A freight train, traveling at a speed of 18.0 m/s, begins braking as it approaches a train yard. The train's acceleration while braking is –0.33 m/s2 . What is the train's speed after 23s?

v = 10.41 m/s

Step-by-step explanation:

using one of the equations of motion:

v = u + at

where v= final velocity

u= initial velocity

a = acceleration

t = time

u = 18.0 m/s

v = ?

a = –0.33 m/s2

t = 23s

v = 18.0 + (–0.33 x  23)

v = 18.0 + (-7.59)

v = 10.41 m/s

answer: idk

step-by-step explanation:

he used the algebra tiles correctly.

step-by-step explanation:

given : 3x(2x-1)

solution :

since he used algebra tiles .

armando's solution is the same in the solution shown in figure

as he distributed 2x into x and x and wrote them vertically

and also write   -1 vertically

and 3x into x and x and x and wrote it horizontally

performing the multiplication in the algebra tiles we get

and solving the given expression :

3x(2x-1)

since solving the expression using tiles   and solving the expression algebraically we get the same answer

thus   we can choose the first statement as the answer to this problem which is "he used the algebra tiles correctly".