A CHICKEN COOP HOLDS A TOTAL OF 10 HENS OR 20 CHICKS. IF 2/5THS OF THE HENS IN A FULL COOP ARE REMOVED, HOW MANY NEW CHICKS COUL
1 answer:
Answer:
8 new chicks can be fitted in the coop.
Step-by-step explanation:
A chicken coop holds 10 hens or 20 chicks.
That means space in the coop for 10 hens = space for 20 chicks
Or space for 1 hen = space for 2 chicks
Now th of the hens were removed from the coop.
So, number of hens removed from the coop =
= 4 hens
And space for 4 hens in the coop = space for the 8 chicks
Therefore, 8 new chicks can be fitted in the coop.
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<h3>Answer:</h3>
- -100x +100; 3
- 5x +81; 162
The distributive property is your friend. It tells you ...
a(b + c) = ab + ac
It can also be used to simplify the product of binomials (or other polynomials).
(a +b)^2 = (a +b)(a +b) = a(a +b) + b(a +b) = a^2 +ab +ab +b^2
(a +b)^2 = a^2 +2ab + b^2 . . . . . . . worth remembering
1. (x -10)^2 -x(x +80) = (x^2 -20x +100) +(-x^2 -80x)
= -100x +100 . . . . . . simplified form
For the purposes of calculation, it can be easier to factor out 100:
= 100 (1 -x)
Then for x = 0.97
= 100(1 -0.97) = 100(0.03) = 3
___
2. (2x +9)^2 -x(4x +31) = (4x^2 +36x +81) -4x^2 -31x = 5x +81
For x = 16.2, this is ...
5(16.2) +81 = 81 +81 = 162
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The purpose of the attachment is to show the evaluation is correct.