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Making the usual substitutions (x=r·cos(θ), y=r·sin(θ)), you have
(r·cos(θ))² +4r·cos(θ) +4 + (r·sin(θ))² -4r·sin(θ) +4 = 8
r(r +4·cos(θ) -4·sin(θ)) = 0
Dividing by r then subtracting the non-r terms gives
r = 4(sin(θ) -cos(θ))
(r·cos(θ))² +4r·cos(θ) +4 + (r·sin(θ))² -4r·sin(θ) +4 = 8
r(r +4·cos(θ) -4·sin(θ)) = 0
Dividing by r then subtracting the non-r terms gives
r = 4(sin(θ) -cos(θ))
Answer:
n = 59
Step-by-step explanation:
Since the difference between consecutive terms is constant, then the sequence is arithmetic.
The n th term of an arithmetic sequence is
= a₁ + (n - 1)d
Given
= 40, then
a₁ + 9d = 40 → (1)
Given
= 20, then
a₁ + 19d = 20 → (2)
Subtract (1) from (2) term by term
10d = - 20 ( divide both sides by 10 )
d = - 2
Substitute d = - 2 in (1)
a₁ - 18 = 40 ( add 18 to both sides )
a₁ = 58
The sum to n terms of an arithmetic sequence is
=
[2a₁ + (n - 1)d ]
Here a₁ = 58, d = - 2 and = 0, thus
[ ( 2 × 58) - 2(n - 1)] = 0
( 116 - 2n + 2) = 0
Multiply through by 2
n(118 - 2n) = 0
Equate each factor to zero and solve for n
n = 0
118 - 2n = 0 ⇒ - 2n = - 118 ⇒ n = 59
However, n > 0 ⇒ n = 59