Answer:
Deceleration = 
units/sec^2
Step-by-step explanation:
Given that a car is traveling at when the driver sees an accident m ahead and slams on the brakes.
i.e. if he is coming with speed v units/sec say, he has to make his velocity 0.
Distance travelled = m units
Initial velocity = v
Final velocity =0
Use the third formula as
So he has to apply an acceleration of 
units/sec^2 to stop the car in time to avoid a pile up.
Deceleration = units/sec^2
You can factor this by taking out the GCF (greatest common factor).
In the expression, the greatest common factor between the two terms is 5.
Take out 5:
5(2b + 5)
Is it possible to factor 2b + 5 even more?
No. So, your answer is D. 5(2b + 5).
Answer:
(3x^2-5x-5)/(x^2+x) <--- answer
Step-by-step explanation:
3x/(x+1) - 5/x = (3x^2 - 5(x+1) )/[ x(x+1) ]
= (3x^2 - 5x - 5)/(x^2 + x)
F(x) = 3x+1
G(x) = X^2 - 6
F(G(x)) = F(X^2 - 6) = 3(X^2 - 6) + 1 = 3X^2 - 18 + 1 = 3X^2 -17
F(G(x)) = 3X^2 - 17
Im sorry im in sixgh grade i was so close to solving it but that didnt work, i tried
