2007-10-25 05:04:25 · 14 answers · asked by clappinhands 1 in Science & Mathematics ➔ Mathematics
30
2007-10-25 05:06:29 · answer #1 · answered by emj 4 · 0⤊ 0⤋
The difference between -14 and 16 = 30.
2007-10-25 05:07:27 · answer #2 · answered by kja63 7 · 0⤊ 0⤋
30
2007-10-25 05:07:56 · answer #3 · answered by tarot_frog 3 · 0⤊ 0⤋
30
2007-10-25 05:07:36 · answer #4 · answered by yash 2 · 0⤊ 0⤋
30
2007-10-25 05:06:44 · answer #5 · answered by Anonymous · 0⤊ 0⤋
30
2007-10-25 05:06:39 · answer #6 · answered by wizjp 7 · 0⤊ 0⤋
30 = the difference of the two numbers.
16-(-14)=30
2007-10-25 05:12:00 · answer #7 · answered by cidyah 7 · 0⤊ 0⤋
0 to 16 =16
-14 to 0 = 14
range = 16+14=30 but depends on whether you include the endpoints
2007-10-25 05:07:50 · answer #8 · answered by Jim M 3 · 0⤊ 0⤋
I 2d what Ephesians 2:8 mentioned. All radiometric dating approaches are according to assumptions and have been shown to be incorrect many cases. sometimes they'll date an merchandise many cases till they get the outcomes they choose. If blunders are made while dating issues of primary age, how are you able to assume the reliability of issues of unknown age? If a newspaper from August 5, 1914 carbon dates at 500,000 years previous, there are right here thoughts: a million. The date on the newspaper is incorrect. 2. The dating approach is defective. 3. The tester made a mistake. while dating a fossil without date stamped on it, there's no thank you to tell if the tester made a mistake or the recommendations-set is defective. The assumptions made in radiometric dating: a million. the quantity of preliminary verify product would nicely be primary. 2. The decay fee has remained consistent. 3. The pattern has by no potential been contaminated. that may no longer saying something approximately faith. those are in uncomplicated terms clinical criticisms.
2016-11-09 10:47:21 · answer #9 · answered by ? 4 · 0⤊ 0⤋
Take the difference: 16 - (-14) = 16 + 14 = 30 << tadah!
2007-10-25 05:11:27 · answer #10 · answered by Shh! Be vewy, vewy quiet 6 · 0⤊ 0⤋