Answer:
X is 160 degrees
Step-by-step explanation:
Angles on a straight line add up to 180 degrees.
43.586 rounded to the nearest ones place is 43.587. To the nearest tenth is 43.590 and to the nearest hundreds is 43.600
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.
Answer:
Mean = 3
Step-by-step explanation:
0 + 0 + 1 + 1 + 1 + 2 + 3 + 4 + 4 + 5 + 6 + 6 + 6 = 39
39/13 = 3
Make the whole number what ever the whole number is over one