Answer:

• let f(x) be m:

• make x the subject of the function:
![{ \rm{8m = {x}^{3} + 128}} \\ \\ { \rm{ {x}^{3} = 8m - 128 }} \\ \\ { \rm{ {x}^{3} = 8(m - 16) }} \\ \\ { \rm{x = \sqrt[3]{8} \times \sqrt[3]{(m - 16)} }} \\ \\ { \rm{x = 2 \sqrt[3]{(m - 16)} }}](https://tex.z-dn.net/?f=%7B%20%5Crm%7B8m%20%3D%20%20%7Bx%7D%5E%7B3%7D%20%20%2B%20128%7D%7D%20%5C%5C%20%20%5C%5C%20%7B%20%5Crm%7B%20%7Bx%7D%5E%7B3%7D%20%3D%208m%20-%20128%20%7D%7D%20%5C%5C%20%20%5C%5C%20%7B%20%5Crm%7B%20%7Bx%7D%5E%7B3%7D%20%3D%208%28m%20-%2016%29%20%7D%7D%20%5C%5C%20%20%5C%5C%20%7B%20%5Crm%7Bx%20%3D%20%20%5Csqrt%5B3%5D%7B8%7D%20%20%5Ctimes%20%20%5Csqrt%5B3%5D%7B%28m%20-%2016%29%7D%20%7D%7D%20%5C%5C%20%20%5C%5C%20%7B%20%5Crm%7Bx%20%3D%202%20%5Csqrt%5B3%5D%7B%28m%20-%2016%29%7D%20%7D%7D)
• therefore:

It is true that the product of two consecutive even integers are always one less than the square of their average.
<u>Step-by-step explanation</u>:
Let the two consecutive odd integers be 1 and 3.
- The product of 1 and 3 is (1
3)=3 - The average of 1 and 3 is (1+3)/2 =4/2 = 2
- The square of their average is (2)² = 4
∴ The product 3 is one less than the square of their average 4.
Let the two consecutive even integers be 2 and 4.
- The product of 2 and 4 is (2
4)=8 - The average of 2 and 4 is (2+4)/2 =6/2 = 3
- The square of their average is (3)² = 9
∴ The product 8 is one less than the square of their average 9.
Thus, It is true that the product of two consecutive even integers are always one less than the square of their average.
The problem with the question is that 51% of 1009 adults could not have said that their biggest fear was losing vision. The issue with 51% of 1009 is that you wont be left with a whole number, but a decimal instead. You cant have a decimal of a person.
C. they are worth the same amount of money, 1/2 and 2/4 are both 50% or 50 cents