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Allisa [31]
2 years ago
10

Solve the system of equations. 5x - 7y = 58 y = -x + 2 х = y =

Mathematics
1 answer:
Romashka-Z-Leto [24]2 years ago
3 0

Answer:

the x and y is 6 and -4 respectively

Step-by-step explanation:

The computation of solving the system of equations is given below:

Given that

5x - 7y = 58

And,

y = -x + 2

Put y value in the first equation

5x - 7(-x + 2) = 58

5x + 7x - 14 = 58

12x = 58 + 14

12x = 72

x = 6

So y = -6 + 2

= -4

Hence, the x and y is 6 and -4 respectively

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Answer:

b.  \displaystyle \frac{1}{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<u> </u>

<u>Algebra I</u>

  • Functions
  • Function Notation
  • Exponential Rule [Rewrite]:                                                                              \displaystyle b^{-m} = \frac{1}{b^m}
  • Exponential Rule [Root Rewrite]:                                                                     \displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}<u> </u>

<u>Calculus</u>

Derivatives

Derivative Notation

Basic Power Rule:

  • f(x) = cxⁿ
  • f’(x) = c·nxⁿ⁻¹

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

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

<em />\displaystyle H(x) = \sqrt[3]{F(x)}<em />

<em />

<u>Step 2: Differentiate</u>

  1. Rewrite function [Exponential Rule - Root Rewrite]:                                      \displaystyle H(x) = [F(x)]^\bigg{\frac{1}{3}}
  2. Chain Rule:                                                                                                        \displaystyle H'(x) = \frac{d}{dx} \bigg[ [F(x)]^\bigg{\frac{1}{3}} \bigg] \cdot \frac{d}{dx}[F(x)]
  3. Basic Power Rule:                                                                                             \displaystyle H'(x) = \frac{1}{3}[F(x)]^\bigg{\frac{1}{3} - 1} \cdot F'(x)
  4. Simplify:                                                                                                             \displaystyle H'(x) = \frac{F'(x)}{3}[F(x)]^\bigg{\frac{-2}{3}}
  5. Rewrite [Exponential Rule - Rewrite]:                                                              \displaystyle H'(x) = \frac{F'(x)}{3[F(x)]^\bigg{\frac{2}{3}}}

<u>Step 3: Evaluate</u>

  1. Substitute in <em>x</em> [Derivative]:                                                                              \displaystyle H'(5) = \frac{F'(5)}{3[F(5)]^\bigg{\frac{2}{3}}}
  2. Substitute in function values:                                                                          \displaystyle H'(5) = \frac{6}{3(8)^\bigg{\frac{2}{3}}}
  3. Exponents:                                                                                                        \displaystyle H'(5) = \frac{6}{3(4)}
  4. Multiply:                                                                                                             \displaystyle H'(5) = \frac{6}{12}
  5. Simplify:                                                                                                             \displaystyle H'(5) = \frac{1}{2}

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

Unit: Derivatives

Book: College Calculus 10e

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