namely, let's rationalize the denominator in the fraction, for which case we'll be using the <u>conjugate</u> of that denominator, so we'll multiply top and bottom by its <u>conjugate</u>.
so the denominator is 5 + i, simply enough, its conjugate is just 5 - i, recall that same/same = 1, thus (5-i)/(5-i) = 1, and any expression multiplied by 1 is just itself, so we're not really changing the fraction per se.
There are 3 different numbers
slot method
how many 3 digit numbers?
321
312
123
132
213
231
6 numbes
how many 2 digit numbers?
12
21
13
31
23
32
6 numbers
how many 1 digit numbers
1
2
3
3 of them
6+6+3=15
15 different number, yah you ar right
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:
Because I know what a translation is
Answer:
Step-by-step explanation:
