1) how many significant digits are there in the sum of 45.10 and 102.203?

1 answer:

7 0

Oi :)

1) There are 3 significant digits
$45.10+102.203= 147.303$

2)
15 Mi/h = 22 ft/s

3)
$C= 72*m* \frac{3}{4} \\ \\ C= 72*81* \frac{3}{4} \\ \\ C= \frac{17496}{4} \\ \\ C=4374$

4)
There are 4 significant digitis. Leter C

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Answer:

See proof below

Step-by-step explanation:

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