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: So True
Step-by-step explanation:
Answer:
C)
Step-by-step explanation:
Just see the length of the R line, A and B are almost the same large when you add them.
X² - 14x + 33 = 0 is in the form ax² + bx + c = 0. We will need this for the second step.
Subtract 33 from both sides
x² - 14x = - 33
Divide the b term by 2 and then square the result. Then add that to both sides.
- 14 / 2 is - 7. (- 7)² = 49. Add that to both sides
x² - 14x + 49 = - 33 + 49
Combine the constants and factor the trinomial
(x - 7)² = 16
Square root both sides
x - 7 = +/- 4
or
x - 7 = - 4 and x - 7 = 4
You have to place a plus-minus statement because (- 4)² and 4² both equal 16.
Now we solve for x in each equation. Add 7 to both sides to isolate our variable.
x = 3 and x = 11
The answer is 6 because u have to divided it
