22:6 is one of them. Do you need more than one?
I believe <span>2 3 5 7 11 13 17 19 23 29 31</span>
Before the driver applies the brakes ( with the reaction time ):
d 1 = v0 · t = 20 m/s · 0.53 s = 10.6 m
After that:
v = v0 - a · t1
0 = 20 m/s - 7 · t1
7 · t1 = 20
t1 = 2.86 s
d 2 = v 0 · t1 - a · t1² / 2
d 2 = 20 m/s · 2.86 s - 7 m/s² · (2.86 s)²/2 = 57.2 m - 28.6 m = 28.6 m
d = d 1 + d 2 = 10.6 m + 28.6 m = 39.2 m
Answer: the stopping distance of a car is 39.2 m.
You would start by saying his budget is $80 so you start with 80= he pays an initial fee of $59.50 every month plus $5 per gigabyte so you would set it up like
80=59.50+5x
X being how many gigabytes he uses then you solve 80-59.50 is 20.5 divided by 5 gives you 4.1 but you round down cause you don’t want to pass the limit so therefor the most gigabytes he can use is 4
Answer:
<h3>18</h3>
Step-by-step explanation:
From the given diagram;
FGH = 49
GH = 31
Using the expression to get FG;
FG + GH = FGH
FG + 31 = 49
FG = 49-31
FG = 18
Hence the measure of FG is 18
