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Musya8 [376]
3 years ago
13

For the following geometric sequence find the recursive formula: {-1, 3, -9, ...}.

Mathematics
1 answer:
vredina [299]3 years ago
3 0
First term (a1) is -1

recursive formula goes like this

a_n is the nth term
a_{n-1} is the term before that

we normally have a_n=f(a_{n-1})

we see each term is multipying by -3 to get next one


so that would be
a_n=-3*a_{n-1} where a1=-1


the 3rd option is correct except that it is the explicit formula

so answer is 2nd one
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The coordinates of HF are (1, 4)

Step-by-step explanation:

The parameters of the line are;

The coordinate of the end points are H = (-11, 7), and J = (5, 3)

The ratio by which the point F divides the line = 3:1

The segments in the line are HF, and FJ

Therefore;

The fraction of the length of HJ that is represented by HF = 3/(3 + 1) × HJ = 3/4 × HJ

HF = 3/4 × HJ

Which gives the coordinates of the point F as follows;

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The coordinates of F are (1, 4)

We check the length of HF, from the equation for the length to of a line to get;

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

l_{HF} = \sqrt{\left (4-7  \right )^{2}+\left (1-(-11)  \right )^{2}} = \sqrt{\left (-3  \right )^{2}+\left (12  \right )^{2}} = 3\cdot \sqrt{17}

Similarly, we check the length of HJ, to get;

l_{HF} = \sqrt{\left (3-7  \right )^{2}+\left (5-(-11)  \right )^{2}} = \sqrt{\left (-4  \right )^{2}+\left (16  \right )^{2}} = 4\cdot \sqrt{17}

The length of HF = 3·√(17)

The length of HJ = 4·√(17)

Therefore, from HF = 3/4× HJ, we have;

HF = 3/4 × 4·√(17) = 3·√(17)

Therefore, the coordinates of HF are (1, 4)

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