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.
Step-by-step explanation:
Absolute error=14.7-13.0
=1.7m
Percentage error = 1.7/14.7 x 100%
=11.56462
=11.6%
Heres a answer for your question
Answer: C
explanation: can i have brainilestv