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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Answer: A
Step-by-step explanation:
Nothing you a bum
Answer:
Number 7: 40%
Number 8: 75%
Step-by-step explanation: Hope i helped mate
Answer:
v=6
Step-by-step explanation:
First let's simplify the equation so it will be easier to PEMDAS the equation later.
-7v+4(2-5v)=-154
-7v+8-20v=-154
-27v+8=-154
Now you must isolate the v variable. You do this by subtracting 8 first and then divide by -27.
-27v+8=-154
-27v=-162
v = 6
Answer:
8.85
Step-by-step explanation:
