This is called the median.
Answer:
See below for answers and explanations
Step-by-step explanation:
<u>Part A</u>

<u>Part B</u>

<u>Part C</u>

Answer:
(5x + 7y)(25x² - 35xy + 49y²)
Step-by-step explanation:
125x³ + 343y³ ← is a sum of cubes and factors in general as
a³ + b³ = (a + b)(a² - ab + b² )
125x³ = (5x)³ ⇒ a = 5x
343y³ = (7y)³ ⇒ b = 7y
125x³ + 343y³
= (5x + 7y)((5x)² - (5x × 7y) + (7y)²)
= (5x + 7y)(25x² - 35xy + 49y²) ← in factored form
Absolutely not. The least mixed numbers are 1 and 1/2, and 1 1/2 + 1 1/2 is 3, so no, it is not possible. Hope I helped!