Answer:
B
Step-by-step explanation:
We are given that the position of a particle is modeled by the function:
![s(t)=2t^3-24t^2+90t+7](https://tex.z-dn.net/?f=s%28t%29%3D2t%5E3-24t%5E2%2B90t%2B7)
And we want to find the times for which the <em>speed</em> of our particle is <em>increasing. </em>
In other words, we want to find the times for which our <em>acceleration</em> is positive (a(t)>0).
So first, we will find our acceleration function. We can differentiate twice.
By taking the derivative of both sides with respect to <em>t</em>, we acquire:
![\displaystyle s^\prime(t)=v(t)=\frac{d}{dt}\big[2t^3-24t^2+90t+7\big]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20s%5E%5Cprime%28t%29%3Dv%28t%29%3D%5Cfrac%7Bd%7D%7Bdt%7D%5Cbig%5B2t%5E3-24t%5E2%2B90t%2B7%5Cbig%5D)
Differentiate:
![v(t)=6t^2-48t+90](https://tex.z-dn.net/?f=v%28t%29%3D6t%5E2-48t%2B90)
This is our velocity function. We can differentiate once more to acquire the acceleration function. Therefore:
![\displaystyle v^\prime(t)=a(t)=\frac{d}{dt}\big[6t^2-48t+90\big]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20v%5E%5Cprime%28t%29%3Da%28t%29%3D%5Cfrac%7Bd%7D%7Bdt%7D%5Cbig%5B6t%5E2-48t%2B90%5Cbig%5D)
Differentiate:
![a(t)=12t-48](https://tex.z-dn.net/?f=a%28t%29%3D12t-48)
If our speed is increasing, our acceleration must be positive. So:
![a(t)>0](https://tex.z-dn.net/?f=a%28t%29%3E0)
By substitution:
![12t-48>0](https://tex.z-dn.net/?f=12t-48%3E0)
Now, we can solve for <em>t:</em>
<em />
<em />
<em />
Therefore, the only interval for which the speed of the particle is increasing (i.e. the acceleration is positive) is for all times t>4.
So, our answer is B.
The answer is 24 two ways you can find it is one by dividing or two my multiplying the reciprocal
The answer is D.Age
Age of birth does not affect your credit limit
Answer:
Yes she is correct because 5*10 is 50 when you cross multiply but 25*50 doesn't give you 50 so they are not equal to each other, hence, they are not a proportional relationship.
Step-by-step explanation: