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Art [367]
4 years ago
11

Use the quadrant system to find the area of the polygon

Mathematics
1 answer:
AfilCa [17]4 years ago
3 0

Answer:

The area of the polygon is 64 units²

Step-by-step explanation:

* Lets explain how to solve the problem

- The polygon has 5 sides and 5 vertices

- To find its area by easy way split it into two trapezoid by a horizontal

  line y = 1

- The vertices of trapezoid (1) are (-5 , 1) , (4 , 1) , (-5 , 5) , (0 , 5)

- The vertices of one of the parallel bases are (-5 , 1) , (4 , 1)

- The length of this base = 4 - (-5) = 4 + 5 = 9 units

- The vertices of the other parallel bases are (-5 , 5) , (0 , 5)

- The length of this base = 0 - (-5) = 0 + 5 = 5 units

- The length of its height = 5 - 1 = 4 units

- The area of the trapezoid = 1/2 (b1 + b2) × h

∴ Area of trapezoid 1 = 1/2 (9 + 5) × 4 = 28 units²

- The vertices of trapezoid (2) are (4 , 1) , (-5 , 1) , (-5 , -5) , (-2 , -5)

- The vertices of one of the parallel bases are (4 , 1) , (-5 , 1)

- The length of this base = 4 - (-5) = 4 + 5 = 9 units

- The vertices of the other parallel bases are (-5 , -5) , (-2 , -5)

- The length of this base = (-2) - (-5) = -2 + 5 = 3 units

- The length of its height = 1 - (-5) = 1 + 5 = 6 units

∴ Area of trapezoid 1 = 1/2 (9 + 3) × 6 = 36 units²

∵ Area of the polygon = Area trapezoid (1) + Area trapezoid (2)

∵ Area trapezoid (1) = 28 units²

∵ Area trapezoid (2) = 36 units²

∴ Area of polygon = 28 + 36 = 64 units²

∴ The area of the polygon is 64 units²

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Answer:

\displaystyle \frac{dy}{dx} \bigg| \limit_{(1, 4)} = 2

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

Equality Properties

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  • Division Property of Equality
  • Addition Property of Equality
  • Subtraction Property of Equality

<u>Algebra I</u>

  • Coordinates (x, y)
  • Exponential Rule [Root Rewrite]:                                                                 \displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}
  • Exponential Rule [Rewrite]:                                                                           \displaystyle b^{-m} = \frac{1}{b^m}

<u>Calculus</u>

Derivatives

Derivative Notation

Derivative of a constant is 0

Implicit Differentiation

Basic Power Rule:

  • f(x) = cxⁿ
  • f’(x) = c·nxⁿ⁻¹

Step-by-step explanation:

<u>Step 1: Define</u>

\displaystyle \sqrt{x} - \sqrt{y} = -1

Point (1, 4)

<u>Step 2: Differentiate</u>

  1. [Function] Rewrite [Exponential Rule - Root Rewrite]:                               \displaystyle x^{\frac{1}{2}} - y^{\frac{1}{2}} = -1
  2. [Implicit Differentiation] Basic Power Rule:                                                 \displaystyle \frac{1}{2}x^{\frac{1}{2} - 1} - \frac{1}{2}y^{\frac{1}{2} - 1}\frac{dy}{dx} = 0
  3. [Implicit Differentiation] Simplify Exponents:                                               \displaystyle \frac{1}{2}x^{\frac{-1}{2}} - \frac{1}{2}y^{\frac{-1}{2}}\frac{dy}{dx} = 0
  4. [Implicit Differentiation] Rewrite [Exponential Rule - Rewrite]:                   \displaystyle \frac{1}{2x^{\frac{1}{2}}} - \frac{1}{2y^{\frac{1}{2}}}\frac{dy}{dx} = 0
  5. [Implicit Differentiation] Isolate <em>y</em> terms:                                                       \displaystyle -\frac{1}{2y^{\frac{1}{2}}}\frac{dy}{dx} = -\frac{1}{2x^{\frac{1}{2}}}
  6. [Implicit Differentiation] Isolate \displaystyle \frac{dy}{dx}:                                                               \displaystyle \frac{dy}{dx} = \frac{2y^{\frac{1}{2}}}{2x^{\frac{1}{2}}}
  7. [Implicit Differentiation] Simplify:                                                                 \displaystyle \frac{dy}{dx} = \frac{y^{\frac{1}{2}}}{x^{\frac{1}{2}}}

<u>Step 3: Evaluate</u>

  1. Substitute in point [Derivative]:                                                                     \displaystyle \frac{dy}{dx} = \frac{(4)^{\frac{1}{2}}}{(1)^{\frac{1}{2}}}
  2. Exponents:                                                                                                     \displaystyle \frac{dy}{dx} = \frac{2}{1}
  3. Division:                                                                                                         \displaystyle \frac{dy}{dx} = 2

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

Unit: Implicit Differentiation

Book: College Calculus 10e

6 0
3 years ago
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