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2 years ago

Help please with this

Mathematics

1 answer:

Elena-2011 [213]2 years ago

3 0

Negative reciprocals of each other

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2 years ago

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If a₁ = 4 and an = 5an-1 then find the value of a5.

igor_vitrenko [27]

The value of $a_{5}$ is 2500, when $a_{1}=4$ and $5a_{n-1}$ .

Given that, $a_{1}=4$ and $5a_{n-1}$ .

We need to find the value of $a_{5}$ .

<h3>What is an arithmetic sequence?</h3>

An arithmetic progression or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant.

Now, to find the value of $a_{5}$ :

$a_{2} =5a_{2-1}=5a_{1}=5 \times4=20$

$a_{3} =5a_{3-1}=5a_{2}=5 \times20=100$

$a_{4} =5a_{4-1}=5a_{3}=5 \times100=500$

$a_{5} =5a_{5-1}=5a_{4}=5 \times500=2500$

Therefore, the value of $a_{5}$ is 2500.

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2 years ago

PLEASE HELP! 50 POINTS

Alex73 [517]

5.1 ANSWERS.pdf

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3 years ago

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OLEGan [10]

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3 years ago

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Solve (1/81)^x*1/243=(1/9)^−3−1 by rewriting each side with a common base.

elena55 [62]

Answer:

$x=(243)log_{\frac{1}{81}}[(\frac{1}{81})-1]$

Step-by-step explanation:

you have the following formula:

$(\frac{1}{81})^{\frac{x}{243}}=(\frac{1}{9})^{-3}-1$

To solve this equation you use the following properties:

$log_aa^x=x$

Thne, by using this propwerty in the equation (1) you obtain for x

$log_{(\frac{1}{81})}(\frac{1}{81})^{\frac{x}{243}}=log_{\frac{1}{81}}[(\frac{1}{81})-1]\\\\\frac{x}{243}=log_{\frac{1}{81}}[(\frac{1}{81})-1]\\\\x=(243)log_{\frac{1}{81}}[(\frac{1}{81})-1]$

8 0

3 years ago