<span>Let us start with the schnauzers, the easiest one to imagine. Let us assume that there were x number of schnauzers.
Scottie's are 3 more than schnauzers. So their number is x+3
Wire haired terriers are 5 less than twice the number of schnauzers.
So their number is 2x -5 (2x for twice the number of schnauzers)
Now add all these numbers.
That is x + x+3 + 2x-5 = 4x -2
The total number of dogs, 78, is given in the question
Now we know that 4x -2 =78
4x = 78 +2
= 80
Therefore x = 80/4
= 20.
So there were 20 schnauzers, 23 Scottie’s and 35 wire haired terriers</span>
9514 1404 393
Answer:
5 seconds
Step-by-step explanation:
Suppose the front parts of the trains meet at point A. Since both are the same length and traveling the same speed, each will pass point A in time ...
time = distance/speed
time = (1/18 mi)/(40 mi/h) = (1/720 h) × (3600 s)/(1 h) = 5 s
That is, the rear part of each train will be at point A 5 seconds after the front part.
The rear parts will pass each other 5 seconds after the front parts meet.
Answer:
-2 is a solution
Step-by-step explanation:
-2-5<8
-7<8
Ans: About 2.30
-15 x 3.16 / 9 = x sqrt
-15 x 3.16 = - 47.4
-47.4 / 9 = -5.27
-5.27 sqrt = About 2.30
Hope this helped :)
