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3 years ago

How do I solve this. Log 5x = log (2x+9)

1 answer:

3 0

$D:5x\ \textgreater \ 0 \wedge 2x+9\ \textgreater \ 0\\
D:x\ \textgreater \ 0 \wedge 2x\ \textgreater \ -9\\
D:x\ \textgreater \ 0 \wedge x\ \textgreater \ -\dfrac{9}{2}\\
D:x\ \textgreater \ 0\\\\
\log5x=\log(2x+9)\\
5x=2x+9\\
3x=9\\
x=3$

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Calculate the arithmetic sequence in which a9=17 and the common difference is d=-2.1

jek_recluse [69]

Answer:

$S_{31}=71.3$

Step-by-step explanation:

The nth term of an arithmetic sequence is given by the formula,

$U_n=a_1+(n-1)d$

We were given that the 9th term is $17$ .

$\Rightarrow 17=a_1+(9-1)(-2.1)$

$\Rightarrow 17=a_1+(8)\times(-2.1)$

$\Rightarrow 17=a_1-\frac{84}{5}$

$\Rightarrow 17+\frac{84}{5}=a_1$

$\Rightarrow a_1=\frac{169}{5}$

The sum of the first n-terms is given by the formula,

$S_n=\frac{n}{2}(2a_1+(n-1)d)$

To find $S_{31}$ , we substitute $n=31$ , $a_1=\frac{169}{5}$ and $d=-2.1$ .

$\Rightarrow S_{31}=\frac{31}{2}(2(\frac{169}{5}+(31-1)(-2.1))$

$\Rightarrow S_{31}=\frac{31}{2}(2(\frac{169}{5}+(30)(-2.1))$

$\Rightarrow S_{31}=\frac{31}{2}(\frac{23}{5})$

$\Rightarrow S_{31}=\frac{713}{10}$

$\Rightarrow S_{31}=71.3$

The correct answer is D

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nataly862011 [7]

Answer:

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Arte-miy333 [17]

Answer:

x < -14

Step-by-step explanation:

Step 1: Subtract 6 from both sides.

$-3x+6-6 > 48-6$
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Step 2: Divide both sides by -3 and flip the inequality sign.

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Therefore, the answer is x < -14.

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Answer:

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