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valentinak56 [21]
3 years ago
8

TIMED, need the answer

Mathematics
2 answers:
Kay [80]3 years ago
7 0
It’s the last option, D. Adjacent means next to or adjoining, and 8 is a complementary angle to 7 (: hope this helps
jeyben [28]3 years ago
4 0
See the answer to brainly.com/question/10046378 for a full discussion of the answer choices.

The appropriate selection is ...
  ◉  ∠1 and ∠2
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A=1/2b(c+d) solve for c
TiliK225 [7]

Answer:

c = \frac{2A}{b} - d

Step-by-step explanation:

Multiply both sides by 2 to eliminate the fraction

2A = b(c + d) ← divide both sides by b

\frac{2A}{b} = c + d ← subtract d from both sides

\frac{2A}{b} - d = c

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3(x+1)=5(x−2)+7<br> solve for x
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2 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:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Algebra I</u>

  • Functions
  • Function Notation

<u>Calculus</u>

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.

<u />

<u>Step 1: Define</u>

<em>Identify</em>

f(x) = ln(1 - x)

Center: x = 0

<em>n</em> = 3

<u>Step 2: Differentiate</u>

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

<u>Step 3: Evaluate Functions</u>

  1. Substitute in center <em>x</em> [Function]:                                                                     \displaystyle f(0) = ln(1 - 0)
  2. Simplify:                                                                                                             \displaystyle f(0) = 0
  3. Substitute in center <em>x</em> [1st Derivative]:                                                             \displaystyle f'(0) = \frac{1}{0 - 1}
  4. Simplify:                                                                                                             \displaystyle f'(0) = -1
  5. Substitute in center <em>x</em> [2nd Derivative]:                                                           \displaystyle f''(0) = \frac{-1}{(0 - 1)^2}
  6. Simplify:                                                                                                             \displaystyle f''(0) = -1
  7. Substitute in center <em>x</em> [3rd Derivative]:                                                            \displaystyle f'''(0) = \frac{2}{(0 - 1)^3}
  8. Simplify:                                                                                                             \displaystyle f'''(0) = -2

<u>Step 4: Write Taylor Polynomial</u>

  1. 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
  2. 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
I need help with there questions
Charra [1.4K]
Number six is c number five is c too
6 0
3 years ago
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