into (1)
x + 6 = 13
x = 13 - 6
x = 7
x = 7, y = 6
pigs = 7, chickens = 6
Answer:
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Step-by-step explanation:
What grade is this for?
The answer is 5 good luck
Unless unusually unusual unification under
Answer:
![-16x^3+9x^2](https://tex.z-dn.net/?f=-16x%5E3%2B9x%5E2)
Step-by-step explanation:
![-4x^3-12x^3+9x^2](https://tex.z-dn.net/?f=-4x%5E3-12x%5E3%2B9x%5E2)
![(-4-12)x^3+9x^2](https://tex.z-dn.net/?f=%28-4-12%29x%5E3%2B9x%5E2)
![-16x^3+9x^2](https://tex.z-dn.net/?f=-16x%5E3%2B9x%5E2)