10x^2-10x+20 is the answer to this
Answer:
E
Step-by-step explanation:
We are given that a particle's position along the x-axis at time <em>t </em>is modeled by:

And we want to determine at which time(s) <em>t</em> is the particle at rest.
If the particle is at rest, this suggests that its velocity at that time is 0.
Since are we given the position function, we can differentiate it to find the velocity function.
So, by differentiating both sides with respect to <em>t</em>, we acquire:
![\displaystyle x^\prime(t)=v(t)=\frac{d}{dt}\big[2t^3-21t^2+72t-53\big]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20x%5E%5Cprime%28t%29%3Dv%28t%29%3D%5Cfrac%7Bd%7D%7Bdt%7D%5Cbig%5B2t%5E3-21t%5E2%2B72t-53%5Cbig%5D)
Differentiate. So, our velocity function is:

So, we will set the velocity to 0 and solve for <em>t</em>. Hence:

We can divide both sides by 6:

Factoring yields:

By the Zero Product Property:

Hence:

Therefore, at the 3rd and 4th seconds, the velocity of the particle is 0, impling that the particle is at rest.
Our answer is E.
Answer:
Step-by-step explanation:
1-x²=-(x²-1)=-(x-1)(x+1)
L.C.M.=(x+1)(x-1)³
Answer:
<em>2 times
</em>
<em></em>
Step-by-step explanation:
Please refer to the image attached of our <em>solar system </em>in which Saturn and Uranus are clearly visible.
Distance of Saturn from Sun, 
Distance of Uranus from Sun,
We have to find, how many times the Uranus is far from Sun than Saturn is far from Sun.
Let us begin by multiplying the distance of Saturn from Sun by 2:
From above, we can say that
is approximately <em>twice </em>of
.
In other words, we can say that Uranus is at approximately twice distance from Sun than that of Saturn is.
A sideways opening parabola is in the form

, so we know from the process of elimination that it will either be b or c. Next we have to realize that if the parabola opens to the left it is a negative parabola, just like if a parabola opens upside down it is a negative parabola. So the one that has the negative out front is b.