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3 years ago

Which sequence is geometric and has 14 as its fifth term and 12 as the common ratio?

1 answer:

7 0

Geometric sequence is
an=a1(r)^(n-1)
an=nth erm
a1=first term
r=common ratio

5th term=14
a5=a1(12)^(5-1)=15

a1(12)^4=15
a1(20736)=15
divide both sides by 20736
a1=15/20736=5/6912
geometric sequence is
an=(5/6912)(12)^(n-1)

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Which is equivalent to RootIndex 4 StartRoot 9 EndRoot Superscript one-half x?

nata0808 [166]

Answer:

${9}^ {\frac{1}{8}x}$

Step-by-step explanation:

We want to find an equivalent expression for

$(\sqrt[4]{9})^{ \frac{1}{2}x}$

To find an equivalent expression, we need to apply the following property of exponents:

${a}^{ \frac{m}{n}}=( \sqrt[n]{ {a}} )^{m}$

We let a=9, n=4 and m=½x

Then :

${9}^{ \frac{ \frac{1}{2}x}{4}}=( \sqrt[4]{ {9}} )^{ \frac{1}{2}x}$

Simplify the left hand side to get:

${9}^ {\frac{1}{8}x} =( \sqrt[4]{ {9}} )^{ \frac{1}{2}x}$

Therefore the correct answer is:

${9}^ {\frac{1}{8}x}$

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