(3y^7)^3 3^3*y^(7*3)
------------ = ------------------
9y^9 9*y^9
(3y^7)^3 3^3*y^(7*3) 27*y^21 (mult. together the exponents 3 and 7)
------------ = ------------------ = --------------
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.
Hi! So what this is asking for is how LONG the line is, not how tall. So you see B. It is on 0. And you go to A, it is on 6. Zero from 6 is 6. So 6 is your answer.
There aren't any solutions. I recommend using MathPapa algebraic calculator for problems like this. Here's the link: https://www.mathpapa.com/algebra-calculator.html
Answer:
j^13
Step-by-step explanation:
Its 9 because it includes f(x)
