The expression \[b^{-3\2} , b>0 \] is the equivalent to;
=1/b3/2=1/b3−−√
Answer:
standard form is 6x−11y=13
Answer:
<h2>cos90° = 0°, tan0° = 0°,</h2><h2>cos(-90°) = 0°, cot270° = 0°</h2>
Step-by-step explanation:

Using Sin^2 and Cos^2 identities,
= Cos^2(2A) + 4[(1/2)(1-Cos(2A)][(1/2)(1+Cos(2A)] = 1
= Cos^2(2A) + 1 -Cos^2(2A) = 1
0 = 0
There is nothing to solve for as it is an identity of sorts.
Answer:
I'm just guessing but I'm 1000000000%sure out of 10000000000000000000000000000