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3 years ago

12

Consider the following relation:(1,12) (3.8) (3,11) (6,9) (7,11) . Which ordered pair could be removed so that the relation is a

function.

1 answer:

Y_Kistochka [10]3 years ago

6 0

You could remove 3,11)

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3x – 2y = 7<br> y = x + 2<br><br> What the answer to this?

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Answer:

{x,y}={−3,−1}

Step-by-step explanation:

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AnnyKZ [126]

Y=5x-1
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Manny makes $8.75 an hour working at the video store. His paycheck shows that he worked 37.5 hours over the past week. How much

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Y=8.75x
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3 years ago

Complete the statement below.

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3 years ago

a cell phone company plans to market a new smartphone. they have already sold 612 units durning the first week of the campaign.

Vadim26 [7]

The first term is 612.

The common ratio is 1.08 and

The recursive rule is $a_{n} = a^{n-1} \times r$

<u>Step-by-step explanation:</u>

the question to the problem is to write the values of the first term, common ratio, and expression for the recursive rule.

<u>The first term :</u>

In geometric sequence, the first term is given as $a_{1}$ .

⇒ $a_{1} = 612$

Now, the geometric sequence follows as 612, 661, ........

<u>The common ratio (r) :</u>

It is the ratio between two consecutive numbers in the sequence.

Therefore, to determine the common ratio, you just divide the number from the number preceding it in the sequence.

⇒ r = 661 divided by 612

⇒ r = 1.08

<u>To find the recursive rule :</u>

A geometric series is of the form a,ar,ar2,ar3,ar4,ar5........

Here, first term $a_{1} = a$ and other terms are obtained by multiplying by r.

Observe that each term is r times the previous term.
Hence to get nth term we multiply (n−1)th term by r .

The recursive rule is of the form $a_{n} = a^{n-1} \times r$

This is called recursive formula for geometric sequence.

We know that r = 1.08 and $a_{1}$ = 612.

To find the second term $a_{2}$ , use the recursive rule $a_{n} = a^{n-1} \times r$

⇒ $a_{2} = a^{2-1}\times r$

⇒ $a_{2} = a^{1}\times r$

⇒ $a_{2} = 612\times 1.08$

⇒ $a_{2} = 661$

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3 years ago