<h2>
Step-by-step explanation:</h2>
As per the question,
Let a be any positive integer and b = 4.
According to Euclid division lemma , a = 4q + r
where 0 ≤ r < b.
Thus,
r = 0, 1, 2, 3
Since, a is an odd integer, and
The only valid value of r = 1 and 3
So a = 4q + 1 or 4q + 3
<u>Case 1 :-</u> When a = 4q + 1
On squaring both sides, we get
a² = (4q + 1)²
= 16q² + 8q + 1
= 8(2q² + q) + 1
= 8m + 1 , where m = 2q² + q
<u>Case 2 :-</u> when a = 4q + 3
On squaring both sides, we get
a² = (4q + 3)²
= 16q² + 24q + 9
= 8 (2q² + 3q + 1) + 1
= 8m +1, where m = 2q² + 3q +1
Now,
<u>We can see that at every odd values of r, square of a is in the form of 8m +1.</u>
Also we know, a = 4q +1 and 4q +3 are not divisible by 2 means these all numbers are odd numbers.
Hence , it is clear that square of an odd positive is in form of 8m +1
Y=31.2
X=15.6
I rounded both to the nearest tenth place.
-m-4=-13
-m=-13+4=-9
m=9
☺☺☺☺
Answer:
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Step-by-step explanation:
