For proof of 3 divisibility, abc is a divisible by 3 if the sum of abc (a + b + c) is a multiple of 3.
<h3>
Integers divisible by 3</h3>
The proof for divisibility of 3 implies that an integer is divisible by 3 if the sum of the digits is a multiple of 3.
<h3>Proof for the divisibility</h3>
111 = 1 + 1 + 1 = 3 (the sum is multiple of 3 = 3 x 1) (111/3 = 37)
222 = 2 + 2 + 2 = 6 (the sum is multiple of 3 = 3 x 2) (222/3 = 74)
213 = 2 + 1 + 3 = 6 ( (the sum is multiple of 3 = 3 x 2) (213/3 = 71)
27 = 2 + 7 = 9 (the sum is multiple of 3 = 3 x 3) (27/3 = 9)
Thus, abc is a divisible by 3 if the sum of abc (a + b + c) is a multiple of 3.
Learn more about divisibility here: brainly.com/question/9462805
#SPJ1
7/6
it’s positive because it’s going up
Answer:
x^2 - 2x - 12 with remainder 12
Step-by-step explanation:
Synthetic division is the fastest way in which to carry out this division.
The divisor (x - 1) from long division corresponds to the divisor 1 in synthetic division. Setting up synthetic division, we get:
1 / 1 -3 -10 24
1 -2 -12
--------------------------------
1 -2 -12 12
The first three digits {1, -2, -12} are the coefficients of the quotient, and 12 represents the remainder:
The quotient is 1x^2 - 2x - 12 and the remainder is 12.
Answer:
15
Step-by-step explanation:
first you have to subtract 4 from 19 to get what k is. that should be your answer. good luck!
