If ' y ' is greater than ' 1 ', then [ 3y to the second power ] is. If ' y ' is less than ' 1 ', then 3 is.
Oh, what a happy little question! When we see 3y to the 2nd power, it means we are multiplying 3 by y two times. So the answer is 3y * y, which simplifies to 3y^2. Just imagine those y's dancing together on the canvas of mathematics!
This is like a² - b², with a = 9x² and b = 9y². This can be factored as (a - b)(a + b), then substitute for a & b: (9x² - 9y²)(9x² + 9y²), then (9x² - 9y²) can factor to (3x - 3y)(3x + 3y) so we have: (3x - 3y)(3x + 3y)(9x² + 9y²)
x = 3y x - 3y = 3y - 3y x - 3y = 3y - 3y x - 3y = 0
3y +3y = 6y
(3y2)4*3y2*3y2*3y2*3y2 = 38*y16 = 6561y16
If ' y ' is greater than ' 1 ', then [ 3y to the second power ] is. If ' y ' is less than ' 1 ', then 3 is.
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Oh, what a happy little question! When we see 3y to the 2nd power, it means we are multiplying 3 by y two times. So the answer is 3y * y, which simplifies to 3y^2. Just imagine those y's dancing together on the canvas of mathematics!
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This is like a² - b², with a = 9x² and b = 9y². This can be factored as (a - b)(a + b), then substitute for a & b: (9x² - 9y²)(9x² + 9y²), then (9x² - 9y²) can factor to (3x - 3y)(3x + 3y) so we have: (3x - 3y)(3x + 3y)(9x² + 9y²)
I'm not sure why this is labeled as calculus; it's just algebra. You take the first coefficient and the 2nd coefficient and add them. 1y x 3y = (1+3)y = 4y
x = 3y x - 3y = 3y - 3y x - 3y = 3y - 3y x - 3y = 0
6xy(2x2 - 3y)
2y2(3y - 1)
3y +3y = 6y
3y + 4z + 12y = 3y -3y -3y ------------------------- 4z + 12y = 0 divide everything by 4 z + 3y = 0 -3y -3y z = -3y