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iren [92.7K]
3 years ago
13

Would you round 14.47 to 14 or 15??

Mathematics
2 answers:
zhuklara [117]3 years ago
6 0

Answer:

15

Step-by-step explanation:

If the first digit is 4 or under then you either leave it or round it down, if the digit is 5 or over then round up no matter what. If there is a digit in the hundreds place then the same rule applies as well, in this example you can round the .47 to a .5 which is 14.5 and since the digit is 5 we round it to 15.

torisob [31]3 years ago
5 0

Answer:

14

Step-by-step explanation:

When rounding to a whole number, you round down if the decimal in the 10th's place is below 5, you round up if the decimal is 5 or above.  

Ex:  12.4  rounds down to 12 because there is a 4 in the 10th's

       12.5 rounds up to 13 because there is a 5 in the 10th's place

        14.47 rounds down because there is a 4 in the 10th's place, the 7 in the hundredth's place would only affect the 4 in the 10th's place when rounding.  It wouldn't affect the 14.  

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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

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larisa86 [58]

Answer:

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Step-by-step explanation:

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Eduardwww [97]

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To find the y-intercept, you have to substitute the point (8,2) into y = mx + b

2 = 1(8) + b

b = -6

The equation is y = x - 6

8 0
3 years ago
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