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Ostrovityanka [42]
3 years ago
14

A painter leans a 34-ft ladder against a building. The base of the ladder is 16 ft from the building. To the nearest foot, how h

igh on the building does the ladder reach?
A) 30 feet
B) 38 feet
C) 26 feet
D) 18 feet
Mathematics
2 answers:
Gnesinka [82]3 years ago
4 0

Answer:

The height reached by ladder on the building is 30 ft.

Step-by-step explanation:

Let x be the height reached by ladder on the building.

Since, the building must be perpendicular with respect to surface,

Thus,

(The length of the ladder)² = (The  base of the ladder from the building)² + ( The height reached by ladder on the building)²

Given,

The length of the ladder = 34 ft,

The  base of the ladder from the building = 16 ft.

So,

34^2 = 16^2 + x^2

1156 = 256 + x^2

1156 - 256 = x^2

900 = x^2

\implies x = 30

Hence, the height reached by ladder on the building is 30 ft.

dybincka [34]3 years ago
3 0
Answer = A). 34 squared minus 16 squared = 900. Square-root that to get 30.

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\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
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