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Nikolay [14]
3 years ago
9

The bigger

Mathematics
1 answer:
Arte-miy333 [17]3 years ago
5 0

Answer:

See Explanation

Step-by-step explanation:

Required

Prove that Dexter is wrong

To prove that Dexter is wrong, we use the following numbers.

Number = 4

The factors are:

Factor = 1,2\ and\ 4

3 factors

Using a greater number:

Number = 5

The factors are:

Factor = 1\ and\ 5

2 factors

From the above analysis:

4 has 3 factors and 5 has 2 factors; which shows that 5 has fewer factors

This shows that Dexter is wrong

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PLEASE HELP 100 POINTS GUARANTEED! PLEASE PUT THE ANSWER IN AN ALGEBRAIC EXPRESSION!!! ANSWER ALL AND YOU WILL RECEIVE BRAINLIES
zlopas [31]

Answer:

1.

-2 = x/4 + 1

Subtract 1 from both sides:

-2 - 1 = x/4 + 1 - 1

-3 = x/4

Multiply both sides by 4:

-3 * 4 = x/4 * 4

-12 = x

2.

3x+5

3.

X= 12 or 18

This could be read two different ways.

0=2/3x-12

12=2/3x

12÷2/3=x

12*3/2=x

18=x

—-OR—-

0=2/3(x-12)

0=2/3x-8

8=2/3x

8÷2/3=x

8*3/2=x

12=x

7 0
3 years ago
Where do I plot 1.25 and -1.25
slavikrds [6]

Answer:

I'd need to see more, but because they are the same number just positive and negative they'd be on opposite ends of zero

7 0
3 years ago
Determine the values of xfor which the function can be replaced by the Taylor polynomial if the error cannot exceed 0.001.(Enter
MrMuchimi

Answer:

The values of x for which the function can be replaced by the Taylor polynomial if the error cannot exceed 0.001 is 0 < x < 0.3936.

Step-by-step explanation:

Note: This question is not complete. The complete question is therefore provided before answering the question as follows:

Determine the values of x for which the function can be replaced by the Taylor polynomial if the error cannot exceed 0.001. f(x) = e^x ≈ 1 + x + x²/2! + x³/3!, x < 0

The explanation of the answer is now provided as follows:

Given:

f(x) = e^x ≈ 1 + x + x²/2! + x³/3!, x < 0 …………….. (1)

R_{3} = (x) = (e^z /4!)x^4

Since the aim is R_{3}(x) < 0.001, this implies that:

(e^z /4!)x^4 < 0.0001 ………………………………….. (2)

Multiply both sided of equation (2) by (1), we have:

e^4x^4 < 0.024 ……………………….......……………. (4)

Taking 4th root of both sided of equation (4), we have:

|xe^(z/4) < 0.3936 ……………………..........…………(5)

Dividing both sides of equation (5) by e^(z/4) gives us:

|x| < 0.3936 / e^(z/4) ……………….................…… (6)

In equation (6), when z > 0, e^(z/4) > 1. Therefore, we have:

|x| < 0.3936 -----> 0 < x < 0.3936

Therefore, the values of x for which the function can be replaced by the Taylor polynomial if the error cannot exceed 0.001 is 0 < x < 0.3936.

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It would be answer D your welcome
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