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2 years ago

Please help ASAP for the brainliest answer

Mathematics

2 answers:

Wittaler [7]2 years ago

7 0

Answer:

arithmetic sequence

Step-by-step explanation:

The next number is 21

Ronch [10]2 years ago

5 0

Answer:

Triangular Number Sequence

Step-by-step explanation:

This sequence is called the Triangular Number Sequence.

To find the next term, add 6 to 15.

The next term in the sequence is 21.

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Help me on this math question

Blababa [14]

Answer:

$\huge \boxed{X=30}$

B. 30

Step-by-step explanation:

First, you multiply by 5 from both sides of equation.

$\displaystyle \frac{x}{5}=6\times5$

Simplify, to find the answer.

$\displaystyle 6\times5=30$

$\huge \boxed{X=30}$ , which is our answer.

6 0

2 years ago

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How can you tell that a system of linear equations will have infinitely many solutions without computation?

Contact [7]

If you can manipulate the two equations so that they have exactly the same coefficients (e. g., 3x + 4y = 8 and 2x + 4y = 8), then you conclude that the two lines coincide (overlap), and that there are thus infinitely many solutions.

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2 years ago

What is the perimeter and are of rectangle for ABCD?

Rainbow [258]

Answer:

P=40 A=91 is the answer

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3 years ago

Reduce 72:81 to its simplist form

rewona [7]

$\frac { 72 }{ 81 } \\ \\ =\frac { 8 }{ 9 } \cdot \frac { 9 }{ 9 } \\ \\ =\frac { 8 }{ 9 } \cdot 1\\ \\ =\frac { 8 }{ 9 }$

8 0

2 years ago

Read 2 more answers

The third-degree Taylor polynomial about x = 0 of In(1 - x) is

gizmo_the_mogwai [7]

Answer:

$\displaystyle P_3(x) = -x - \frac{x^2}{2} - \frac{x^3}{3}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Functions
Function Notation

Calculus

Derivatives

Derivative Notation

Derivative Rule [Quotient Rule]: $\displaystyle \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}$

Derivative Rule [Chain Rule]: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

MacLaurin/Taylor Polynomials

Approximating Transcendental and Elementary functions
MacLaurin Polynomial: $\displaystyle P_n(x) = \frac{f(0)}{0!} + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + ... + \frac{f^{(n)}(0)}{n!}x^n$
Taylor Polynomial: $\displaystyle P_n(x) = \frac{f(c)}{0!} + \frac{f'(c)}{1!}(x - c) + \frac{f''(c)}{2!}(x - c)^2 + \frac{f'''(c)}{3!}(x - c)^3 + ... + \frac{f^{(n)}(c)}{n!}(x - c)^n$

Step-by-step explanation:

*Note: I will not be showing the work for derivatives as it is relatively straightforward. If you request for me to show that portion, please leave a comment so I can add it. I will also not show work for elementary calculations.

Step 1: Define

Identify

f(x) = ln(1 - x)

Center: x = 0

n = 3

Step 2: Differentiate

[Function] 1st Derivative: $\displaystyle f'(x) = \frac{1}{x - 1}$
[Function] 2nd Derivative: $\displaystyle f''(x) = \frac{-1}{(x - 1)^2}$
[Function] 3rd Derivative: $\displaystyle f'''(x) = \frac{2}{(x - 1)^3}$

Step 3: Evaluate Functions

Substitute in center x [Function]: $\displaystyle f(0) = ln(1 - 0)$
Simplify: $\displaystyle f(0) = 0$
Substitute in center x [1st Derivative]: $\displaystyle f'(0) = \frac{1}{0 - 1}$
Simplify: $\displaystyle f'(0) = -1$
Substitute in center x [2nd Derivative]: $\displaystyle f''(0) = \frac{-1}{(0 - 1)^2}$
Simplify: $\displaystyle f''(0) = -1$
Substitute in center x [3rd Derivative]: $\displaystyle f'''(0) = \frac{2}{(0 - 1)^3}$
Simplify: $\displaystyle f'''(0) = -2$

Step 4: Write Taylor Polynomial

Substitute in derivative function values [MacLaurin Polynomial]: $\displaystyle P_3(x) = \frac{0}{0!} + \frac{-1}{1!}x + \frac{-1}{2!}x^2 + \frac{-2}{3!}x^3$
Simplify: $\displaystyle P_3(x) = -x - \frac{x^2}{2} - \frac{x^3}{3}$

Topic: AP Calculus BC (Calculus I/II)

Unit: Taylor Polynomials and Approximations

Book: College Calculus 10e

5 0

3 years ago