Formula for the geometric sequence 1/2, 1, 2, 4

Mathematics

1 answer:

zmey [24]1 year ago

5 0

Answer:

$a_{n}$ = $\frac{1}{2}$ $(2)^{n-1}$

Step-by-step explanation:

the nth term of a geometric sequence is

$a_{n}$ = a₁ $(r)^{n-1}$

where a₁ is the first term and r the common ratio

here a₁ = $\frac{1}{2}$ and r = $\frac{a_{2} }{a_{1} }$ = $\frac{1}{\frac{1}{2} }$ = 2 , then

$a_{n}$ = $\frac{1}{2}$ $(2)^{n-1}$

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algol [13]

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2 years ago

A line passes through the points (0, 4) and (1, 9). What is its equation in slope-intercept

Simora [160]

Answer:

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Step-by-step explanation:

First get slope:

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2 years ago

Factor completely. <br> <img src="https://tex.z-dn.net/?f=x%5E%7B8%7D-%5Cfrac%7B1%7D%7B81%7D" id="TexFormula1" title="x^{8}-\fra

Eduardwww [97]

We have 3⁴ = 81, so we can factorize this as a difference of squares twice:

$x^8 - \dfrac1{81} = \left(x^2\right)^4 - \left(\dfrac13\right)^4 \\\\ x^8 - \dfrac1{81} = \left(\left(x^2\right)^2 - \left(\dfrac13\right)^2\right) \left(\left(x^2\right)^2 + \left(\dfrac13\right)^2\right) \\\\ x^8 - \dfrac1{81} = \left(x^2 - \dfrac13\right) \left(x^2 + \dfrac13\right) \left(\left(x^2\right)^2 + \left(\dfrac13\right)^2\right) \\\\ x^8 - \dfrac1{81} = \left(x^2 - \dfrac13\right) \left(x^2 + \dfrac13\right) \left(x^4 + \dfrac19\right)$

Depending on the precise definition of "completely" in this context, you can go a bit further and factorize $x^2-\frac13$ as yet another difference of squares: