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pickupchik [31]
3 years ago
5

ACDE is reflected over the y-axis.

Mathematics
1 answer:
GenaCL600 [577]3 years ago
5 0

Answer:

The Answer is C

Step-by-step explanation:

E flipped over the Y axis is 2,4 and C is the only option with 2,4 for E

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So I guess this would look like this originally 113/10, then as a mixed number 11 3/10
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The volume of a rectangular prism is (x4 + 4x3 + 3x2 + 8x + 4), and the area of its base is (x3 + 3x2 + 8). If the volume of a r
olga55 [171]
Dividing the volume by the base area, you find the height to be ...
\frac{x^{4}+4x^{3}+3x^{2}+8x+4}{x^{3}+3x^{2}+8}=x+1-\frac{4}{x^{3}+3x^{2}+8}

The height of the prism is x+1-\frac{4}{x^{3}+3x^{2}+8}

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2. Each of the following lines has a slope that can be determined by examining the graph. Use another point
Natali [406]

The slope, y-intercept and the equation of the following graph are as follows:

1(a)

slope : 1

y-intercept: 2

Equation: y = x + 2

(b)

slope : -  1 / 3

y-intercept: - 1

Equation: y = - 1 / 3x - 1

2.

(a)

slope : 3 / 4

y-intercept: 5 / 4

Equation: y = 3 / 4 x + 5 / 4

(b)

slope : - 3 / 2

y-intercept: 1 / 2

Equation: y = - 3 / 2 x + 1 / 2  

<h3 />

<h3>Slope intercept equation</h3>
  • y = mx + b

where

m = slope

b = y-intercept

Therefore lets find the slope, y-intercept and equation of the following graph.

1.

(a)

(0, 2)(1, 3)

m = 3 - 2 / 1 - 0 = 1

b = 2

y = x + 2

(b)

(0, -1)(-3, 0)

m = 0 + 1 / -3 - 0 = - 1 / 3

b = -1

y = - 1 / 3x - 1

2.

(a)

(1, 2)(-3, -1)

m = -1 - 2 / -3 - 1 = 3 / 4

2 = 3 / 4 (1) + b

b = 2 - 3 / 4  = 5 / 4

y = 3 / 4 x + 5 / 4

3.

(b)

(-3, 5)(1, -1)

m = - 1 - 5 / 1 + 3 = - 6 / 4 = - 3 / 2

-1 = - 3 /2 (1) + b

b = -1 + 3 / 2 = 1 /2

y = - 3 / 2 x + 1 / 2  

learn more on y-intercept here: brainly.com/question/2833377?referrer=searchResults

6 0
3 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
2 units/ 3 units if you triple the recipe would the rate change
Sergeeva-Olga [200]
No if you tripled the recipe the rate wouldn't change. The only difference would be that you're values would be tripled.
7 0
3 years ago
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