To solve this problem, let us first assign some
variables. Let us say that:
x = pigs
y = chickens
z = ducks
From the problem statement, we can formulate the
following equations:
1. y + z = 30 --->
only chicken and ducks have feathers
2. 4 x + 2 y + 2 z = 120 --->
pig has 4 feet, while chicken and duck has 2 each
3. 2 x + 2 y + 2 z = 90 --->
each animal has 2 eyes only
Rewriting equation 1 in terms of y:
y = 30 – z
Plugging this in equation 2:
4 x + 2 (30 – z) + 2 z = 120
4 x + 60 – 2z + 2z = 120
4 x = 120 – 60
4 x = 60
x = 15
From the given choices, only one choice has 15 pigs. Therefore
the answers are:
She has 15 pigs, 12 chickens, and 18 ducks.
Answer:
The answer final
x1= -5; x2=9
Step-by-step explanation:
If you want me to expand on the equation leave me a comment :)
Answer:
(1,2)
Step-by-step explanation:
The other intervals contain local minimums, local maximums are also known as maximum points and look like upwards parabolas
15. 5 times 3 is 15.
If you check, 15 divided by 3 is 5(red cars).
