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coldgirl [10]
3 years ago
6

What is the median??

Mathematics
2 answers:
prisoha [69]3 years ago
8 0

Answer: 8

Step-by-step explanation: The median is the middle number in the data set when the data set is written from least to greatest.

For example, 8 would be the median here since it's in the middle.

stellarik [79]3 years ago
7 0
The "middle" of a sorted list of numbers. To find the Median, place the numbers in value order and find the middle number. ... The middle number is 15, so the median is 15. (When there are two middle numbers we average them.)
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Answer:

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Step-by-step explanation:

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The angle we want to calculate here is the angle at circumference but we were given the angle at the center

So the value of the angle at the circumference would be 108/2 = 54

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4 years ago
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Step-by-step explanation:

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Hi, how do we do this question?​
Nutka1998 [239]

Answer:

\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{-2(ln|3x + 1| - 3x)}{9} + C

General Formulas and Concepts:

<u>Algebra I</u>

  • Terms/Coefficients
  • Factoring

<u>Algebra II</u>

  • Polynomial Long Division

<u>Calculus</u>

Differentiation

  • Derivatives
  • Derivative Notation

Derivative Property [Multiplied Constant]:                                                           \displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)

Derivative Property [Addition/Subtraction]:                                                         \displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]  

Basic Power Rule:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

Integration

  • Integrals
  • Integration Constant C
  • Indefinite Integrals

Integration Rule [Reverse Power Rule]:                                                               \displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C

Integration Property [Multiplied Constant]:                                                         \displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx

Integration Property [Addition/Subtraction]:                                                       \displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx

Logarithmic Integration

U-Substitution

Step-by-step explanation:

*Note:

You could use u-solve instead of rewriting the integrand to integrate this integral.

<u>Step 1: Define</u>

<em>Identify</em>

\displaystyle \int {\frac{2x}{3x + 1}} \, dx

<u>Step 2: Integrate Pt. 1</u>

  1. [Integrand] Rewrite [Polynomial Long Division (See Attachment)]:           \displaystyle \int {\frac{2x}{3x + 1}} \, dx = \int {\bigg( \frac{2}{3} - \frac{2}{3(3x + 1)} \bigg)} \, dx
  2. [Integral] Rewrite [Integration Property - Addition/Subtraction]:               \displaystyle \int {\frac{2x}{3x + 1}} \, dx = \int {\frac{2}{3}} \, dx - \int {\frac{2}{3(3x + 1)}} \, dx
  3. [Integrals] Rewrite [Integration Property - Multiplied Constant]:               \displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{2}{3}\int {} \, dx - \frac{2}{3}\int {\frac{1}{3x + 1}} \, dx
  4. [1st Integral] Reverse Power Rule:                                                               \displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{2}{3}x - \frac{2}{3}\int {\frac{1}{3x + 1}} \, dx

<u>Step 3: Integrate Pt. 2</u>

<em>Identify variables for u-substitution.</em>

  1. Set <em>u</em>:                                                                                                             \displaystyle u = 3x + 1
  2. [<em>u</em>] Differentiate [Basic Power Rule]:                                                             \displaystyle du = 3 \ dx

<u>Step 4: Integrate Pt. 3</u>

  1. [Integral] Rewrite [Integration Property - Multiplied Constant]:                 \displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{2}{3}x - \frac{2}{9}\int {\frac{3}{3x + 1}} \, dx
  2. [Integral] U-Substitution:                                                                               \displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{2}{3}x - \frac{2}{9}\int {\frac{1}{u}} \, du
  3. [Integral] Logarithmic Integration:                                                               \displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{2}{3}x - \frac{2}{9}ln|u| + C
  4. Back-Substitute:                                                                                            \displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{2}{3}x - \frac{2}{9}ln|3x + 1| + C
  5. Factor:                                                                                                           \displaystyle \int {\frac{2x}{3x + 1}} \, dx = -2 \bigg( \frac{1}{9}ln|3x + 1| - \frac{x}{3}  \bigg) + C
  6. Rewrite:                                                                                                         \displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{-2(ln|3x + 1| - 3x)}{9} + C

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

Unit: Integration

Book: College Calculus 10e

8 0
3 years ago
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