Any number raised to the power of 0 is 1.
(z^3)^4(q^3)^0 = (z^3)^4 = z^(3 x 4) = z^12
When n = 1, 2n - 1 = 2(1) - 1 = 1
When n = 2, 2n - 1 = 2(2) - 1 = 3
When n = 3, 2n - 1 = 2(3) - 1 = 5
...
So: the total will be :
It would have an infinite amount of solutions.
Change each equation into y=mx+b form and then do system of equations.
The first shape is 60
The second shape is 15
Answer:
hope you understand .............