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OleMash [197]
2 years ago
9

Do the side lengths 4, 8, and 10 form a triangle? Select the correct answer and reasoning. Question 9 options: A) No, because 4

+ 8 is greater than 10. B) No, because the third side length is greater than the sum of the other two side lengths. C) Yes, because the sum of any two side lengths is greater than the third side length. D) Yes, because the sum of all three side lengths is greater than 20.
Mathematics
1 answer:
Arada [10]2 years ago
5 0

Answer:C

Step-by-step explanation:The side lengths of the triangle is 4, 8 and 10. To check if the sides form a triangle, see if the following inequalities hold true:

4

+

8

?

>

10

4

+

10

?

>

8

10

+

8

?

>

4

12

✓

>

10

14

✓

>

8

18

✓

>

4

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See explanation.

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>

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<u>Calculus</u>

Limits

  • Right-Side Limit:                                                                                             \displaystyle  \lim_{x \to c^+} f(x)
  • Left-Side Limit:                                                                                               \displaystyle  \lim_{x \to c^-} f(x)

Limit Rule [Variable Direct Substitution]:                                                             \displaystyle \lim_{x \to c} x = c

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

\displaystyle f(x) = \left \{ {{\sqrt{x + 1}, \ x < 3} \atop {5 - x, \ x \geq 3}} \right.

<u>Step 2: Find Right Limit</u>

  1. Substitute in variables [Right-Side Limit]:                                                      \displaystyle  \lim_{x \to 3^+} 5 - x
  2. Evaluate limit [Limit Rule - Variable Direct Substitution]:                           \displaystyle  \lim_{x \to 3^+} 5 - x = 5 - 3
  3. Subtract:                                                                                                         \displaystyle  \lim_{x \to 3^+} 5 - x = 2

∴ the right-side limit equals 2.

<u>Step 3: Find Left Limit</u>

  1. Substitute in variables [Left-Side Limit]:                                                       \displaystyle  \lim_{x \to 3^-} \sqrt{x + 1}
  2. Evaluate limit [Limit Rule - Variable Direct Substitution]:                           \displaystyle  \lim_{x \to 3^-} \sqrt{x + 1} = \sqrt{3 + 1}
  3. [√Radical] Add:                                                                                             \displaystyle  \lim_{x \to 3^-} \sqrt{x + 1} = \sqrt{4}
  4. [√Radical] Evaluate:                                                                                       \displaystyle  \lim_{x \to 3^-} \sqrt{x + 1} = 2

∴ the left-side limit equals 2.

<u>Step 4: Find Limit</u>

<em>The right and left-side limits are equal.</em>

∴  \displaystyle  \lim_{x \to 3} f(x) = 2

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

Unit: Limits

Book: College Calculus 10e

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