$( \arcsin{x})'= \frac{1}{ \sqrt{1-x^2} } 
\\
\\ (\arcsin{(2x+1)})'= \frac{1}{ \sqrt{1-(2x+1)^2} } (2x+1)'= \frac{2}{ \sqrt{1-(4x^2+4x+1)} } =
\\
\\= \frac{2}{ \sqrt{-4x^2-4x} } =\frac{2}{ \sqrt{4x(-x-1} )} =\frac{2}{ 2\sqrt{x(-x-1} )} =\frac{1}{ \sqrt{x(-x-1} )}$

3 0

3 years ago

( Algebra)plz can you do it?

Naya [18.7K]

I'm pretty sure the answer to this is x=65 but I could be wrong.Not very good with this kinda Algebra.

6 0

4 years ago

A. If the two premises fit the pattern of Law of Detachment, Law of Contrapositive, or Law of Syllogism, determine the conclusio

Sunny_sXe [5.5K]

Answer:

YTUNDRXFTXURTUX

Step-by-step explanation:

ERVETYTRNRTHS RTH

5 0

3 years ago