Answer:
Let first no. be x
Therefore,
2nd no. is x+1
Third no. is x+2
<em><u>SUM</u></em><em><u> </u></em><em><u>OF</u></em><em><u> </u></em><em><u>NO</u></em><em><u>.</u></em><em><u>'</u></em><em><u>s</u></em><em><u>=</u></em><em><u>x</u></em><em><u>+</u></em><em><u>x</u></em><em><u>+</u></em><em><u>1</u></em><em><u>+</u></em><em><u>x</u></em><em><u>+</u></em><em><u>2</u></em>
<em><u>S</u></em><em><u>=</u></em><em><u>3x</u></em><em><u>+</u></em><em><u>3</u></em>
<em><u>S</u></em><em><u>=</u></em><em><u>3</u></em><em><u>(</u></em><em><u>x</u></em><em><u>+</u></em><em><u>1</u></em><em><u>)</u></em>
-z³ + 5k^6 + z³ -10k^6
(-z³ cancels out with z³)
5k^6 -10k^6
(then subtract)
Answer is -5k^6
The probability of loss is 0.2 because they must add up to 1.
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:
See Step by step explanation
Step-by-step explanation:
