2y - x = 5
x^2 + y^2 - 25 = 0
x = 2y - 5
(2y-5)^2 + y^2 - 25 = 0
(2y-5)(2y-5) + y^2 - 25 = 0
4y^2 - 20y + 25 + y^2 - 25 = 0
5y^2 - 20y = 0
y = 0 , y = 4
x = 2y - 5 , when y = 0
x = - 5
x = 2y - 5 , when y = 4
x = 8 - 5
x = 3
Answer:
Step-by-step explanation:
In 
, 
represents a constant related to the period of the function. Here's how it's related:
, where 
is the period of the function.
We're given 
, so solving for :
8.77 x 10^1 i believe but would double check
Answer:
Im in 7th grade and I need some help on a math problem. can I post it?
