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
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The first option is correct
Answer:
Facing Right
Step-by-step explanation:
Given inequalities are:
4-x≤-1
Subtracting 4 from both sides
4-x-4≤-1-4
-x≤-5
Multiplying both sides with -1. Multiplying with a negative number changes the sign of the inequality
So,
x≥5
Second Inequality:
2+3x≥17
Subtracting 2 from both sides
2+3x-2≥17-2
3x≥15
Dividing both sides by 3
x≥5
Union of both solutions:
x≥5 ∩ x≥5
=> x≥5
Hence the solution will be facing right on the number line towards all numbers greater than or equals to 5 ..
Answer:
(-1,4)
Step-by-step explanation:
mp=xi+xii/2,yi+yii/2
7+-9/2=-1
0+8/2=4
=(-1,4)
