I'm pretty sure it's the second option.
m= -1/3; b= -6
Answer:
The second city is 54 miles behind the first city
Step-by-step explanation:
The first bus is assumed to go ahead of the second bus and is moving at 54 mph. The second bus is moving at a speed such as 54 mph is 60% (0.6) of its speed.
The speed of the second bus is then 54/0.6 = 90 mph
When 1 and a half hour has passed, the first bus has moved a distance of
= 81 miles
The second bus (behind the first bus) has moved
= 135 miles
The problem states they both meet in that time, it can only be possible if the second bus departed a distance 135 - 81 = 54 miles behind the first city
So, the second city is 54 miles behind the first city
Yfiyfidy. It’s soot ioydotdi
It’s Bc. H. Dey h. Yes so I’m hehe what crazy now
Answer:
6x^2 ( x^2 -2) ( x^2 +2)(x^2+2x+2)(x^2-2x+2)
Step-by-step explanation:
6x^10 − 96x^2
Factor out 6x^2
6x^2 ( x^8 - 16)
Notice that inside the parentheses we have the difference of squares
6x^2 ( x^4 ^2 - 4^2) a^2 - b^2 = (a-b) (a+b)
6x^2 ( x^4 -4) (x^4 +4)
Notice that x^4-4 is also the difference of squares
6x^2 ( x^2^2 -2^2) (x^4 +4)
6x^2 ( x^2 -2) ( x^2 +2) (x^4 +4)
Note also that x^4 + 4 can be factored into (x^2+2x+2)(x^2-2x+2)
6x^2 ( x^2 -2) ( x^2 +2)(x^2+2x+2)(x^2-2x+2)
Answer:
i think it is the last one
Step-by-step explanation:
-2/25^3x^3