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melamori03 [73]
3 years ago
9

Determine whether each equation defines y as function of x, 2x+y=8​

Mathematics
1 answer:
solong [7]3 years ago
6 0

Answer:

Yes. This equation defines y as function of x.

Step-by-step explanation:

Step 1:

Solve the equation for y (this puts the equations in terms of y)

2x+y=8​     becomes

 y = 2x - 8    

This is a linear equation, which is a funtion.  


*Another way to solve this is to look at the graph.  Graphing this equation gives us a line.  Use the vertical line test to see if it's a function.  

To pass the vertical line test (VLT), there can only be one y value to each x value along the graph.  If it passes the VLT, the equation is a function.

If it fails the VRT, the equation is not a function.

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Find the discriminant of x2-6x -10 = 0, and determine the number of real solutions of the equation.​
sergiy2304 [10]

Answer:

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2 real solutions

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>

Standard Form: ax² + bx + c = 0

Discriminant: b² - 4ac

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  • Equal to 0 - 1 solution
  • Negative - No solutions/Imaginary

Step-by-step explanation:

<u>Step 1: Define</u>

x² - 6x - 10 = 0

<u>Step 2: Identify Variables</u>

<em>Compare quadratic.</em>

x² - 6x - 10 = 0 ↔ ax² + bx + c = 0

a = 1, b = -6, c = -10

<u>Step 3: Find Discriminant</u>

  1. Substitute in variables [Discriminant]:                                                            (-6)² - 4(1)(-10)
  2. [Discriminant] Evaluate exponents:                                                                36 - 4(1)(-10)
  3. [Discriminant] Multiply:                                                                                    36 + 40
  4. [Discriminant] Add:                                                                                          76

This tells us that our quadratic has 2 real solutions.

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