Answer:
procedure always produces 6
Step-by-step explanation:
Let 'n' be the unknown number
Add 4 to the number : n+4
multiply the sum by 3.
multiply the sum n+4 by 3
Now subtract 6, so we subtract 6 from 3n+12
finally decrease the difference by the tripe of the original number
triple of original number is 3n
so the procedure always produces 6
Answer:
The answer might be c
Step-by-step explanation:
Answer:
88
Step-by-step explanation:
Given:
(h⁴ + h² – 2) ÷ (h + 3).
We could obtain the remainder using the remainder theorem :
That is the remainder obtained when (h⁴ + h² – 2) is divided by (h + 3).
Using the reminder theorem,
Equate h+3 to 0 and obtain the value of h at h+3 = 0
h + 3 = 0 ; h = - 3
Substituting h = - 3 into (h⁴ + h² – 2) to obtain the remainder
h⁴ + h² – 2 = (-3)⁴ + (-3)² - 2 = 81 + 9 - 2 = 88
Hence, remainder is 88
