Answer:
![\frac{dy}{dx} =\frac{1-x}{y+2}](https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdx%7D%20%3D%5Cfrac%7B1-x%7D%7By%2B2%7D)
Step-by-step explanation:
We begin with the equation ![x^2+y^2-2x+4y-4=0](https://tex.z-dn.net/?f=x%5E2%2By%5E2-2x%2B4y-4%3D0)
Next, we need to differentiate with respect to x
![2x+2y\frac{dy}{dx} -2+4\frac{dy}{dx} =0](https://tex.z-dn.net/?f=2x%2B2y%5Cfrac%7Bdy%7D%7Bdx%7D%20-2%2B4%5Cfrac%7Bdy%7D%7Bdx%7D%20%3D0)
Now we need to solve for ![\frac{dy}{dx}](https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdx%7D)
To do this, we need to factor out the
and then isolate it.
![2x+2y\frac{dy}{dx} -2+4\frac{dy}{dx} =0\\\\2y\frac{dy}{dx}+4\frac{dy}{dx}=2-2x\\\\\frac{dy}{dx}(2y+4)=2-2x\\\\\frac{dy}{dx}=\frac{2-2x}{2y+4}\\\\\frac{dy}{dx}=\frac{1-x}{y+2}](https://tex.z-dn.net/?f=2x%2B2y%5Cfrac%7Bdy%7D%7Bdx%7D%20-2%2B4%5Cfrac%7Bdy%7D%7Bdx%7D%20%3D0%5C%5C%5C%5C2y%5Cfrac%7Bdy%7D%7Bdx%7D%2B4%5Cfrac%7Bdy%7D%7Bdx%7D%3D2-2x%5C%5C%5C%5C%5Cfrac%7Bdy%7D%7Bdx%7D%282y%2B4%29%3D2-2x%5C%5C%5C%5C%5Cfrac%7Bdy%7D%7Bdx%7D%3D%5Cfrac%7B2-2x%7D%7B2y%2B4%7D%5C%5C%5C%5C%5Cfrac%7Bdy%7D%7Bdx%7D%3D%5Cfrac%7B1-x%7D%7By%2B2%7D)
I don’t know the answer but i need to help people before i get help so just act like i helped you
1+2+3= 6 (even)
2+3+4= 9 (odd)
it can be both
Answer:
y= 11 | 19 | 27 |35 | 43 |
Answer:
<h2>See the explanation.</h2>
Step-by-step explanation:
a.
The initial length of the candle is 16 inch. It also given that, it burns with a constant rate of 0.8 inch per hour.
After one hour since the candle was lit, the length of the candle will be (16 - 0.8) = 15.2 inch.
After two hour since the candle was lit, the length of the candle will be (15.2 - 0.8) = 14.4 inch. The length of the candle after two hours can also be represented by {16 - 2(0.8)}.
Hence, the length of the candle after t hours when it was lit can be represented by the function,
.
at t = 20.
b.
The domain of the function is 0 to 20.
c.
The range is 0 to 16.