assuming you means k = log_2(3) [as log(2)3 is the same thing as 3log(2) due to multiplication being commutative]
given log(ab) = log(a) + log(b)
log_2(48) = log_2(3) + log_2(16)
Answer:
3.5 hours
Step-by-step explanation:
Speed of train X=30 mph
Speed of train Y=40 mph
Relative speed When the two trains travelling in same direction
Relative speed=40-30=10 mph
Total distance =25+10=35 miles
We have to find the time when train Y is 10 miles ahead of train X.
We know that
Time=
Using the formula
Then, we get
Time=
Hence, it will be 3.5 hours until train Y is 10 miles ahead of train X.
The mistake Shawna made was in her answer the 54 repeats so the bar should be over the .54 not the .545
Step-by-step explanation:
in newton rapson method

Let 


Answer:
Well if it’s a regular box I’m sure it will fit, because a box of tissues is at least 10 inches long in diameter.