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3 years ago

The first 5 terms of a number pattern are shown below.

1 answer:

4 0

$\bf 4~~,~~\stackrel{4+5}{9}~~,~~\stackrel{9+5}{14}~~,~~\stackrel{14+5}{19}~~,~~\stackrel{19+5}{24}\qquad \impliedby \stackrel{\textit{common difference}}{d=5} \\\\[-0.35em] ~\dotfill\\\\ n^{th}\textit{ term of an arithmetic sequence} \\\\ a_n=a_1+(n-1)d\qquad \begin{cases} a_n=n^{th}\ term\\ n=\textit{term position}\\ a_1=\textit{first term}\\ d=\textit{common difference}\\ \cline{1-1} a_1=4\\ d=5 \end{cases} \\\\\\ a_n=4+(n-1)5\implies a_n=4+5n-5\implies a_n=5n-1$

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What is triangle angle sum theorem

liubo4ka [24]

The triangle angle sum theorem states that the 3 angles within the triangle must have a sum that equals 180 degrees.

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3 years ago

Marianna [84]

Answer:

B. {16, 19, 20}

Step-by-step explanation:

The triangle inequality requires for any sides a, b, c you must have ...

a + b > c

b + c > a

c + a > b

The net result of those requirements are ...

the sum of the two shortest sides must be greater than the longest side
the length of the third side lies between the difference and sum of the other two sides

If we look at the offered side length choices, we see ...

A: 8+11 = 19 . . . not > 19; not a triangle

B: 16+19 = 35 > 20; could be a triangle

C: 3+4 = 7 . . . not > 8; not a triangle

D: 5+5 = 10 . . . not > 11; not a triangle

The side lengths {16, 19, 20} could represent the sides of a triangle.

_____

Additional comment

The version of triangle inequality shown above ensures that a triangle will have non-zero area.

The alternative version of the triangle inequality uses ≥ instead of >. Triangles where a+b=c will look like a line segment--they will have zero area. Many authors disallow this case. (If it were allowed, then {8, 11, 19} would also be a "triangle.")

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2 years ago

Graph the equation (x - 3)2 + (y + 4)2 = 144

Anna007 [38]

It will be a circle with a center of (3, -4) and a radius of 12.

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3 years ago

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Paul [167]

$\red{answer}$

$-150

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3 years ago

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