PLZ ANSWER. UR THE BEST IF U DO. I’LL GIVE BRAINLIEST

You basically can solve for hypothetical sides of this triangle using sohcahtoa backwards. In the first column, it says that adjacent/hypotenuse = .97. We can come up with two values that satisfy this equation. For example, the adjacent side can be .97 and the hypotenuse can be 1. These values don't matter as long as adjacent/hypotenuse is .97.

The second column tells us that opposite/hypotenuse = .26. At this point, we already have a hypothetical hypotenuse with a length of 1, so we can plug that in and find that the opposite side of C is .26.

Now that we have a hypothetical triangle with sides .97, .26, and 1, you can use these values as well as sohcahtoa to solve for the rest with respect to angle A. Keep in mind that the line AB is .26 and that the line BC is .97.

use the power rule  

( a b ) n = a n b n

to distribute the exponent.

apply the product rule to  

3 x 2 y 3 . ( 3 x 2 ) 2 ( y 3 ) 2

apply the product rule to  

3 x 2 . 3 2 ( x 2 ) 2 ( y 3 ) 2

raise  

3

to the power of  

2 . 9 ( x 2 ) 2 ( y 3 ) 2 multiply the exponents in   ( x 2 ) 2

.

apply the power rule and multiply exponents,  

( a m ) n = a m n . 9 x 2 ⋅ 2 ( y 3 ) 2

multiply  

2   by   2 . 9 x 4 ( y 3 ) 2

multiply the exponents in  

( y 3 ) 2 .

apply the power rule and multiply exponents,  

( a m ) n = a m n . 9 x 4 y 3

⋅

2

multiply  

3   by   2 . 9 x 4 y 6

step-by-step explanation: