Answer:
The answer is 12
Step-by-step explanation:
The answer to this question is d
Answer:
<u>Solution</u><u> </u><u>given</u><u>:</u>
<u>one</u><u> </u><u>side</u><u> </u><u>of</u><u> </u><u>a</u><u> </u><u>triangle</u><u> </u><u>must</u><u> </u><u>be</u><u> </u><u>less</u><u> </u><u>than</u><u> </u><u>the</u><u> </u><u>sum</u><u> </u><u>of</u><u> </u><u>two</u><u> </u><u>other</u><u> </u><u>side</u><u> </u><u>.</u>
<u>by</u><u> </u><u>making</u><u> </u><u>these</u><u> </u><u>sense</u><u>:</u>
3<10+20
10<20+3
20<10+3not true
these three side are not possible.
again
10<20+25
20<25+10
25<20+10
these three side are true so
required side are:
Side 1:10
Side 2:25
Side 3:20
So if you think about it, this problem isn't that hard. So say it's 12:59, the total sum of those digits would be 71. If it was 1:11, then the sum would be 3. If it was 3:47, the sum would be 50. So now, you have to find the time where it would be the least possible sum. that time would be 1:00 because 1+0+0=1