Question

Real Numbers
If 4" ends with 0 then it must have 5 as a factor.
But. 4" = (22)" = 22 shows that 2 is the only prime factor
of 4".
Also, we know from the fundamental theorem of arithmetic
that the prime factorization of each number is unique.
So, 5 is not a factor of 4".
Hence, 4" can never end with the digit 0.
Show that any number of the form 6", where ne N can never end
with the digit 0.
If 6" ends with then it must have 5 as a factor.
But 6" = (2x 3) = (2" x 3") shows that 2 and 3 are the only​

Answer 1

Answer:

Thank you for asking this question. Here is your answer:

The quadrilateral angle's circle meet at the point of the supplementary angles and that is at the center.

ABCD will meet at a point where it won't cut it but it will just pass it, and that would be the center at O.

∠AOB + ∠COD = 180°

125° + ∠COD = 180°

∠COD = 180° - 125°

∠COD = 55°

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