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3 years ago

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Which recursive formula can be used to generate the sequence below, where f(1) = 3 and n ≥ 1? 3, –6, 12, –24, 48, … f (n + 1) =

–3 f(n ) f (n + 1) = 3 f(n ) f (n + 1) = –2 f(n ) f (n + 1) = 2 f(n)

2 answers:

tatuchka [14]3 years ago

8 0

Which recursive formula can be used to generate the sequence below, where f(1) = 3 and n ≥ 1?

3, –6, 12, –24, 48

The recursive formula for this sequence is

f (n + 1) = –2 f(n)

at n=1 f(n)= 3

at n = 2

f(2) = -2 (3) = -6

n = 3

f(3) = -2 (-6) = 12 and so on

Flauer [41]3 years ago

4 0

The recursive formula for the sequence is:
f(n+1) = -2f(n)

Let's take an example:
Let f(n) = 12 and we need to calculate f(n+1) which should be -24
If we apply the formula than the answer is:
f(n+1) = -2*12 = -24.

I hope that this is the answer that you were looking for and it has helped you.

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Given the center of the circle (-3,4) and a point on the circle (-6,2), (10,4) is on the circle

Anastasy [175]

Answer:

Part 1) False

Part 2) False

Step-by-step explanation:

we know that

The equation of the circle in standard form is equal to

$(x-h)^{2} +(y-k)^{2}=r^{2}$

where

(h,k) is the center and r is the radius

In this problem the distance between the center and a point on the circle is equal to the radius

The formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

Part 1) given the center of the circle (-3,4) and a point on the circle (-6,2), (10,4) is on the circle.

true or false

substitute the center of the circle in the equation in standard form

$(x+3)^{2} +(y-4)^{2}=r^{2}$

Find the distance (radius) between the center (-3,4) and (-6,2)

substitute in the formula of distance

$r=\sqrt{(2-4)^{2}+(-6+3)^{2}}$

$r=\sqrt{(-2)^{2}+(-3)^{2}}$

$r=\sqrt{13}\ units$

The equation of the circle is equal to

$(x+3)^{2} +(y-4)^{2}=(\sqrt{13}){2}$

$(x+3)^{2} +(y-4)^{2}=13$

Verify if the point (10,4) is on the circle

we know that

If a ordered pair is on the circle, then the ordered pair must satisfy the equation of the circle

For x=10,y=4

substitute

$(10+3)^{2} +(4-4)^{2}=13$

$(13)^{2} +(0)^{2}=13$

$169=13$ -----> is not true

therefore

The point is not on the circle

The statement is false

Part 2) given the center of the circle (1,3) and a point on the circle (2,6), (11,5) is on the circle.

true or false

substitute the center of the circle in the equation in standard form

$(x-1)^{2} +(y-3)^{2}=r^{2}$

Find the distance (radius) between the center (1,3) and (2,6)

substitute in the formula of distance

$r=\sqrt{(6-3)^{2}+(2-1)^{2}}$

$r=\sqrt{(3)^{2}+(1)^{2}}$

$r=\sqrt{10}\ units$

The equation of the circle is equal to

$(x-1)^{2} +(y-3)^{2}=(\sqrt{10}){2}$

$(x-1)^{2} +(y-3)^{2}=10$

Verify if the point (11,5) is on the circle

we know that

If a ordered pair is on the circle, then the ordered pair must satisfy the equation of the circle

For x=11,y=5

substitute

$(11-1)^{2} +(5-3)^{2}=10$

$(10)^{2} +(2)^{2}=10$

$104=10$ -----> is not true

therefore

The point is not on the circle

The statement is false

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answer

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Answer:

A dependent variable is a variable (often denoted by y ) whose value depends on that of another. so it is the variable that changes depending on the numbers beside it.

I hope this helps if not check out Khan academy

https://www.khanacademy.org/math/algebra/introduction-to-algebra/alg1-dependent-independent/v/dependent-and-independent-variables-exercise-example-3

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