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vichka [17]
3 years ago
12

Which point is a solution to y<_3x-4?

Mathematics
2 answers:
dalvyx [7]3 years ago
5 0

Answer:

D. (3,1)

Step-by-step explanation:

Plug the X and Y values into the expression y<3x-4

Substitute 3 for X and 1 for Y and simplify

y < 3x - 4

1 < 3(3) - 4

1< 9 - 4

1< 5

In this case 1 is less than 5, that is a correct statement, so your answer is C.

GenaCL600 [577]3 years ago
3 0

Answer:

D

Step-by-step explanation:

Replace by the coordinates

3*3 -1>1

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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

3 0
4 years ago
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