Answer:
i think it is C
Step-by-step explanation
are the only equivalent answer to y and v
Answer:
![y=-2x+3](https://tex.z-dn.net/?f=y%3D-2x%2B3)
Step-by-step explanation:
Given: ![8x+4y=12](https://tex.z-dn.net/?f=8x%2B4y%3D12)
To solve for y, let us first isolate y on one side of the equation. We can do this by subtracting
from the left and move it to the right. We then get:
![4y=12-8x](https://tex.z-dn.net/?f=4y%3D12-8x)
Now, we have y isolated. We now have to remove the 4 from the y. The only way to do this is divide both sides by 4. We then get our final answer:
![y=3-2x](https://tex.z-dn.net/?f=y%3D3-2x)
We can clean this up by putting it in a common form, slope-intercept form:
![y=-2x+3](https://tex.z-dn.net/?f=y%3D-2x%2B3)
Taking a wild guess here but I think it is D...
If wrong, let me know.
Answer:
A: 4 miles
B: 3.5 (not sure with this one)
C: 4.5 miles
Answer:
C
Step-by-step explanation:
Remember that if s(t) is a position function then:
is the velocity function and
is the acceleration function.
So, to find the acceleration, we need to solve for the second derivative of our original function. Our original function is:
![s(t)=t^2+4t+10](https://tex.z-dn.net/?f=s%28t%29%3Dt%5E2%2B4t%2B10)
So, let's take the first derivative first with respect to t:
![\frac{d}{dt}[s(t)]=\frac{d}{dt}[t^2+4t+10]](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdt%7D%5Bs%28t%29%5D%3D%5Cfrac%7Bd%7D%7Bdt%7D%5Bt%5E2%2B4t%2B10%5D)
Expand on the right:
![s'(t)=\frac{d}{dt}[t^2]+\frac{d}{dt}[4t]+\frac{d}{dt}[10]](https://tex.z-dn.net/?f=s%27%28t%29%3D%5Cfrac%7Bd%7D%7Bdt%7D%5Bt%5E2%5D%2B%5Cfrac%7Bd%7D%7Bdt%7D%5B4t%5D%2B%5Cfrac%7Bd%7D%7Bdt%7D%5B10%5D)
Use the power rule. Remember that the derivative of a constant is 0. So, our derivative is:
![v(t)=s'(t)=2t+4](https://tex.z-dn.net/?f=v%28t%29%3Ds%27%28t%29%3D2t%2B4)
This is also our velocity function.
To find acceleration, we want to second derivative. So, let's take the derivative of both sides again:
![\frac{d}{dt}[s'(t)]=\frac{d}{dt}[2t+4]](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdt%7D%5Bs%27%28t%29%5D%3D%5Cfrac%7Bd%7D%7Bdt%7D%5B2t%2B4%5D)
Again, expand the right:
![s''(t)=\frac{d}{dt}[2t]+\frac{d}{dt}[4]](https://tex.z-dn.net/?f=s%27%27%28t%29%3D%5Cfrac%7Bd%7D%7Bdt%7D%5B2t%5D%2B%5Cfrac%7Bd%7D%7Bdt%7D%5B4%5D)
Power rule. This yields:
![a(t)=s''(t)=2](https://tex.z-dn.net/?f=a%28t%29%3Ds%27%27%28t%29%3D2)
So, our answer is C.
And we're done!