Since the buses travel in opposite directions, the speed at which they distance themselves is the sum of their speeds.
One bus travels at speed s.
The other bus travels at speed s + 15.
The sum of the speeds is s + s + 15 = 2s + 15
speed = distance/time
distance = speed * time
366 = (2s + 15) * 2
183 = 2s + 15
168 = 2s
s = 84
The slower bus travels at 84 km/h.
s + 15 = 84 + 15 = 99
The faster bus travels at 99 km/h.
Check:
In 2 hours, the slower bus travels 2 * 84 km = 168 km
In 2 hours, the faster bus travels 2 * 99 km = 198 km
In 2 hours, the buses are 198 km + 168 km = 366 km apart.
Our answer is correct.
A = 1/2 b h
5 = 1/2 (5) h
10 = 5h
h = 10/5
h = 2
answer
2
Answer:
See proof below
Step-by-step explanation:
An equivalence relation R satisfies
- Reflexivity: for all x on the underlying set in which R is defined, (x,x)∈R, or xRx.
- Symmetry: For all x,y, if xRy then yRx.
- Transitivity: For all x,y,z, If xRy and yRz then xRz.
Let's check these properties: Let x,y,z be bit strings of length three or more
The first 3 bits of x are, of course, the same 3 bits of x, hence xRx.
If xRy, then then the 1st, 2nd and 3rd bits of x are the 1st, 2nd and 3rd bits of y respectively. Then y agrees with x on its first third bits (by symmetry of equality), hence yRx.
If xRy and yRz, x agrees with y on its first 3 bits and y agrees with z in its first 3 bits. Therefore x agrees with z in its first 3 bits (by transitivity of equality), hence xRz.
-46
Hope this helps!
May I please have brainliest? :D
