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marshall27 [118]
3 years ago
15

∠A and ∠B are vertical angles. If m∠A=(7x-20)° and m∠B=(6x-1)°, find the measure of ∠A. (Just type answer as a number.)

Mathematics
1 answer:
WINSTONCH [101]3 years ago
6 0

Answer:

x=19 so ∠A=113

Step-by-step explanation:

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If -y-2x^3=Y^2 then find D^2y/dx^2 at the point (-1,-2) in simplest form
algol13

Answer:

\frac{d^2y}{dx^2} = \frac{-4}{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>Algebra I</u>

  • Factoring

<u>Calculus</u>

Implicit Differentiation

The derivative of a constant is equal to 0

Basic Power Rule:

  • f(x) = cxⁿ
  • f’(x) = c·nxⁿ⁻¹

Product Rule: \frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)

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

Quotient Rule: \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}

Step-by-step explanation:

<u>Step 1: Define</u>

-y - 2x³ = y²

Rate of change of tangent line at point (-1, -2)

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

<em>Find 1st Derivative</em>

  1. Implicit Differentiation [Basic Power Rule]:                                                  -y'-6x^2=2yy'
  2. [Algebra] Isolate <em>y'</em> terms:                                                                              -6x^2=2yy'+y'
  3. [Algebra] Factor <em>y'</em>:                                                                                       -6x^2=y'(2y+1)
  4. [Algebra] Isolate <em>y'</em>:                                                                                         \frac{-6x^2}{(2y+1)}=y'
  5. [Algebra] Rewrite:                                                                                           y' = \frac{-6x^2}{(2y+1)}

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

<em>Find 2nd Derivative</em>

  1. Differentiate [Quotient Rule/Basic Power Rule]:                                          y'' = \frac{-12x(2y+1)+6x^2(2y')}{(2y+1)^2}
  2. [Derivative] Simplify:                                                                                       y'' = \frac{-24xy-12x+12x^2y'}{(2y+1)^2}
  3. [Derivative] Back-Substitute <em>y'</em>:                                                                     y'' = \frac{-24xy-12x+12x^2(\frac{-6x^2}{2y+1} )}{(2y+1)^2}
  4. [Derivative] Simplify:                                                                                      y'' = \frac{-24xy-12x-\frac{72x^4}{2y+1} }{(2y+1)^2}

<u>Step 4: Find Slope at Given Point</u>

  1. [Algebra] Substitute in <em>x</em> and <em>y</em>:                                                                     y''(-1,-2) = \frac{-24(-1)(-2)-12(-1)-\frac{72(-1)^4}{2(-2)+1} }{(2(-2)+1)^2}
  2. [Pre-Algebra] Exponents:                                                                                      y''(-1,-2) = \frac{-24(-1)(-2)-12(-1)-\frac{72(1)}{2(-2)+1} }{(2(-2)+1)^2}
  3. [Pre-Algebra] Multiply:                                                                                   y''(-1,-2) = \frac{-48+12-\frac{72}{-4+1} }{(-4+1)^2}
  4. [Pre-Algebra] Add:                                                                                         y''(-1,-2) = \frac{-36-\frac{72}{-3} }{(-3)^2}
  5. [Pre-Algebra] Exponents:                                                                               y''(-1,-2) = \frac{-36-\frac{72}{-3} }{9}
  6. [Pre-Algebra] Divide:                                                                                      y''(-1,-2) = \frac{-36+24 }{9}
  7. [Pre-Algebra] Add:                                                                                          y''(-1,-2) = \frac{-12}{9}
  8. [Pre-Algebra] Simplify:                                                                                    y''(-1,-2) = \frac{-4}{3}
6 0
2 years ago
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