<h3>
Answer: (x-10)^2 + (y-1)^2 = 25</h3>
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Explanation:
The general template of a circle is
(x-h)^2 + (y-k)^2 = r^2
Where,
- (h,k) is the center
- r is the radius
We are given (10,1) as the center so (h,k) = (10,1) meaning that h = 10 and k = 1 pair up together.
r = 5 is the given radius.
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Plug the values h = 10, k = 1, r = 5 into the equation to get:
(x-h)^2 + (y-k)^2 = r^2
(x-10)^2 + (y-1)^2 = 5^2
(x-10)^2 + (y-1)^2 = 25
Get it to equal 0
minus 17x from both sides
2x²-17x+35=0
tricky part
get something like
(2x-a)(x-b)
we know the signs are negative since the middle term is negative and last term is positive
so a and b are either 1 and 35 or 5 and 7
(2x-1)(x-35), nope
(2x-35)(x-1) nope
(2x-5)(x-7), nope
(2x-7)(x-5), yep
(2x-7)(x-5)=0
set to zero
2x-7=0
2x=7
x=7/2
x-5=0
x=5
x=7/2 and 5
The nth term of the geometric sequence is:
an=ar^(n-1)
where
a=first term
r=common ratio
n=nth term
from the question:
120=ar(3-1)
120=ar^2
a=120/(r^2)....i
also:
76.8=ar^(5-1)
76.8=ar^4
a=76.8/r^4.....i
thus from i and ii
120/r^2=76.8/r^4
from above we can have:
120=76.8/r²
120r²=76.8
r²=76.8/120
r²=0.64
r=√0.64
r=0.8
hence:
a=120/(0.64)=187.5
therefore the formula for the series will be:
an=187.5r^0.8
