Answer:
[tex]t^2+1[/tex]
Step-by-step explanation:
We are given that
Distance traveled by a toy car
[tex]D(t)=t^3-3t^2+t-3[/tex]
Where t>3
Time taken by car =t-3 seconds
We have to find the expression which represents the speed of the toy car.
We know that
[tex]Speed=\frac{Distance}{time}[/tex]
Using the formula
[tex]Speed=\frac{(t^3-3t^2+t-3)}{t-3}[/tex]
[tex]Speed=\frac{t^2(t-3)+(t-3)}{t-3}[/tex]
[tex]Speed=\frac{(t-3)(t^2+1)}{t-3}[/tex]
[tex]Speed=t^2+1[/tex]
Hence, the expression which represents the speed of the toy car
[tex]t^2+1[/tex]
The expression that represents the speed of the toy car is [tex]t^2 + 1[/tex].
Given that,
The distance traveled should be [tex]D(t) = t^3 - 3t^2 + t - 3[/tex]
Here t>3
Time taken by car = t -3 seconds
Now the expression should be
[tex]Speed = Distance \div time\\\\= ( t^3 - 3t^2 + t - 3) \div (t - 3)\\\\= t^2(t - 3) + (t - 3) \div (t - 3)\\\\= (t - 3) (t^2 + 1) \div (t - 3)\\\\= t^2 + 1[/tex]
Learn more: brainly.com/question/16911495
