Answer:
1.6962 = 1.70 (2 dp)
0.4247 = 0.425 to 3 dp
0.007395 = 0.007 to 3 dp
0.007395 = 0.0074 to 4 dp
32549 = 32500 to 3 significant figures
32549 = 32550 to 4 significant figures
909520 = 910000 to 3 significant figures
909520 = 909500 to 4 significant figures
Step-by-step explanation:
1.6962 = 1.70 (2 dp)
0.4247 = 0.425 to 3 dp
0.007395 = 0.007 to 3 dp
0.007395 = 0.0074 to 4 dp
32549 = 32500 to 3 significant figures
32549 = 32550 to 4 significant figures
909520 = 910000 to 3 significant figures
909520 = 909500 to 4 significant figures
Answer:
Distance between Montpelier and Columbia is 1020
Step-by-step explanation:
51 mi/hr average:
d = 51 t
60 mi/hr average:
d t = 60*(t-3)
Substitute (1) into (2)
51 t = 60*(t-3)
51 t = 60 t + 180
9 t = 180
t = 20
d = 51 t
= 51 (20)
= 1020
let see distance is 1020 lets check
51 t = 60 * (t-3)
51 (20)= 60* (20-3)
1020 = 60*17
1020 = 1020
Answer:
I look it up in go.o.gle and found this answer
1e-47
23.4 ft2
Step-by-step explanation:
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