Give each of the kids a full field then have them share or split the other field
I would say true. Hope this helps :)
Assume 0 < <em>x</em>/2 < <em>π</em>/2. Then
tan²(<em>x</em>/2) + 1 = sec²(<em>x</em>/2) ===> sec(<em>x</em>/2) = √(1 - tan²(<em>x</em>/2))
===> cos(<em>x</em>/2) = 1/√(1 - tan²(<em>x</em>/2))
===> cos(<em>x</em>/2) = 1/√(1 - <em>t</em> ²)
We also know that
sin²(<em>x</em>/2) + cos²(<em>x</em>/2) = 1 ===> sin(<em>x</em>/2) = √(1 - cos²(<em>x</em>/2))
Recall the double angle identities:
cos(<em>x</em>) = 2 cos²(<em>x</em>/2) - 1
sin(<em>x</em>) = 2 sin(<em>x</em>/2) cos(<em>x</em>/2)
Then
cos(<em>x</em>) = 2/(1 - <em>t</em> ²) - 1 = (1 + <em>t</em> ²)/(1 - <em>t</em> ²)
sin(<em>x</em>) = 2 √(1 - 1/(1 - <em>t</em> ²)) / √(1 - <em>t</em> ²) = 2<em>t</em>/(1 - <em>t</em> ²)
Answer:
The radius is 10
Step-by-step explanation:
Given
Required
The radius
Rewrite as:
Subtract 81 from both sides
Expand
Factorize
Factor out y - 9
Express as squares
The equation of a circle is:
By comparison:
We know that both groups feed grass hay in the morning, so that's 100/100.
First group feeds alfalfa and grass in the evening
Second group feeds alfalfa and oat hay in the evening.
The estimated probability that horse owners feed grass hay the morning (AM) is 100%.
The estimated probability that horse owners feed alfalfa + oat hay in the pm is 28/100 or 28% (C)
