for the implicit integration I got:
![1+3y^2y'-y'=0\\y'=-\frac{1}{3y^2-1}](https://tex.z-dn.net/?f=1%2B3y%5E2y%27-y%27%3D0%5C%5Cy%27%3D-%5Cfrac%7B1%7D%7B3y%5E2-1%7D)
the vertical tangent would at points at which y' goes to infinity.
that happen when the right-hand side has singularities, that is, the denominator is 0:
![3y^2-1 =0\\y^2=1/3\\|y|=\sqrt{1/3}\\y=\pm\sqrt{1/3}](https://tex.z-dn.net/?f=3y%5E2-1%20%3D0%5C%5Cy%5E2%3D1%2F3%5C%5C%7Cy%7C%3D%5Csqrt%7B1%2F3%7D%5C%5Cy%3D%5Cpm%5Csqrt%7B1%2F3%7D)
so the two points y above have a vertical tangent.
Answer:
3/2 or 1.5
Step-by-step explanation:
A negative divided by a negative makes the quotient a positive, 9 divided by 6 is 1.5 and the gcf of 9 and 6 is 3, so you would simplify it by doing 9 divided by 3 and 6 divided by3 which is 3/2
Answer:
60 watts i believe double check my work
Step-by-step explanation:
10 feet is the one i hop i help
17. 2x<16 x=7 or lower
All I can help you with is #17,
sorry that I can't answer the other 3.
I hope this helped! :-)