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

3 0

3 years ago

Graph y = (x - 1) 2 + 2. Click on the graph until the graph of y = (x - 1) 2 + 2 appears.

strojnjashka [21]

The vertex at (1,2) and from there use this to make the graph.

over 1, up 1. over 2 up 4, over 3 up 9. and if you know its right side, you know its left side too.

6 0

3 years ago

Read 2 more answers

Add.

Viktor [21]

16-3x+4x-2+x^2

14+x+x^2

4 0

3 years ago

Factor the expression.<br> 32b + 12<br> 32b + 12 =<br> (Factor completely

grigory [225]

32b+12

32 and 12 is both divisible by 4 so divide them by 4 and the answers in parentheses