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3 years ago

In a geometric sequence, a2=108 and as 256. Write the explicit formula for this sequence.

Mathematics

2 answers:

torisob [31]3 years ago

8 0

Answer:

The first choice is correct.

Step-by-step explanation:

First work out the common ratio r:-

a2 = a1r and a5 = a1 r^4

So r^3 = a1r^4 / a1r = 256/ 108

r = cube root (256/-108) = 4/3

So the explicit formula is 108(4/3)^(n-1)

mariarad [96]3 years ago

4 0

Answer:

C. 81[-4/3]^n-1

Step-by-step explanation:

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Step-by-step explanation:

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3 years ago

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Zanzabum

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6 0

3 years ago

What is the value of the expression below when y = 3? 10y - 8

ycow [4]

Answer:

Step-by-step explanation:

<h3>Given, y = 3 </h3>

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3 0

2 years ago

Find the nth degree polynomial function with real coefficients satisfying the given conditions.

RUDIKE [14]

$\bf (x+1)(x-3)(x-2-4i)(x-2+4i)
\\\\
\textit{now, let's take a peek at }(x-2-4i)(x-2+4i)\\
\textit{and recall our }\textit{difference of squares}
\\ \quad \\
(a-b)(a+b) = a^2-b^2\qquad \qquad 
a^2-b^2 = (a-b)(a+b)\\\\
-----------------------------\\\\
(x-2-4i)(x-2+4i)\implies [(x-2)-(4i)][(x-2)+(4i)]
\\\\\
[(x-2)^2-(4i)^2]\implies [(x^2-2x+4)-(4^2\boxed{i^2})]
\\\\\
[(x^2-2x+4)-(16\cdot \boxed{-1})]\implies [(x^2-2x+4)+16]
\\\\
(x^2-2x+20)\\\\
-----------------------------$

$\bf thus
\\\\

\begin{array}{llll}
(x+1)(x-3)\\(x-2-4i)(x-2+4i)
\end{array}\implies (x+1)(x-3)(x^2-2x+20)
\\\\\\
(x^2-2x-3)(x^2-2x+20)\implies 
\begin{array}{llll}
x^4-2x^3+20x^2-2x^3\\+4x^2-40x+3x^2-6x+60
\end{array}
\\\\\\
x^4-4x^3+27x^2-46x+60$

4 0

3 years ago