possible error on each rod is .02*125 = 2.5 cm
So you want n rods such that
122.5n >=39 690
n >= 324
A plastic rod has a length of 125 cm with a possible error of 2% the number of these rods that must be placed end-to-end to ensure a total length of at least 39 690 cm is
4 answers
324
SO U PUT THE 125 then dIVIDE IS BI THE 5 AT DA END OF THE SCREEN WITH IS 24 TEHN U HAVE THAT 65 WE WERE TALKING ABOUT THEN PUT IT IN THE 24 THEN IT IS THE ANSWER IS 423 CUZ THEN U FLIP DA DIGITIS IT NOW 321 CUZ YA FLIPPER THE DIGIT LOL IM SMART.
p.s OOBLECK is A poopy like me
p.s OOBLECK is A poopy like me
for some reason no one got it, im sure oobleck knew how to do it but he didnt include the answer >.<
1. 1216 = 2^6 * 19 (A*B)
2. 2812 = 2^2 * 19 * 37 (B*C)
3. 2368 = 2^6 * 37 (A*C)
in 1. and 2. both had "B" and both also had 19 and 2^2. so B=19*2^2/2, which is just 19*2 = 38
in 2. and 3., both at "C" and "37" and 2^2. so C=37*2^2/2, which is 37*2 which is 74
and in 1. and 3. both had "C" and 2^6, but as you know, all of them had 2^2, so really they both shared 2^4. so A=2*2^4 which is 2^5 which is 32.
A=32 B=38 C=74
i suck at explaining, so go read oobleck's answer
1. 1216 = 2^6 * 19 (A*B)
2. 2812 = 2^2 * 19 * 37 (B*C)
3. 2368 = 2^6 * 37 (A*C)
in 1. and 2. both had "B" and both also had 19 and 2^2. so B=19*2^2/2, which is just 19*2 = 38
in 2. and 3., both at "C" and "37" and 2^2. so C=37*2^2/2, which is 37*2 which is 74
and in 1. and 3. both had "C" and 2^6, but as you know, all of them had 2^2, so really they both shared 2^4. so A=2*2^4 which is 2^5 which is 32.
A=32 B=38 C=74
i suck at explaining, so go read oobleck's answer