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Talja [164]
3 years ago
6

What are the points on the graph?

Mathematics
1 answer:
scoray [572]3 years ago
8 0

Answer:

69 ;)

Step-by-step explanation:

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Find the limits<br>The answer is - 3/4
Kipish [7]

Hello from MrBillDoesMath!

Answer:

This is a puzzling one! The answer is indeed -3/4 but I was unable to prove it using mathematics appropriate to the "middle school" level. Sorry I couldn't be more helpful.

(To prove it I generated the first few terms of its Laurent series, a Calculus topic. The first few terms are  -(3/4) + (41/64x) +  and as x approaches  minus infinity the term with "x" in the denominator goes to zero, leaving the constant value -3/4 as the answer. )


Thank you,

MrB


8 0
4 years ago
What shape has 6 square faces​
diamong [38]

Answer: A Hexagon

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Write the domain AND range of the following graph.
Effectus [21]

Answer:

Domain: (-∞,∞), Range: [2,∞)

3 0
3 years ago
Tim is playing a game. His score in round 1 is 1.7 points, in round 2 is 5.1 points, in round 3 is 15.3 points, and it continues
tatiyna

The recursive definition for the geometric sequence is given as follows:

a_n = 3a_{n-1}, a_1 = 1.7

<h3>What is a geometric sequence?</h3>

A geometric sequence is a sequence in which the result of the division of consecutive terms is always the same, called common ratio q.

The nth term of a geometric sequence is given by:

a_n = a_1q^{n-1}

In which a_1 is the first term.

The recursive definition of a geometric sequence is given by:

a_n = qa_{n-1}

In this problem, we have that the first term and the common ratio are given, respectively, by:

a_1 = 1.7, q = \frac{15.3}{5.1} = \frac{5.1}{1.7} = 3.

Hence the recursive definition is given by:

a_n = 3a_{n-1}, a_1 = 1.7

More can be learned about geometric sequences at brainly.com/question/11847927

#SPJ1

4 0
2 years ago
Use the line tool to graph the function f(x)=12x−3 Graph the Problem Please
mestny [16]
F(x) = 12x - 3 would look like this graphed.

Hope this helped, you're welcome! :)

7 0
3 years ago
Read 2 more answers
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