The attached figure reprsents a prove that a rectangle has congruent diagonals.
<u>Given:</u> rectangle ABCD
<u>Prove:</u> BD ≅ AC
Without any calculations it's evident it can't be neither B (both numbers are even, so they're divisible by 2) nor C (the numbers end in 0 and 5, so they're divisible by 5).
A.

Both numbers have a factor of 3, so they're not relatively prime.
That means it must be D. But, let's check it.

Indeed, those two numbers are relatively prime.
3,989.237
A thousandth = 1/1000 = 0.001
Round 4th decimal place = 5
5 and over is rounding up and under 5 rounds down so it would be 3,989.237
The answer would be D. Conditional Statement
Answer:
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