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zaharov [31]
3 years ago
12

You run one lap around a mile track every 8 minutes. Your friend runs around the same track every 10 minutes. You both start at

the starting line at the same time.
a: How far have each of you run when you first meet again?

b: How far have each of you run the next time you meet at the starting line?



Can anyone help me solve this?
Mathematics
1 answer:
Sonja [21]3 years ago
6 0
B) it would make more since because it how far you and how much time you meet again at the starting line
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Enter an expression equivalent to (3x^2+2y^2-3x) + (2x^2+y^2-2x) - (x^2 + 3y^2 +x)
11111nata11111 [884]

Answer:

4x² - 6x

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>

Distributive Property

<u>Algebra I</u>

  • Terms/Coefficients

Step-by-step explanation:

<u>Step 1: Define</u>

(3x² + 2y² - 3x) + (2x² + y² - 2x) - (x² + 3y² + x)

<u>Step 2: Simplify</u>

  1. [Distributive Property] Distribute negative:                                                    3x² + 2y² - 3x + 2x² + y² - 2x - x² - 3y² - x
  2. Combine like terms (x²):                                                                                   4x² + 2y² - 3x + y² - 2x - 3y² - x
  3. Combine like terms (y²):                                                                                   4x² - 3x - 2x - x
  4. Combine like terms (x):                                                                                    4x² - 6x
4 0
3 years ago
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3 years ago
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Answer:

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Step-by-step explanation:

Dados dos ángulos vecinos, ambos son complementarios si la suma de sus medidas es igual a 90° y suplementarios si esa suma de medidas es igual a 180°. Puesto que uno de los ángulos es el ángulo agudo mencionado en el enunciado, es decir, un ángulo cuya medida es mayor que 0° y menor que 90°. Entonces, el ángulo complementario debe ser inevitablemente menor que el ángulo suplementario.

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Answer:

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