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3 years ago

What is the 33rd term of this arithmetic sequence? 12, 7, 2, -3, -8, …

1 answer:

5 0

$12,7,2,-3,-8,... \\ \\
a_1=12 \\
d=a_2-a_1=7-12=-5 \\ \\
a_n=a_1+d(n-1) \\
a_{33}=12-5(33-1) \\
a_{33}=12-5 \times 32 \\
a_{33}=12-160 \\
a_{33}=-148$

The answer is C.

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4 years ago

Which expression is equivalent to log Subscript w Baseline StartFraction (x squared minus 6) Superscript 4 Baseline Over RootInd

lora16 [44]

The logarithmic expression $\rm log_w\dfrac{(x^2-6)^4}{\sqrt[3]{x^2-8}}$ can be expressed as $\rm 4log_w{(x^2-6)} - {\dfrac{1}{3}}log_w(x^2-8)$ .

Given to us

$\rm log_w\dfrac{(x^2-6)^4}{\sqrt[3]{x^2-8}}$

<h3>Which expression is equivalent to $\rm log_w\dfrac{(x^2-6)^4}{\sqrt[3]{x^2-8}}$ ?</h3>

To solve the problem we will use the basic logarithmic properties,

$\rm log_w\dfrac{(x^2-6)^4}{\sqrt[3]{x^2-8}}$

Using the logarithmic property $\rm log_a\dfrac{x}{y} = log_ax-log_ay$ ,

$=\rm log_w{(x^2-6)^4} - log_w{\sqrt[3]{x^2-8}}$

Using the exponential property $\sqrt[m]{a^n} = a^{\frac{n}{m}}$ ,

$=\rm log_w{(x^2-6)^4} - log_w(x^2-8)^{\dfrac{1}{3}}$

Using the logarithmic property $\rm log_aB^x = xlog_aB$ ,

$=\rm 4log_w{(x^2-6)} - {\dfrac{1}{3}}log_w(x^2-8)$

Hence, the logarithmic expression $\rm log_w\dfrac{(x^2-6)^4}{\sqrt[3]{x^2-8}}$ can be expressed as $\rm 4log_w{(x^2-6)} - {\dfrac{1}{3}}log_w(x^2-8)$ .

Learn more about Logarithmic Expression:

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4 years ago

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3 years ago

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→ Solutions

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Answer
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