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Nady [450]
3 years ago
7

Which of the following is an example of a nonterminating, nonrepeating decimal?

Mathematics
1 answer:
BARSIC [14]3 years ago
5 0

Answer:

Option B. 0.578942190874654...

Step-by-step explanation:

we know that

A non-terminating, non-repeating decimal is a decimal number that continues endlessly, with no group of digits repeating endlessly

example

the number \pi= 3.1415926535...

In this problem

the number 0.578942190874654... is a non-terminating, non-repeating decimal

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What is the nth term rule of the quadratic sequence below?
Ahat [919]

Answer:

The nth term = n^2 - 2n - 3.

Step-by-step explanation:

Here, we will be finding the nth term of a quadratic number sequence. A quadratic number sequence has nth term = an² + bn + c

Example 1

Write down the nth term of this quadratic number sequence.

-3, 8, 23, 42, 65...

Step 1: Confirm the sequence is quadratic. This is done by finding the second difference.

Sequence = -3, 8, 23, 42, 65

1st difference = 11,15,19,23

2nd difference = 4,4,4,4

Step 2: If you divide the second difference by 2, you will get the value of a.

4 ÷ 2 = 2

So the first term of the nth term is 2n²

Step 3: Next, substitute the number 1 to 5 into 2n².

n = 1,2,3,4,5

2n² = 2,8,18,32,50

Step 4: Now, take these values (2n²) from the numbers in the original number sequence and work out the nth term of these numbers that form a linear sequence.

n = 1,2,3,4,5

2n² = 2,8,18,32,50

Differences = -5,0,5,10,15

Now the nth term of these differences (-5,0,5,10,15) is 5n -10.

So b = 5 and c = -10.

Step 5: Write down your final answer in the form an² + bn + c.

2n² + 5n -10

Example 2

Write down the nth term of this quadratic number sequence.

9, 28, 57, 96, 145...

Step 1: Confirm if the sequence is quadratic. This is done by finding the second difference.

Sequence = 9, 28, 57, 96, 145...

1st differences = 19,29,39,49

2nd differences = 10,10,10

Step 2: If you divide the second difference by 2, you will get the value of a.

10 ÷ 2 = 5

So the first term of the nth term is 5n²

Step 3: Next, substitute the number 1 to 5 into 5n².

n = 1,2,3,4,5

5n² = 5,20,45,80,125

Step 4: Now, take these values (5n²) from the numbers in the original number sequence and work out the nth term of these numbers that form a linear sequence.

n = 1,2,3,4,5

5n² = 5,20,45,80,125

Differences = 4,8,12,16,20

Now the nth term of these differences (4,8,12,16,20) is 4n. So b = 4 and c = 0.

Step 5: Write down your final answer in the form an² + bn + c.

5n² + 4n

5 0
3 years ago
Multiply the fractions:<br> 2 / 5 ∙ 3 / 4.
Bingel [31]

Answer:

\frac{3}{10}

Step-by-step explanation:

\frac{2}{5} * \frac{3}{4} can be solved by just multiplying the numerators together and the denominators together

2 * 3 = 6 and 5 * 4 = 20 so the answer would be \frac{6}{20}

This simplifies to \frac{3}{10}

4 0
4 years ago
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