(3y^7)^3 3^3*y^(7*3)
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9y^9 9*y^9
(3y^7)^3 3^3*y^(7*3) 27*y^21 (mult. together the exponents 3 and 7)
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9y^9 9*y^9 9*y^9
That 27/9 reduces to 3. y^(21) / y^9 reduces to y^12.
Thus, we obtain 3*y^12 (answer)
Please do the next one, asking questions if need be, and showing your work.
In this use SOHCAHTOA and according to this we will use CAH on your calculator do cos(72)times 4= 1.23 should be the answer
I think it’s 9+2 because it say sum