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gizmo_the_mogwai [7]
3 years ago
9

Write a recursive rule for the sequence.

Mathematics
1 answer:
Vadim26 [7]3 years ago
8 0

Answer:

f(n)=f(n-1)+f(n-2)

f(1)=1x

f(2)=1x

Step-by-step explanation:

This is the fibonacci sequence with each term times x.

Notice, you are adding the previous two terms to get the third term per consecutive triples of the sequence.

That is:

1x+1x=2x

1x+2x=3x

2x+3x=5x

3x+5x=8x

So since we need the two terms before the third per each consecutive triple in the sequence, our recursive definition must include two terms of the sequence. People normally go with the first two.

f(1)=1x since first term of f is 1x

f(2)=1x since second term of f is 1x

Yes, I'm naming the sequence f.

So I said a third term in a consecutive triple of the sequence is equal to the sum of it's two prior terms. Example, f(3)=f(2)+f(1) and f(4)=f(3)+f(2) and so on...

Note, the term before the nth term is the (n-1)th term and the term before the (n-1)th term is the (n-2)th term. Just like before the 15th term you have the (15-1)th term and before that one you have the (15-2)th term. That example simplified means before the 15th term you have the 14th and then the 13th.

So in general f(n)=f(n-1)+f(n-2).

So the full recursive definition is:

f(n)=f(n-1)+f(n-2)

f(1)=1x

f(2)=1x

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If the endpoints of the diameter of a circle are (−8, −6) and (−4, −14), what is the standard form equation of the circle?
kondaur [170]

Equation of the circle is (x+6)^{2}+(y+10)^{2}=20.

Solution:

The endpoints of the diameter of a circle are (–8, –6) and (–4, –14).

Center of the circle = Mid point of the diameter

Mid point formula:

$P(x, y)=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)

Here, x_1=-8, y_1=-6, x_2=-4, y_2=-14

$P(x, y) =\left(\frac{-8-4}{2}, \frac{-6-14}{2}\right)

$P(x, y) =\left(\frac{-12}{2}, \frac{-20}{2}\right)

$P(x, y) =(-6, -10)

Center of the circle = (–6, –10)

Radius is the distance between center and any endpoint of the diameter.

To calculate the radius using distance formula.

r=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}

Here, x_1=-6, y_1=-10, x_2=-8, y_2=-6

r=\sqrt{\left(-8-(-6)\right)^{2}+\left(-6-(-10)}\right)^{2}}

r=\sqrt{(-8+6)^{2}+(-6+10)^{2}}

r=\sqrt{(-2)^{2}+(4)^{2}}

r=\sqrt{20} units

The standard form of the equation of a circle is

(x-a)^{2}+(y-b)^{2}=r^{2}, where (a, b) are center and r is the radius.

Here, center = (–6, –10) and r=\sqrt{20}

(x-(-6))^{2}+(y-(-10))^{2}={(\sqrt{20})} ^{2}

(x+6)^{2}+(y+10)^{2}=20

Equation of the circle is (x+6)^{2}+(y+10)^{2}=20.

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Answer:

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