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ExtremeBDS [4]
4 years ago
5

What is the slope of the line y= x/5 + 16

Mathematics
2 answers:
Brilliant_brown [7]4 years ago
6 0
The y intercept to this is 16 and the slope is 1/5. Hope this help

m_a_m_a [10]4 years ago
3 0
Your answer then would be 1/5. The slope is the number that is by the side of teh x. also knows as the x-intercept... the slope is teh rise/run so if you put this on a graph you will have to go up once and run 5 times.. hope I have helped. 

Let me know if you got it rigth ok
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Answer:

(D)  \displaystyle \frac{12}{1 + ln8}

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

Basic Power Rule:

  • f(x) = cxⁿ
  • f’(x) = c·nxⁿ⁻¹

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

Integration

Integration Rule [Fundamental Theorem of Calculus 2]:                                           \displaystyle \frac{d}{dx}[\int\limits^x_a {f(t)} \, dt] = f(x)

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

<em />\displaystyle f(x) = \int\limits^{x^3}_1 {\frac{1}{1 + ln(t)}} \, dt<em />

<em />

<u>Step 2: Differentiate</u>

  1. Chain Rule:                                                                                                       \displaystyle f'(x) = \frac{d}{dx} \bigg[ \int\limits^{x^3}_1 {\frac{1}{1 + ln(t)}} \, dt \bigg] \cdot \frac{d}{dx}[x^3]
  2. Integration Rule - Fundamental Theorem of Calculus 2:                              \displaystyle f'(x) = \frac{1}{1 + ln(x^3)} \cdot \frac{d}{dx}[x^3]
  3. Basic Power Rule:                                                                                             \displaystyle f'(x) = \frac{1}{1 + ln(x^3)} \cdot 3x^{3 - 1}
  4. Simplify:                                                                                                             \displaystyle f'(x) = \frac{3x^2}{1 + ln(x^3)}

<u>Step 3: Evaluate</u>

  1. Substitute in <em>x</em> [Derivative]:                                                                              \displaystyle f'(2) = \frac{3(2)^2}{1 + ln(2^3)}
  2. Exponents:                                                                                                        \displaystyle f'(2) = \frac{3(4)}{1 + ln(8)}
  3. Multiply:                                                                                                             \displaystyle f'(2) = \frac{12}{1 + ln(8)}

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Integration

Book: College Calculus 10e

7 0
3 years ago
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