Let 
be the speed of train A, and let's set the origin in the initial position of train A. The equations of motion are
where 
are the positions of trains A and B respectively, and t is the time in hours.
The two trains meet if and only if 
, and we know that this happens after two hours, i.e. at 
Solving this equation for v we have
So, train A is travelling at 105 km/h. This implies that train B travels at
2k^2+3k-13=k^2+15
subtract k^2 -15 from each side
k^2 +3k -28=0
(k+7)(k-4)=0
k=-7 k=4
Answer: k= -7,4
Answer:
x = 12 , y = 10
Step-by-step explanation:
Let x , y are two numbers.
x > y
1 ) Three times the greater is 18 times their
difference
3x = 18( x - y )
x = 6( x - y )
x = 6x - 6y
6y = 5x
y = 5x/6 ——-( 1 )
2 ) 4 times the smaller is 4 less than twice
the sum of the two
4y + 4 = 2 ( x + y )
2y + 2 = x + y
y = x -2 ——( 2 )
From ( 1 ) and ( 2 ) ,
5x/6 = x -2
( 5x /6 ) - x = -2
( 5x - 6x ) /6 = -2
-x = -12
x = 12
Put x = 12 in equation ( 2 ) , we get
y = 12 - 2
y = 10
Therefore ,
x = 12 , y = 10
Answer:
-0.125
Step-by-step explanation:
- Im learning about this , so im not really sure
