v₀ = initial velocity of the freight train while it approach a road crossing = 16 km/h = 16 (5/18) m/s = 4.44 m/s
v = final velocity of the freight train after it crosses a road crossing = 65 km/h = 65 x 5/18 m/s = 18.06 m/s
t = time to do so = 10 min = 10 x 60 sec = 600 sec
acceleration is given as
a = (v - v₀ )/t
a = (18.06 - 4.44)/(600)
a = 0.023 m/s²
a = 294
Answer: -10x+7.5
Step-by-step explanation:
First multiply -2.5(4x)= -10x and then -2.5x-3= 7.5 it should be -10x+7.5
I think it’s d I’m not sure though
In decimal form=-7.37
In fraction form plzz refer to the attachment
Hello from MrBillDoesMath!
Answer:
See Discussion section below
Discussion:
2x^2 + 5x + 3 = (2x+3)(x+1) which is Choice 3
3x^2+10x+4 =-1/3 (-3 x + sqrt(13) - 5) (3 x + sqrt(13) + 5)
I don't think any of the provided choices are right!
8x^2+10x-3 = (2x+3)(4x-1) which is Choice 2
6x^2-7x-4 = -1/24 (-12 x + sqrt(145) + 7) (12 x + sqrt(145) - 7)
I don't think any of the provided choices are right!
2x^2+x-28 = (2x-7)(x+4) which is Choice 2
Thank you,
MrB
