He can conclude that the system of equations has no solution since one equation in it always results with false equivalency.
Hope this helps.
r3t40
Answer:
0.4 sec
Step-by-step explanation:
Remember that
1 mile=5,280 feet
1 foot=12 inches
1 hour=3,600 seconds
Let
s -----> the speed in ft/sec
d ----> the distance in ft
t -----> the time in sec
s=d/t
Solve for t
t=d/s
step 1
Convert miles/hour to ft/sec
105 mi/h=105(5,280/3,600)=154 ft/sec
Convert 60 ft 6 in to ft
60 ft 6 in=60+(6/12)=60.5 ft
step 2
Find the time
t=d/s
we have
s=154 ft/sec
d=60.5 ft
substitute
t=60.5/154
t=0.39 sec
Round to the nearest tenth
t=0.4 sec
Let's begin by listing the first few multiples of 4: 4, 8, 12, 16, 20, 24, 28, 32, 36, 38, 40, 44. So, between 1 and 37 there are 9 such multiples: {4, 8, 12, 16, 20, 24, 28, 32, 36}. Note that 4 divided into 36 is 9.
Let's experiment by modifying the given problem a bit, for the purpose of discovering any pattern that may exist:
<span>How many multiples of 4 are there in {n; 37< n <101}? We could list and then count them: {40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96, 100}; there are 16 such multiples in that particular interval. Try subtracting 40 from 100; we get 60. Dividing 60 by 4, we get 15, which is 1 less than 16. So it seems that if we subtract 40 from 1000 and divide the result by 4, and then add 1, we get the number of multiples of 4 between 37 and 1001:
1000
-40
-------
960
Dividing this by 4, we get 240. Adding 1, we get 241.
Finally, subtract 9 from 241: We get 232.
There are 232 multiples of 4 between 37 and 1001.
Can you think of a more straightforward method of determining this number? </span>
